3.6.59 \(\int \frac {(d+e x)^{5/2}}{(a-c x^2)^3} \, dx\)

Optimal. Leaf size=279 \[ -\frac {3 \sqrt {\sqrt {c} d-\sqrt {a} e} \left (2 \sqrt {a} \sqrt {c} d e-a e^2+4 c d^2\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a} e}}\right )}{32 a^{5/2} c^{7/4}}+\frac {3 \sqrt {\sqrt {a} e+\sqrt {c} d} \left (-2 \sqrt {a} \sqrt {c} d e-a e^2+4 c d^2\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {a} e+\sqrt {c} d}}\right )}{32 a^{5/2} c^{7/4}}+\frac {3 \sqrt {d+e x} \left (x \left (2 c d^2-a e^2\right )+a d e\right )}{16 a^2 c \left (a-c x^2\right )}+\frac {(d+e x)^{3/2} (a e+c d x)}{4 a c \left (a-c x^2\right )^2} \]

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Rubi [A]  time = 0.46, antiderivative size = 279, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {739, 821, 827, 1166, 208} \begin {gather*} -\frac {3 \sqrt {\sqrt {c} d-\sqrt {a} e} \left (2 \sqrt {a} \sqrt {c} d e-a e^2+4 c d^2\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a} e}}\right )}{32 a^{5/2} c^{7/4}}+\frac {3 \sqrt {\sqrt {a} e+\sqrt {c} d} \left (-2 \sqrt {a} \sqrt {c} d e-a e^2+4 c d^2\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {a} e+\sqrt {c} d}}\right )}{32 a^{5/2} c^{7/4}}+\frac {3 \sqrt {d+e x} \left (x \left (2 c d^2-a e^2\right )+a d e\right )}{16 a^2 c \left (a-c x^2\right )}+\frac {(d+e x)^{3/2} (a e+c d x)}{4 a c \left (a-c x^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(5/2)/(a - c*x^2)^3,x]

[Out]

((a*e + c*d*x)*(d + e*x)^(3/2))/(4*a*c*(a - c*x^2)^2) + (3*Sqrt[d + e*x]*(a*d*e + (2*c*d^2 - a*e^2)*x))/(16*a^
2*c*(a - c*x^2)) - (3*Sqrt[Sqrt[c]*d - Sqrt[a]*e]*(4*c*d^2 + 2*Sqrt[a]*Sqrt[c]*d*e - a*e^2)*ArcTanh[(c^(1/4)*S
qrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(32*a^(5/2)*c^(7/4)) + (3*Sqrt[Sqrt[c]*d + Sqrt[a]*e]*(4*c*d^2 - 2
*Sqrt[a]*Sqrt[c]*d*e - a*e^2)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/(32*a^(5/2)*c^(7/4
))

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rule 739

Int[((d_) + (e_.)*(x_))^(m_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^(m - 1)*(a*e - c*d*x)*(a
+ c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] + Dist[1/((p + 1)*(-2*a*c)), Int[(d + e*x)^(m - 2)*Simp[a*e^2*(m - 1) -
c*d^2*(2*p + 3) - d*c*e*(m + 2*p + 2)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e}, x] && NeQ[c*d^
2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 821

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x)^m*
(a + c*x^2)^(p + 1)*(a*g - c*f*x))/(2*a*c*(p + 1)), x] - Dist[1/(2*a*c*(p + 1)), Int[(d + e*x)^(m - 1)*(a + c*
x^2)^(p + 1)*Simp[a*e*g*m - c*d*f*(2*p + 3) - c*e*f*(m + 2*p + 3)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x
] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && GtQ[m, 0] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rule 827

Int[((f_.) + (g_.)*(x_))/(Sqrt[(d_.) + (e_.)*(x_)]*((a_) + (c_.)*(x_)^2)), x_Symbol] :> Dist[2, Subst[Int[(e*f
- d*g + g*x^2)/(c*d^2 + a*e^2 - 2*c*d*x^2 + c*x^4), x], x, Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0]

Rule 1166

Int[((d_) + (e_.)*(x_)^2)/((a_) + (b_.)*(x_)^2 + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[b^2 - 4*a*c, 2]}, Di
st[e/2 + (2*c*d - b*e)/(2*q), Int[1/(b/2 - q/2 + c*x^2), x], x] + Dist[e/2 - (2*c*d - b*e)/(2*q), Int[1/(b/2 +
q/2 + c*x^2), x], x]] /; FreeQ[{a, b, c, d, e}, x] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - a*e^2, 0] && PosQ[b^
2 - 4*a*c]

