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

Optimal. Leaf size=209 \[ -\frac {\sqrt {\sqrt {c} d-\sqrt {a} e} \left (\sqrt {a} e+2 \sqrt {c} d\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a} e}}\right )}{4 a^{3/2} c^{5/4}}+\frac {\left (2 \sqrt {c} d-\sqrt {a} e\right ) \sqrt {\sqrt {a} e+\sqrt {c} d} \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {a} e+\sqrt {c} d}}\right )}{4 a^{3/2} c^{5/4}}+\frac {\sqrt {d+e x} (a e+c d x)}{2 a c \left (a-c x^2\right )} \]

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Rubi [A]  time = 0.27, antiderivative size = 209, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {739, 827, 1166, 208} \begin {gather*} -\frac {\sqrt {\sqrt {c} d-\sqrt {a} e} \left (\sqrt {a} e+2 \sqrt {c} d\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a} e}}\right )}{4 a^{3/2} c^{5/4}}+\frac {\left (2 \sqrt {c} d-\sqrt {a} e\right ) \sqrt {\sqrt {a} e+\sqrt {c} d} \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {a} e+\sqrt {c} d}}\right )}{4 a^{3/2} c^{5/4}}+\frac {\sqrt {d+e x} (a e+c d x)}{2 a c \left (a-c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(d + e*x)^(3/2)/(a - c*x^2)^2,x]

[Out]

((a*e + c*d*x)*Sqrt[d + e*x])/(2*a*c*(a - c*x^2)) - (Sqrt[Sqrt[c]*d - Sqrt[a]*e]*(2*Sqrt[c]*d + Sqrt[a]*e)*Arc
Tanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d - Sqrt[a]*e]])/(4*a^(3/2)*c^(5/4)) + ((2*Sqrt[c]*d - Sqrt[a]*e)*Sq
rt[Sqrt[c]*d + Sqrt[a]*e]*ArcTanh[(c^(1/4)*Sqrt[d + e*x])/Sqrt[Sqrt[c]*d + Sqrt[a]*e]])/(4*a^(3/2)*c^(5/4))

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

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Mathematica [A]  time = 0.25, size = 218, normalized size = 1.04 \begin {gather*} \frac {\left (c x^2-a\right ) \sqrt {\sqrt {c} d-\sqrt {a} e} \left (\sqrt {a} e+2 \sqrt {c} d\right ) \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {c} d-\sqrt {a} e}}\right )-\left (c x^2-a\right ) \left (2 \sqrt {c} d-\sqrt {a} e\right ) \sqrt {\sqrt {a} e+\sqrt {c} d} \tanh ^{-1}\left (\frac {\sqrt [4]{c} \sqrt {d+e x}}{\sqrt {\sqrt {a} e+\sqrt {c} d}}\right )+2 \sqrt {a} \sqrt [4]{c} \sqrt {d+e x} (a e+c d x)}{4 a^{3/2} c^{5/4} \left (a-c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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IntegrateAlgebraic [A]  time = 1.27, size = 288, normalized size = 1.38 \begin {gather*} -\frac {\left (2 \sqrt {c} d-\sqrt {a} e\right ) \sqrt {-\sqrt {c} \left (\sqrt {a} e+\sqrt {c} d\right )} \tan ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {-\sqrt {a} \sqrt {c} e-c d}}{\sqrt {a} e+\sqrt {c} d}\right )}{4 a^{3/2} c^{3/2}}+\frac {\sqrt {-\sqrt {c} \left (\sqrt {c} d-\sqrt {a} e\right )} \left (\sqrt {a} e+2 \sqrt {c} d\right ) \tan ^{-1}\left (\frac {\sqrt {d+e x} \sqrt {\sqrt {a} \sqrt {c} e-c d}}{\sqrt {c} d-\sqrt {a} e}\right )}{4 a^{3/2} c^{3/2}}+\frac {e \sqrt {d+e x} \left (a e^2-c d^2+c d (d+e x)\right )}{2 a c \left (a e^2-c d^2+2 c d (d+e x)-c (d+e x)^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(e*Sqrt[d + e*x]*(-(c*d^2) + a*e^2 + c*d*(d + e*x)))/(2*a*c*(-(c*d^2) + a*e^2 + 2*c*d*(d + e*x) - c*(d + e*x)^
2)) - ((2*Sqrt[c]*d - Sqrt[a]*e)*Sqrt[-(Sqrt[c]*(Sqrt[c]*d + Sqrt[a]*e))]*ArcTan[(Sqrt[-(c*d) - Sqrt[a]*Sqrt[c
]*e]*Sqrt[d + e*x])/(Sqrt[c]*d + Sqrt[a]*e)])/(4*a^(3/2)*c^(3/2)) + (Sqrt[-(Sqrt[c]*(Sqrt[c]*d - Sqrt[a]*e))]*
(2*Sqrt[c]*d + Sqrt[a]*e)*ArcTan[(Sqrt[-(c*d) + Sqrt[a]*Sqrt[c]*e]*Sqrt[d + e*x])/(Sqrt[c]*d - Sqrt[a]*e)])/(4
*a^(3/2)*c^(3/2))

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fricas [B]  time = 0.43, size = 679, normalized size = 3.25 \begin {gather*} -\frac {{\left (a c^{2} x^{2} - a^{2} c\right )} \sqrt {\frac {a^{3} c^{2} \sqrt {\frac {e^{6}}{a^{3} c^{5}}} + 4 \, c d^{3} - 3 \, a d e^{2}}{a^{3} c^{2}}} \log \left (-{\left (4 \, c d^{2} e^{3} - a e^{5}\right )} \sqrt {e x + d} + {\left (2 \, a^{3} c^{4} d \sqrt {\frac {e^{6}}{a^{3} c^{5}}} + a^{2} c e^{4}\right )} \sqrt {\frac {a^{3} c^{2} \sqrt {\frac {e^{6}}{a^{3} c^{5}}} + 4 \, c d^{3} - 3 \, a d e^{2}}{a^{3} c^{2}}}\right ) - {\left (a c^{2} x^{2} - a^{2} c\right )} \sqrt {\frac {a^{3} c^{2} \sqrt {\frac {e^{6}}{a^{3} c^{5}}} + 4 \, c d^{3} - 3 \, a d e^{2}}{a^{3} c^{2}}} \log \left (-{\left (4 \, c d^{2} e^{3} - a e^{5}\right )} \sqrt {e x + d} - {\left (2 \, a^{3} c^{4} d \sqrt {\frac {e^{6}}{a^{3} c^{5}}} + a^{2} c e^{4}\right )} \sqrt {\frac {a^{3} c^{2} \sqrt {\frac {e^{6}}{a^{3} c^{5}}} + 4 \, c d^{3} - 3 \, a d e^{2}}{a^{3} c^{2}}}\right ) - {\left (a c^{2} x^{2} - a^{2} c\right )} \sqrt {-\frac {a^{3} c^{2} \sqrt {\frac {e^{6}}{a^{3} c^{5}}} - 4 \, c d^{3} + 3 \, a d e^{2}}{a^{3} c^{2}}} \log \left (-{\left (4 \, c d^{2} e^{3} - a e^{5}\right )} \sqrt {e x + d} + {\left (2 \, a^{3} c^{4} d \sqrt {\frac {e^{6}}{a^{3} c^{5}}} - a^{2} c e^{4}\right )} \sqrt {-\frac {a^{3} c^{2} \sqrt {\frac {e^{6}}{a^{3} c^{5}}} - 4 \, c d^{3} + 3 \, a d e^{2}}{a^{3} c^{2}}}\right ) + {\left (a c^{2} x^{2} - a^{2} c\right )} \sqrt {-\frac {a^{3} c^{2} \sqrt {\frac {e^{6}}{a^{3} c^{5}}} - 4 \, c d^{3} + 3 \, a d e^{2}}{a^{3} c^{2}}} \log \left (-{\left (4 \, c d^{2} e^{3} - a e^{5}\right )} \sqrt {e x + d} - {\left (2 \, a^{3} c^{4} d \sqrt {\frac {e^{6}}{a^{3} c^{5}}} - a^{2} c e^{4}\right )} \sqrt {-\frac {a^{3} c^{2} \sqrt {\frac {e^{6}}{a^{3} c^{5}}} - 4 \, c d^{3} + 3 \, a d e^{2}}{a^{3} c^{2}}}\right ) + 4 \, {\left (c d x + a e\right )} \sqrt {e x + d}}{8 \, {\left (a c^{2} x^{2} - a^{2} c\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/8*((a*c^2*x^2 - a^2*c)*sqrt((a^3*c^2*sqrt(e^6/(a^3*c^5)) + 4*c*d^3 - 3*a*d*e^2)/(a^3*c^2))*log(-(4*c*d^2*e^
3 - a*e^5)*sqrt(e*x + d) + (2*a^3*c^4*d*sqrt(e^6/(a^3*c^5)) + a^2*c*e^4)*sqrt((a^3*c^2*sqrt(e^6/(a^3*c^5)) + 4
*c*d^3 - 3*a*d*e^2)/(a^3*c^2))) - (a*c^2*x^2 - a^2*c)*sqrt((a^3*c^2*sqrt(e^6/(a^3*c^5)) + 4*c*d^3 - 3*a*d*e^2)
/(a^3*c^2))*log(-(4*c*d^2*e^3 - a*e^5)*sqrt(e*x + d) - (2*a^3*c^4*d*sqrt(e^6/(a^3*c^5)) + a^2*c*e^4)*sqrt((a^3
*c^2*sqrt(e^6/(a^3*c^5)) + 4*c*d^3 - 3*a*d*e^2)/(a^3*c^2))) - (a*c^2*x^2 - a^2*c)*sqrt(-(a^3*c^2*sqrt(e^6/(a^3
*c^5)) - 4*c*d^3 + 3*a*d*e^2)/(a^3*c^2))*log(-(4*c*d^2*e^3 - a*e^5)*sqrt(e*x + d) + (2*a^3*c^4*d*sqrt(e^6/(a^3
*c^5)) - a^2*c*e^4)*sqrt(-(a^3*c^2*sqrt(e^6/(a^3*c^5)) - 4*c*d^3 + 3*a*d*e^2)/(a^3*c^2))) + (a*c^2*x^2 - a^2*c
)*sqrt(-(a^3*c^2*sqrt(e^6/(a^3*c^5)) - 4*c*d^3 + 3*a*d*e^2)/(a^3*c^2))*log(-(4*c*d^2*e^3 - a*e^5)*sqrt(e*x + d
) - (2*a^3*c^4*d*sqrt(e^6/(a^3*c^5)) - a^2*c*e^4)*sqrt(-(a^3*c^2*sqrt(e^6/(a^3*c^5)) - 4*c*d^3 + 3*a*d*e^2)/(a
^3*c^2))) + 4*(c*d*x + a*e)*sqrt(e*x + d))/(a*c^2*x^2 - a^2*c)