Rubi steps

\begin {align*} \int \frac {(d+e x)^{5/2}}{\left (a-c x^2\right )^3} \, dx &=\frac {(a e+c d x) (d+e x)^{3/2}}{4 a c \left (a-c x^2\right )^2}-\frac {\int \frac {\sqrt {d+e x} \left (-\frac {3}{2} \left (2 c d^2-a e^2\right )-\frac {3}{2} c d e x\right )}{\left (a-c x^2\right )^2} \, dx}{4 a c}\\ &=\frac {(a e+c d x) (d+e x)^{3/2}}{4 a c \left (a-c x^2\right )^2}+\frac {3 \sqrt {d+e x} \left (a d e+\left (2 c d^2-a e^2\right ) x\right )}{16 a^2 c \left (a-c x^2\right )}+\frac {\int \frac {\frac {3}{4} c d \left (4 c d^2-3 a e^2\right )+\frac {3}{4} c e \left (2 c d^2-a e^2\right ) x}{\sqrt {d+e x} \left (a-c x^2\right )} \, dx}{8 a^2 c^2}\\ &=\frac {(a e+c d x) (d+e x)^{3/2}}{4 a c \left (a-c x^2\right )^2}+\frac {3 \sqrt {d+e x} \left (a d e+\left (2 c d^2-a e^2\right ) x\right )}{16 a^2 c \left (a-c x^2\right )}+\frac {\operatorname {Subst}\left (\int \frac {\frac {3}{4} c d e \left (4 c d^2-3 a e^2\right )-\frac {3}{4} c d e \left (2 c d^2-a e^2\right )+\frac {3}{4} c e \left (2 c d^2-a e^2\right ) x^2}{-c d^2+a e^2+2 c d x^2-c x^4} \, dx,x,\sqrt {d+e x}\right )}{4 a^2 c^2}\\ &=\frac {(a e+c d x) (d+e x)^{3/2}}{4 a c \left (a-c x^2\right )^2}+\frac {3 \sqrt {d+e x} \left (a d e+\left (2 c d^2-a e^2\right ) x\right )}{16 a^2 c \left (a-c x^2\right )}+\frac {\left (3 \left (\sqrt {c} d+\sqrt {a} e\right ) \left (4 c d^2-2 \sqrt {a} \sqrt {c} d e-a e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{c d+\sqrt {a} \sqrt {c} e-c x^2} \, dx,x,\sqrt {d+e x}\right )}{32 a^{5/2} c}-\frac {\left (3 \left (\sqrt {c} d-\sqrt {a} e\right ) \left (4 c d^2+2 \sqrt {a} \sqrt {c} d e-a e^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{c d-\sqrt {a} \sqrt {c} e-c x^2} \, dx,x,\sqrt {d+e x}\right )}{32 a^{5/2} c}\\ &=\frac {(a e+c d x) (d+e x)^{3/2}}{4 a c \left (a-c x^2\right )^2}+\frac {3 \sqrt {d+e x} \left (a d e+\left (2 c d^2-a e^2\right ) x\right )}{16 a^2 c \left (a-c x^2\right )}-\frac {3 \sqrt {\sqrt {c} d-\sqrt {a} e} \left (4 c d^2+2 \sqrt {a} \sqrt {c} d e-a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a} e}}\right )}{32 a^{5/2} c^{7/4}}+\frac {3 \sqrt {\sqrt {c} d+\sqrt {a} e} \left (4 c d^2-2 \sqrt {a} \sqrt {c} d e-a e^2\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d+\sqrt {a} e}}\right )}{32 a^{5/2} c^{7/4}}\\ \end {align*}

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Mathematica [A]  time = 0.50, size = 260, normalized size = 0.93 \begin {gather*} \frac {\frac {2 \sqrt {a} c^{3/4} \sqrt {d+e x} \left (a^2 e (7 d+e x)+a c x \left (10 d^2+d e x+3 e^2 x^2\right )-6 c^2 d^2 x^3\right )}{\left (a-c x^2\right )^2}-3 \sqrt {\sqrt {c} d-\sqrt {a} e} \left (2 \sqrt {a} \sqrt {c} d e-a e^2+4 c d^2\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a} e}}\right )+3 \sqrt {\sqrt {a} e+\sqrt {c} d} \left (-2 \sqrt {a} \sqrt {c} d e-a e^2+4 c d^2\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {a} e+\sqrt {c} d}}\right )}{32 a^{5/2} c^{7/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(d + e*x)^(5/2)/(a - c*x^2)^3,x]

[Out]

((2*Sqrt[a]*c^(3/4)*Sqrt[d + e*x]*(-6*c^2*d^2*x^3 + a^2*e*(7*d + e*x) + a*c*x*(10*d^2 + d*e*x + 3*e^2*x^2)))/(
a - c*x^2)^2 - 3*Sqrt[Sqrt[c]*d - Sqrt[a]*e]*(4*c*d^2 + 2*Sqrt[a]*Sqrt[c]*d*e - a*e^2)*ArcTanh[(c^(1/4)*Sqrt[d
+ e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]] + 3*Sqrt[Sqrt[c]*d + Sqrt[a]*e]*(4*c*d^2 - 2*Sqrt[a]*Sqrt[c]*d*e - a*e^2
)*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/(32*a^(5/2)*c^(7/4))

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IntegrateAlgebraic [A]  time = 2.15, size = 454, normalized size = 1.63 \begin {gather*} \frac {3 \left (-a^{3/2} e^3+2 \sqrt {a} c d^2 e-3 a \sqrt {c} d e^2+4 c^{3/2} d^3\right ) \tan ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {-\sqrt {a} \sqrt {c} e-c d}}{\sqrt {a} e+\sqrt {c} d}\right )}{32 a^{5/2} c^{3/2} \sqrt {-\sqrt {c} \left (\sqrt {a} e+\sqrt {c} d\right )}}-\frac {3 \left (a^{3/2} e^3-2 \sqrt {a} c d^2 e-3 a \sqrt {c} d e^2+4 c^{3/2} d^3\right ) \tan ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {\sqrt {a} \sqrt {c} e-c d}}{\sqrt {c} d-\sqrt {a} e}\right )}{32 a^{5/2} c^{3/2} \sqrt {-\sqrt {c} \left (\sqrt {c} d-\sqrt {a} e\right )}}+\frac {\sqrt {d+e x} \left (a^2 e^5 (d+e x)+6 a^2 d e^5-12 a c d^3 e^3+17 a c d^2 e^3 (d+e x)-8 a c d e^3 (d+e x)^2+3 a c e^3 (d+e x)^3+6 c^2 d^5 e-18 c^2 d^4 e (d+e x)+18 c^2 d^3 e (d+e x)^2-6 c^2 d^2 e (d+e x)^3\right )}{16 a^2 c \left (a e^2-c d^2+2 c d (d+e x)-c (d+e x)^2\right )^2} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(d + e*x)^(5/2)/(a - c*x^2)^3,x]