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giac [B]  time = 0.54, size = 423, normalized size = 2.02 \begin {gather*} -\frac {{\left (2 \, \sqrt {a c} a c^{3} d^{3} - 2 \, \sqrt {a c} a^{2} c^{2} d e^{2} - {\left (a c^{2} d^{2} e - a^{2} c e^{3}\right )} {\left | a \right |} {\left | c \right |}\right )} \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-\frac {a c^{2} d + \sqrt {a^{2} c^{4} d^{2} - {\left (a c^{2} d^{2} - a^{2} c e^{2}\right )} a c^{2}}}{a c^{2}}}}\right )}{4 \, {\left (a^{2} c^{3} d - \sqrt {a c} a^{2} c^{2} e\right )} \sqrt {-c^{2} d - \sqrt {a c} c e} {\left | a \right |}} + \frac {{\left (2 \, \sqrt {a c} a c^{3} d^{3} - 2 \, \sqrt {a c} a^{2} c^{2} d e^{2} + {\left (a c^{2} d^{2} e - a^{2} c e^{3}\right )} {\left | a \right |} {\left | c \right |}\right )} \arctan \left (\frac {\sqrt {x e + d}}{\sqrt {-\frac {a c^{2} d - \sqrt {a^{2} c^{4} d^{2} - {\left (a c^{2} d^{2} - a^{2} c e^{2}\right )} a c^{2}}}{a c^{2}}}}\right )}{4 \, {\left (a^{2} c^{3} d + \sqrt {a c} a^{2} c^{2} e\right )} \sqrt {-c^{2} d + \sqrt {a c} c e} {\left | a \right |}} - \frac {{\left (x e + d\right )}^{\frac {3}{2}} c d e - \sqrt {x e + d} c d^{2} e + \sqrt {x e + d} a e^{3}}{2 \, {\left ({\left (x e + d\right )}^{2} c - 2 \, {\left (x e + d\right )} c d + c d^{2} - a e^{2}\right )} a c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*(2*sqrt(a*c)*a*c^3*d^3 - 2*sqrt(a*c)*a^2*c^2*d*e^2 - (a*c^2*d^2*e - a^2*c*e^3)*abs(a)*abs(c))*arctan(sqrt
(x*e + d)/sqrt(-(a*c^2*d + sqrt(a^2*c^4*d^2 - (a*c^2*d^2 - a^2*c*e^2)*a*c^2))/(a*c^2)))/((a^2*c^3*d - sqrt(a*c
)*a^2*c^2*e)*sqrt(-c^2*d - sqrt(a*c)*c*e)*abs(a)) + 1/4*(2*sqrt(a*c)*a*c^3*d^3 - 2*sqrt(a*c)*a^2*c^2*d*e^2 + (
a*c^2*d^2*e - a^2*c*e^3)*abs(a)*abs(c))*arctan(sqrt(x*e + d)/sqrt(-(a*c^2*d - sqrt(a^2*c^4*d^2 - (a*c^2*d^2 -
a^2*c*e^2)*a*c^2))/(a*c^2)))/((a^2*c^3*d + sqrt(a*c)*a^2*c^2*e)*sqrt(-c^2*d + sqrt(a*c)*c*e)*abs(a)) - 1/2*((x
*e + d)^(3/2)*c*d*e - sqrt(x*e + d)*c*d^2*e + sqrt(x*e + d)*a*e^3)/(((x*e + d)^2*c - 2*(x*e + d)*c*d + c*d^2 -
a*e^2)*a*c)