[Out]

(Sqrt[d + e*x]*(6*c^2*d^5*e - 12*a*c*d^3*e^3 + 6*a^2*d*e^5 - 18*c^2*d^4*e*(d + e*x) + 17*a*c*d^2*e^3*(d + e*x)
+ a^2*e^5*(d + e*x) + 18*c^2*d^3*e*(d + e*x)^2 - 8*a*c*d*e^3*(d + e*x)^2 - 6*c^2*d^2*e*(d + e*x)^3 + 3*a*c*e^
3*(d + e*x)^3))/(16*a^2*c*(-(c*d^2) + a*e^2 + 2*c*d*(d + e*x) - c*(d + e*x)^2)^2) + (3*(4*c^(3/2)*d^3 + 2*Sqrt
[a]*c*d^2*e - 3*a*Sqrt[c]*d*e^2 - a^(3/2)*e^3)*ArcTan[(Sqrt[-(c*d) - Sqrt[a]*Sqrt[c]*e]*Sqrt[d + e*x])/(Sqrt[c
]*d + Sqrt[a]*e)])/(32*a^(5/2)*c^(3/2)*Sqrt[-(Sqrt[c]*(Sqrt[c]*d + Sqrt[a]*e))]) - (3*(4*c^(3/2)*d^3 - 2*Sqrt[
a]*c*d^2*e - 3*a*Sqrt[c]*d*e^2 + a^(3/2)*e^3)*ArcTan[(Sqrt[-(c*d) + Sqrt[a]*Sqrt[c]*e]*Sqrt[d + e*x])/(Sqrt[c]
*d - Sqrt[a]*e)])/(32*a^(5/2)*c^(3/2)*Sqrt[-(Sqrt[c]*(Sqrt[c]*d - Sqrt[a]*e))])

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fricas [B]  time = 0.45, size = 1029, normalized size = 3.69 \begin {gather*} \frac {3 \, {\left (a^{2} c^{3} x^{4} - 2 \, a^{3} c^{2} x^{2} + a^{4} c\right )} \sqrt {\frac {a^{5} c^{3} \sqrt {\frac {e^{10}}{a^{5} c^{7}}} + 16 \, c^{2} d^{5} - 20 \, a c d^{3} e^{2} + 5 \, a^{2} d e^{4}}{a^{5} c^{3}}} \log \left (27 \, {\left (16 \, c^{2} d^{4} e^{5} - 12 \, a c d^{2} e^{7} + a^{2} e^{9}\right )} \sqrt {e x + d} + 27 \, {\left (2 \, a^{3} c^{2} d e^{6} + {\left (4 \, a^{5} c^{6} d^{2} - a^{6} c^{5} e^{2}\right )} \sqrt {\frac {e^{10}}{a^{5} c^{7}}}\right )} \sqrt {\frac {a^{5} c^{3} \sqrt {\frac {e^{10}}{a^{5} c^{7}}} + 16 \, c^{2} d^{5} - 20 \, a c d^{3} e^{2} + 5 \, a^{2} d e^{4}}{a^{5} c^{3}}}\right ) - 3 \, {\left (a^{2} c^{3} x^{4} - 2 \, a^{3} c^{2} x^{2} + a^{4} c\right )} \sqrt {\frac {a^{5} c^{3} \sqrt {\frac {e^{10}}{a^{5} c^{7}}} + 16 \, c^{2} d^{5} - 20 \, a c d^{3} e^{2} + 5 \, a^{2} d e^{4}}{a^{5} c^{3}}} \log \left (27 \, {\left (16 \, c^{2} d^{4} e^{5} - 12 \, a c d^{2} e^{7} + a^{2} e^{9}\right )} \sqrt {e x + d} - 27 \, {\left (2 \, a^{3} c^{2} d e^{6} + {\left (4 \, a^{5} c^{6} d^{2} - a^{6} c^{5} e^{2}\right )} \sqrt {\frac {e^{10}}{a^{5} c^{7}}}\right )} \sqrt {\frac {a^{5} c^{3} \sqrt {\frac {e^{10}}{a^{5} c^{7}}} + 16 \, c^{2} d^{5} - 20 \, a c d^{3} e^{2} + 5 \, a^{2} d e^{4}}{a^{5} c^{3}}}\right ) + 3 \, {\left (a^{2} c^{3} x^{4} - 2 \, a^{3} c^{2} x^{2} + a^{4} c\right )} \sqrt {-\frac {a^{5} c^{3} \sqrt {\frac {e^{10}}{a^{5} c^{7}}} - 16 \, c^{2} d^{5} + 20 \, a c d^{3} e^{2} - 5 \, a^{2} d e^{4}}{a^{5} c^{3}}} \log \left (27 \, {\left (16 \, c^{2} d^{4} e^{5} - 12 \, a c d^{2} e^{7} + a^{2} e^{9}\right )} \sqrt {e x + d} + 27 \, {\left (2 \, a^{3} c^{2} d e^{6} - {\left (4 \, a^{5} c^{6} d^{2} - a^{6} c^{5} e^{2}\right )} \sqrt {\frac {e^{10}}{a^{5} c^{7}}}\right )} \sqrt {-\frac {a^{5} c^{3} \sqrt {\frac {e^{10}}{a^{5} c^{7}}} - 16 \, c^{2} d^{5} + 20 \, a c d^{3} e^{2} - 5 \, a^{2} d e^{4}}{a^{5} c^{3}}}\right ) - 3 \, {\left (a^{2} c^{3} x^{4} - 2 \, a^{3} c^{2} x^{2} + a^{4} c\right )} \sqrt {-\frac {a^{5} c^{3} \sqrt {\frac {e^{10}}{a^{5} c^{7}}} - 16 \, c^{2} d^{5} + 20 \, a c d^{3} e^{2} - 5 \, a^{2} d e^{4}}{a^{5} c^{3}}} \log \left (27 \, {\left (16 \, c^{2} d^{4} e^{5} - 12 \, a c d^{2} e^{7} + a^{2} e^{9}\right )} \sqrt {e x + d} - 27 \, {\left (2 \, a^{3} c^{2} d e^{6} - {\left (4 \, a^{5} c^{6} d^{2} - a^{6} c^{5} e^{2}\right )} \sqrt {\frac {e^{10}}{a^{5} c^{7}}}\right )} \sqrt {-\frac {a^{5} c^{3} \sqrt {\frac {e^{10}}{a^{5} c^{7}}} - 16 \, c^{2} d^{5} + 20 \, a c d^{3} e^{2} - 5 \, a^{2} d e^{4}}{a^{5} c^{3}}}\right ) + 4 \, {\left (a c d e x^{2} + 7 \, a^{2} d e - 3 \, {\left (2 \, c^{2} d^{2} - a c e^{2}\right )} x^{3} + {\left (10 \, a c d^{2} + a^{2} e^{2}\right )} x\right )} \sqrt {e x + d}}{64 \, {\left (a^{2} c^{3} x^{4} - 2 \, a^{3} c^{2} x^{2} + a^{4} c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(-c*x^2+a)^3,x, algorithm="fricas")

[Out]

1/64*(3*(a^2*c^3*x^4 - 2*a^3*c^2*x^2 + a^4*c)*sqrt((a^5*c^3*sqrt(e^10/(a^5*c^7)) + 16*c^2*d^5 - 20*a*c*d^3*e^2
+ 5*a^2*d*e^4)/(a^5*c^3))*log(27*(16*c^2*d^4*e^5 - 12*a*c*d^2*e^7 + a^2*e^9)*sqrt(e*x + d) + 27*(2*a^3*c^2*d*
e^6 + (4*a^5*c^6*d^2 - a^6*c^5*e^2)*sqrt(e^10/(a^5*c^7)))*sqrt((a^5*c^3*sqrt(e^10/(a^5*c^7)) + 16*c^2*d^5 - 20
*a*c*d^3*e^2 + 5*a^2*d*e^4)/(a^5*c^3))) - 3*(a^2*c^3*x^4 - 2*a^3*c^2*x^2 + a^4*c)*sqrt((a^5*c^3*sqrt(e^10/(a^5
*c^7)) + 16*c^2*d^5 - 20*a*c*d^3*e^2 + 5*a^2*d*e^4)/(a^5*c^3))*log(27*(16*c^2*d^4*e^5 - 12*a*c*d^2*e^7 + a^2*e
^9)*sqrt(e*x + d) - 27*(2*a^3*c^2*d*e^6 + (4*a^5*c^6*d^2 - a^6*c^5*e^2)*sqrt(e^10/(a^5*c^7)))*sqrt((a^5*c^3*sq
rt(e^10/(a^5*c^7)) + 16*c^2*d^5 - 20*a*c*d^3*e^2 + 5*a^2*d*e^4)/(a^5*c^3))) + 3*(a^2*c^3*x^4 - 2*a^3*c^2*x^2 +
a^4*c)*sqrt(-(a^5*c^3*sqrt(e^10/(a^5*c^7)) - 16*c^2*d^5 + 20*a*c*d^3*e^2 - 5*a^2*d*e^4)/(a^5*c^3))*log(27*(16
*c^2*d^4*e^5 - 12*a*c*d^2*e^7 + a^2*e^9)*sqrt(e*x + d) + 27*(2*a^3*c^2*d*e^6 - (4*a^5*c^6*d^2 - a^6*c^5*e^2)*s
qrt(e^10/(a^5*c^7)))*sqrt(-(a^5*c^3*sqrt(e^10/(a^5*c^7)) - 16*c^2*d^5 + 20*a*c*d^3*e^2 - 5*a^2*d*e^4)/(a^5*c^3
))) - 3*(a^2*c^3*x^4 - 2*a^3*c^2*x^2 + a^4*c)*sqrt(-(a^5*c^3*sqrt(e^10/(a^5*c^7)) - 16*c^2*d^5 + 20*a*c*d^3*e^
2 - 5*a^2*d*e^4)/(a^5*c^3))*log(27*(16*c^2*d^4*e^5 - 12*a*c*d^2*e^7 + a^2*e^9)*sqrt(e*x + d) - 27*(2*a^3*c^2*d
*e^6 - (4*a^5*c^6*d^2 - a^6*c^5*e^2)*sqrt(e^10/(a^5*c^7)))*sqrt(-(a^5*c^3*sqrt(e^10/(a^5*c^7)) - 16*c^2*d^5 +
20*a*c*d^3*e^2 - 5*a^2*d*e^4)/(a^5*c^3))) + 4*(a*c*d*e*x^2 + 7*a^2*d*e - 3*(2*c^2*d^2 - a*c*e^2)*x^3 + (10*a*c
*d^2 + a^2*e^2)*x)*sqrt(e*x + d))/(a^2*c^3*x^4 - 2*a^3*c^2*x^2 + a^4*c)