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maple [B]  time = 0.09, size = 432, normalized size = 2.07 \begin {gather*} \frac {c \,d^{2} e \arctanh \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{2 \sqrt {a c \,e^{2}}\, \sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}\, a}+\frac {c \,d^{2} e \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{2 \sqrt {a c \,e^{2}}\, \sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}\, a}-\frac {e^{3} \arctanh \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{4 \sqrt {a c \,e^{2}}\, \sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}-\frac {e^{3} \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{4 \sqrt {a c \,e^{2}}\, \sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}+\frac {\sqrt {e x +d}\, d^{2} e}{2 \left (c \,e^{2} x^{2}-a \,e^{2}\right ) a}+\frac {d e \arctanh \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{4 \sqrt {\left (c d +\sqrt {a c \,e^{2}}\right ) c}\, a}-\frac {d e \arctan \left (\frac {\sqrt {e x +d}\, c}{\sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}}\right )}{4 \sqrt {\left (-c d +\sqrt {a c \,e^{2}}\right ) c}\, a}-\frac {\sqrt {e x +d}\, e^{3}}{2 \left (c \,e^{2} x^{2}-a \,e^{2}\right ) c}-\frac {\left (e x +d \right )^{\frac {3}{2}} d e}{2 \left (c \,e^{2} x^{2}-a \,e^{2}\right ) a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*e/(c*e^2*x^2-a*e^2)*d/a*(e*x+d)^(3/2)-1/2*e^3/(c*e^2*x^2-a*e^2)/c*(e*x+d)^(1/2)+1/2*e/(c*e^2*x^2-a*e^2)/a
*(e*x+d)^(1/2)*d^2-1/4*e^3/(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)+1/2*e/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)*c*d^2+1/4*e/a/((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-1/4*e^3/(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)+1/2*e/a/(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)*c*d^2-1/4*e/a/((-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