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giac [B]  time = 0.64, size = 617, normalized size = 2.21 \begin {gather*} -\frac {3 \, {\left (4 \, \sqrt {a c} c^{4} d^{4} - 3 \, \sqrt {a c} a c^{3} d^{2} e^{2} - {\left (2 \, \sqrt {a c} a c d^{2} e^{2} - \sqrt {a c} a^{2} e^{4}\right )} c^{2} - 2 \, {\left (a c^{3} d^{3} e - a^{2} c^{2} d e^{3}\right )} {\left | c \right |}\right )} \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-\frac {a^{2} c^{2} d + \sqrt {a^{4} c^{4} d^{2} - {\left (a^{2} c^{2} d^{2} - a^{3} c e^{2}\right )} a^{2} c^{2}}}{a^{2} c^{2}}}}\right )}{32 \, {\left (a^{3} c^{4} d - \sqrt {a c} a^{3} c^{3} e\right )} \sqrt {-c^{2} d - \sqrt {a c} c e}} + \frac {3 \, {\left (4 \, \sqrt {a c} c^{4} d^{4} - 3 \, \sqrt {a c} a c^{3} d^{2} e^{2} - {\left (2 \, \sqrt {a c} a c d^{2} e^{2} - \sqrt {a c} a^{2} e^{4}\right )} c^{2} + 2 \, {\left (a c^{3} d^{3} e - a^{2} c^{2} d e^{3}\right )} {\left | c \right |}\right )} \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-\frac {a^{2} c^{2} d - \sqrt {a^{4} c^{4} d^{2} - {\left (a^{2} c^{2} d^{2} - a^{3} c e^{2}\right )} a^{2} c^{2}}}{a^{2} c^{2}}}}\right )}{32 \, {\left (a^{3} c^{4} d + \sqrt {a c} a^{3} c^{3} e\right )} \sqrt {-c^{2} d + \sqrt {a c} c e}} - \frac {6 \, {\left (x e + d\right )}^{\frac {7}{2}} c^{2} d^{2} e - 18 \, {\left (x e + d\right )}^{\frac {5}{2}} c^{2} d^{3} e + 18 \, {\left (x e + d\right )}^{\frac {3}{2}} c^{2} d^{4} e - 6 \, \sqrt {x e + d} c^{2} d^{5} e - 3 \, {\left (x e + d\right )}^{\frac {7}{2}} a c e^{3} + 8 \, {\left (x e + d\right )}^{\frac {5}{2}} a c d e^{3} - 17 \, {\left (x e + d\right )}^{\frac {3}{2}} a c d^{2} e^{3} + 12 \, \sqrt {x e + d} a c d^{3} e^{3} - {\left (x e + d\right )}^{\frac {3}{2}} a^{2} e^{5} - 6 \, \sqrt {x e + d} a^{2} d e^{5}}{16 \, {\left ({\left (x e + d\right )}^{2} c - 2 \, {\left (x e + d\right )} c d + c d^{2} - a e^{2}\right )}^{2} a^{2} c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(-c*x^2+a)^3,x, algorithm="giac")

[Out]

-3/32*(4*sqrt(a*c)*c^4*d^4 - 3*sqrt(a*c)*a*c^3*d^2*e^2 - (2*sqrt(a*c)*a*c*d^2*e^2 - sqrt(a*c)*a^2*e^4)*c^2 - 2
*(a*c^3*d^3*e - a^2*c^2*d*e^3)*abs(c))*arctan(sqrt(x*e + d)/sqrt(-(a^2*c^2*d + sqrt(a^4*c^4*d^2 - (a^2*c^2*d^2
- a^3*c*e^2)*a^2*c^2))/(a^2*c^2)))/((a^3*c^4*d - sqrt(a*c)*a^3*c^3*e)*sqrt(-c^2*d - sqrt(a*c)*c*e)) + 3/32*(4
*sqrt(a*c)*c^4*d^4 - 3*sqrt(a*c)*a*c^3*d^2*e^2 - (2*sqrt(a*c)*a*c*d^2*e^2 - sqrt(a*c)*a^2*e^4)*c^2 + 2*(a*c^3*
d^3*e - a^2*c^2*d*e^3)*abs(c))*arctan(sqrt(x*e + d)/sqrt(-(a^2*c^2*d - sqrt(a^4*c^4*d^2 - (a^2*c^2*d^2 - a^3*c
*e^2)*a^2*c^2))/(a^2*c^2)))/((a^3*c^4*d + sqrt(a*c)*a^3*c^3*e)*sqrt(-c^2*d + sqrt(a*c)*c*e)) - 1/16*(6*(x*e +
d)^(7/2)*c^2*d^2*e - 18*(x*e + d)^(5/2)*c^2*d^3*e + 18*(x*e + d)^(3/2)*c^2*d^4*e - 6*sqrt(x*e + d)*c^2*d^5*e -
3*(x*e + d)^(7/2)*a*c*e^3 + 8*(x*e + d)^(5/2)*a*c*d*e^3 - 17*(x*e + d)^(3/2)*a*c*d^2*e^3 + 12*sqrt(x*e + d)*a
*c*d^3*e^3 - (x*e + d)^(3/2)*a^2*e^5 - 6*sqrt(x*e + d)*a^2*d*e^5)/(((x*e + d)^2*c - 2*(x*e + d)*c*d + c*d^2 -
a*e^2)^2*a^2*c)

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maple [B]  time = 0.10, size = 792, normalized size = 2.84 \begin {gather*} -\frac {3 \sqrt {e x +d}\, d^{3} e^{3}}{4 \left (c \,e^{2} x^{2}-a \,e^{2}\right )^{2} a}+\frac {3 \sqrt {e x +d}\, c \,d^{5} e}{8 \left (c \,e^{2} x^{2}-a \,e^{2}\right )^{2} a^{2}}+\frac {3 \sqrt {e x +d}\, d \,e^{5}}{8 \left (c \,e^{2} x^{2}-a \,e^{2}\right )^{2} c}+\frac {17 \left (e x +d \right )^{\frac {3}{2}} d^{2} e^{3}}{16 \left (c \,e^{2} x^{2}-a \,e^{2}\right )^{2} a}-\frac {9 d \,e^{3} \arctanh \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{32 \sqrt {a c \,e^{2}}\, \sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}\, a}-\frac {9 d \,e^{3} \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{32 \sqrt {a c \,e^{2}}\, \sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}\, a}-\frac {9 \left (e x +d \right )^{\frac {3}{2}} c \,d^{4} e}{8 \left (c \,e^{2} x^{2}-a \,e^{2}\right )^{2} a^{2}}+\frac {3 c \,d^{3} e \arctanh \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{8 \sqrt {a c \,e^{2}}\, \sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}\, a^{2}}+\frac {3 c \,d^{3} e \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{8 \sqrt {a c \,e^{2}}\, \sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}\, a^{2}}+\frac {\left (e x +d \right )^{\frac {3}{2}} e^{5}}{16 \left (c \,e^{2} x^{2}-a \,e^{2}\right )^{2} c}-\frac {\left (e x +d \right )^{\frac {5}{2}} d \,e^{3}}{2 \left (c \,e^{2} x^{2}-a \,e^{2}\right )^{2} a}+\frac {9 \left (e x +d \right )^{\frac {5}{2}} c \,d^{3} e}{8 \left (c \,e^{2} x^{2}-a \,e^{2}\right )^{2} a^{2}}-\frac {3 e^{3} \arctanh \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{32 \sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}\, a c}+\frac {3 e^{3} \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{32 \sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}\, a c}+\frac {3 \left (e x +d \right )^{\frac {7}{2}} e^{3}}{16 \left (c \,e^{2} x^{2}-a \,e^{2}\right )^{2} a}-\frac {3 \left (e x +d \right )^{\frac {7}{2}} c \,d^{2} e}{8 \left (c \,e^{2} x^{2}-a \,e^{2}\right )^{2} a^{2}}+\frac {3 d^{2} e \arctanh \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{16 \sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}\, a^{2}}-\frac {3 d^{2} e \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{16 \sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}\, a^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*x+d)^(5/2)/(-c*x^2+a)^3,x)

[Out]

3/16*e^3/(c*e^2*x^2-a*e^2)^2/a*(e*x+d)^(7/2)-3/8*e/(c*e^2*x^2-a*e^2)^2/a^2*(e*x+d)^(7/2)*c*d^2-1/2*e^3/(c*e^2*
x^2-a*e^2)^2*d/a*(e*x+d)^(5/2)+9/8*e/(c*e^2*x^2-a*e^2)^2*d^3/a^2*(e*x+d)^(5/2)*c+1/16*e^5/(c*e^2*x^2-a*e^2)^2/
c*(e*x+d)^(3/2)+17/16*e^3/(c*e^2*x^2-a*e^2)^2/a*(e*x+d)^(3/2)*d^2-9/8*e/(c*e^2*x^2-a*e^2)^2/a^2*c*(e*x+d)^(3/2
)*d^4+3/8*e^5/(c*e^2*x^2-a*e^2)^2*d/c*(e*x+d)^(1/2)-3/4*e^3/(c*e^2*x^2-a*e^2)^2*d^3/a*(e*x+d)^(1/2)+3/8*e/(c*e
^2*x^2-a*e^2)^2*d^5/a^2*c*(e*x+d)^(1/2)-9/32*e^3/a/(a*c*e^2)^(1/2)/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctanh((e*
x+d)^(1/2)/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)*c)*d+3/8*e/a^2*c/(a*c*e^2)^(1/2)/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)*ar
ctanh((e*x+d)^(1/2)/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)*c)*d^3-3/32*e^3/a/c/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctan
h((e*x+d)^(1/2)/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)*c)+3/16*e/a^2/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctanh((e*x+d)^
(1/2)/((c*d+(a*c*e^2)^(1/2))*c)^(1/2)*c)*d^2-9/32*e^3/a/(a*c*e^2)^(1/2)/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arcta
n((e*x+d)^(1/2)/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2)*c)*d+3/8*e/a^2*c/(a*c*e^2)^(1/2)/((-c*d+(a*c*e^2)^(1/2))*c)^(
1/2)*arctan((e*x+d)^(1/2)/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2)*c)*d^3+3/32*e^3/a/c/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2
)*arctan((e*x+d)^(1/2)/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2)*c)-3/16*e/a^2/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2)*arctan(
(e*x+d)^(1/2)/((-c*d+(a*c*e^2)^(1/2))*c)^(1/2)*c)*d^2

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} -\int \frac {{\left (e x + d\right )}^{\frac {5}{2}}}{{\left (c x^{2} - a\right )}^{3}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)^(5/2)/(-c*x^2+a)^3,x, algorithm="maxima")

[Out]

-integrate((e*x + d)^(5/2)/(c*x^2 - a)^3, x)

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mupad [B]  time = 0.52, size = 1015, normalized size = 3.64 \begin {gather*} \frac {\frac {3\,e\,\left (a\,e^2-2\,c\,d^2\right )\,{\left (d+e\,x\right )}^{7/2}}{16\,a^2}+\frac {{\left (d+e\,x\right )}^{3/2}\,\left (a^2\,e^5+17\,a\,c\,d^2\,e^3-18\,c^2\,d^4\,e\right )}{16\,a^2\,c}-\frac {d\,\left (4\,a\,e^3-9\,c\,d^2\,e\right )\,{\left (d+e\,x\right )}^{5/2}}{8\,a^2}+\frac {3\,\sqrt {d+e\,x}\,\left (a^2\,d\,e^5-2\,a\,c\,d^3\,e^3+c^2\,d^5\,e\right )}{8\,a^2\,c}}{c^2\,{\left (d+e\,x\right )}^4+a^2\,e^4+c^2\,d^4+\left (6\,c^2\,d^2-2\,a\,c\,e^2\right )\,{\left (d+e\,x\right )}^2-\left (4\,c^2\,d^3-4\,a\,c\,d\,e^2\right )\,\left (d+e\,x\right )-4\,c^2\,d\,{\left (d+e\,x\right )}^3-2\,a\,c\,d^2\,e^2}-2\,\mathrm {atanh}\left (\frac {9\,e^8\,\sqrt {d+e\,x}\,\sqrt {\frac {9\,d^5}{256\,a^5\,c}+\frac {45\,d\,e^4}{4096\,a^3\,c^3}-\frac {45\,d^3\,e^2}{1024\,a^4\,c^2}-\frac {9\,e^5\,\sqrt {a^{15}\,c^7}}{4096\,a^{10}\,c^7}}}{32\,\left (\frac {27\,e^{11}}{2048\,a\,c^2}+\frac {27\,d^4\,e^7}{512\,a^3}-\frac {135\,d^2\,e^9}{2048\,a^2\,c}-\frac {27\,d\,e^{10}\,\sqrt {a^{15}\,c^7}}{1024\,a^9\,c^5}+\frac {27\,d^3\,e^8\,\sqrt {a^{15}\,c^7}}{1024\,a^{10}\,c^4}\right )}+\frac {9\,d\,e^7\,\sqrt {a^{15}\,c^7}\,\sqrt {d+e\,x}\,\sqrt {\frac {9\,d^5}{256\,a^5\,c}+\frac {45\,d\,e^4}{4096\,a^3\,c^3}-\frac {45\,d^3\,e^2}{1024\,a^4\,c^2}-\frac {9\,e^5\,\sqrt {a^{15}\,c^7}}{4096\,a^{10}\,c^7}}}{32\,\left (\frac {27\,a^7\,c\,e^{11}}{2048}+\frac {27\,a^5\,c^3\,d^4\,e^7}{512}-\frac {135\,a^6\,c^2\,d^2\,e^9}{2048}-\frac {27\,d\,e^{10}\,\sqrt {a^{15}\,c^7}}{1024\,a\,c^2}+\frac {27\,d^3\,e^8\,\sqrt {a^{15}\,c^7}}{1024\,a^2\,c}\right )}\right )\,\sqrt {-\frac {9\,\left (e^5\,\sqrt {a^{15}\,c^7}-16\,a^5\,c^6\,d^5-5\,a^7\,c^4\,d\,e^4+20\,a^6\,c^5\,d^3\,e^2\right )}{4096\,a^{10}\,c^7}}-2\,\mathrm {atanh}\left (\frac {9\,e^8\,\sqrt {d+e\,x}\,\sqrt {\frac {9\,d^5}{256\,a^5\,c}+\frac {45\,d\,e^4}{4096\,a^3\,c^3}-\frac {45\,d^3\,e^2}{1024\,a^4\,c^2}+\frac {9\,e^5\,\sqrt {a^{15}\,c^7}}{4096\,a^{10}\,c^7}}}{32\,\left (\frac {27\,e^{11}}{2048\,a\,c^2}+\frac {27\,d^4\,e^7}{512\,a^3}-\frac {135\,d^2\,e^9}{2048\,a^2\,c}+\frac {27\,d\,e^{10}\,\sqrt {a^{15}\,c^7}}{1024\,a^9\,c^5}-\frac {27\,d^3\,e^8\,\sqrt {a^{15}\,c^7}}{1024\,a^{10}\,c^4}\right )}-\frac {9\,d\,e^7\,\sqrt {a^{15}\,c^7}\,\sqrt {d+e\,x}\,\sqrt {\frac {9\,d^5}{256\,a^5\,c}+\frac {45\,d\,e^4}{4096\,a^3\,c^3}-\frac {45\,d^3\,e^2}{1024\,a^4\,c^2}+\frac {9\,e^5\,\sqrt {a^{15}\,c^7}}{4096\,a^{10}\,c^7}}}{32\,\left (\frac {27\,a^7\,c\,e^{11}}{2048}+\frac {27\,a^5\,c^3\,d^4\,e^7}{512}-\frac {135\,a^6\,c^2\,d^2\,e^9}{2048}+\frac {27\,d\,e^{10}\,\sqrt {a^{15}\,c^7}}{1024\,a\,c^2}-\frac {27\,d^3\,e^8\,\sqrt {a^{15}\,c^7}}{1024\,a^2\,c}\right )}\right )\,\sqrt {\frac {9\,\left (e^5\,\sqrt {a^{15}\,c^7}+16\,a^5\,c^6\,d^5+5\,a^7\,c^4\,d\,e^4-20\,a^6\,c^5\,d^3\,e^2\right )}{4096\,a^{10}\,c^7}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(5/2)/(a - c*x^2)^3,x)

[Out]

((3*e*(a*e^2 - 2*c*d^2)*(d + e*x)^(7/2))/(16*a^2) + ((d + e*x)^(3/2)*(a^2*e^5 - 18*c^2*d^4*e + 17*a*c*d^2*e^3)
)/(16*a^2*c) - (d*(4*a*e^3 - 9*c*d^2*e)*(d + e*x)^(5/2))/(8*a^2) + (3*(d + e*x)^(1/2)*(a^2*d*e^5 + c^2*d^5*e -
2*a*c*d^3*e^3))/(8*a^2*c))/(c^2*(d + e*x)^4 + a^2*e^4 + c^2*d^4 + (6*c^2*d^2 - 2*a*c*e^2)*(d + e*x)^2 - (4*c^
2*d^3 - 4*a*c*d*e^2)*(d + e*x) - 4*c^2*d*(d + e*x)^3 - 2*a*c*d^2*e^2) - 2*atanh((9*e^8*(d + e*x)^(1/2)*((9*d^5
)/(256*a^5*c) + (45*d*e^4)/(4096*a^3*c^3) - (45*d^3*e^2)/(1024*a^4*c^2) - (9*e^5*(a^15*c^7)^(1/2))/(4096*a^10*
c^7))^(1/2))/(32*((27*e^11)/(2048*a*c^2) + (27*d^4*e^7)/(512*a^3) - (135*d^2*e^9)/(2048*a^2*c) - (27*d*e^10*(a
^15*c^7)^(1/2))/(1024*a^9*c^5) + (27*d^3*e^8*(a^15*c^7)^(1/2))/(1024*a^10*c^4))) + (9*d*e^7*(a^15*c^7)^(1/2)*(
d + e*x)^(1/2)*((9*d^5)/(256*a^5*c) + (45*d*e^4)/(4096*a^3*c^3) - (45*d^3*e^2)/(1024*a^4*c^2) - (9*e^5*(a^15*c
^7)^(1/2))/(4096*a^10*c^7))^(1/2))/(32*((27*a^7*c*e^11)/2048 + (27*a^5*c^3*d^4*e^7)/512 - (135*a^6*c^2*d^2*e^9
)/2048 - (27*d*e^10*(a^15*c^7)^(1/2))/(1024*a*c^2) + (27*d^3*e^8*(a^15*c^7)^(1/2))/(1024*a^2*c))))*(-(9*(e^5*(
a^15*c^7)^(1/2) - 16*a^5*c^6*d^5 - 5*a^7*c^4*d*e^4 + 20*a^6*c^5*d^3*e^2))/(4096*a^10*c^7))^(1/2) - 2*atanh((9*
e^8*(d + e*x)^(1/2)*((9*d^5)/(256*a^5*c) + (45*d*e^4)/(4096*a^3*c^3) - (45*d^3*e^2)/(1024*a^4*c^2) + (9*e^5*(a
^15*c^7)^(1/2))/(4096*a^10*c^7))^(1/2))/(32*((27*e^11)/(2048*a*c^2) + (27*d^4*e^7)/(512*a^3) - (135*d^2*e^9)/(
2048*a^2*c) + (27*d*e^10*(a^15*c^7)^(1/2))/(1024*a^9*c^5) - (27*d^3*e^8*(a^15*c^7)^(1/2))/(1024*a^10*c^4))) -
(9*d*e^7*(a^15*c^7)^(1/2)*(d + e*x)^(1/2)*((9*d^5)/(256*a^5*c) + (45*d*e^4)/(4096*a^3*c^3) - (45*d^3*e^2)/(102
4*a^4*c^2) + (9*e^5*(a^15*c^7)^(1/2))/(4096*a^10*c^7))^(1/2))/(32*((27*a^7*c*e^11)/2048 + (27*a^5*c^3*d^4*e^7)
/512 - (135*a^6*c^2*d^2*e^9)/2048 + (27*d*e^10*(a^15*c^7)^(1/2))/(1024*a*c^2) - (27*d^3*e^8*(a^15*c^7)^(1/2))/
(1024*a^2*c))))*((9*(e^5*(a^15*c^7)^(1/2) + 16*a^5*c^6*d^5 + 5*a^7*c^4*d*e^4 - 20*a^6*c^5*d^3*e^2))/(4096*a^10
*c^7))^(1/2)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**(5/2)/(-c*x**2+a)**3,x)

[Out]

Timed out

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