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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 {3}{2}}}{{\left (c x^{2} - a\right )}^{2}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.43, size = 704, normalized size = 3.37 \begin {gather*} 2\,\mathrm {atanh}\left (\frac {2\,c\,e^6\,\sqrt {d+e\,x}\,\sqrt {\frac {d^3}{16\,a^3\,c}-\frac {3\,d\,e^2}{64\,a^2\,c^2}-\frac {e^3\,\sqrt {a^9\,c^5}}{64\,a^6\,c^5}}}{\frac {d\,e^7}{2\,a}-\frac {c\,d^3\,e^5}{2\,a^2}+\frac {e^8\,\sqrt {a^9\,c^5}}{4\,a^5\,c^3}-\frac {d^2\,e^6\,\sqrt {a^9\,c^5}}{4\,a^6\,c^2}}+\frac {2\,d\,e^5\,\sqrt {a^9\,c^5}\,\sqrt {d+e\,x}\,\sqrt {\frac {d^3}{16\,a^3\,c}-\frac {3\,d\,e^2}{64\,a^2\,c^2}-\frac {e^3\,\sqrt {a^9\,c^5}}{64\,a^6\,c^5}}}{\frac {e^8\,\sqrt {a^9\,c^5}}{4\,c^2}-\frac {a^3\,c^2\,d^3\,e^5}{2}+\frac {a^4\,c\,d\,e^7}{2}-\frac {d^2\,e^6\,\sqrt {a^9\,c^5}}{4\,a\,c}}\right )\,\sqrt {-\frac {e^3\,\sqrt {a^9\,c^5}-4\,a^3\,c^4\,d^3+3\,a^4\,c^3\,d\,e^2}{64\,a^6\,c^5}}-\frac {\frac {\left (a\,e^3-c\,d^2\,e\right )\,\sqrt {d+e\,x}}{2\,a\,c}+\frac {d\,e\,{\left (d+e\,x\right )}^{3/2}}{2\,a}}{c\,{\left (d+e\,x\right )}^2-a\,e^2+c\,d^2-2\,c\,d\,\left (d+e\,x\right )}+2\,\mathrm {atanh}\left (\frac {2\,c\,e^6\,\sqrt {d+e\,x}\,\sqrt {\frac {d^3}{16\,a^3\,c}-\frac {3\,d\,e^2}{64\,a^2\,c^2}+\frac {e^3\,\sqrt {a^9\,c^5}}{64\,a^6\,c^5}}}{\frac {d\,e^7}{2\,a}-\frac {c\,d^3\,e^5}{2\,a^2}-\frac {e^8\,\sqrt {a^9\,c^5}}{4\,a^5\,c^3}+\frac {d^2\,e^6\,\sqrt {a^9\,c^5}}{4\,a^6\,c^2}}+\frac {2\,d\,e^5\,\sqrt {a^9\,c^5}\,\sqrt {d+e\,x}\,\sqrt {\frac {d^3}{16\,a^3\,c}-\frac {3\,d\,e^2}{64\,a^2\,c^2}+\frac {e^3\,\sqrt {a^9\,c^5}}{64\,a^6\,c^5}}}{\frac {e^8\,\sqrt {a^9\,c^5}}{4\,c^2}+\frac {a^3\,c^2\,d^3\,e^5}{2}-\frac {a^4\,c\,d\,e^7}{2}-\frac {d^2\,e^6\,\sqrt {a^9\,c^5}}{4\,a\,c}}\right )\,\sqrt {\frac {e^3\,\sqrt {a^9\,c^5}+4\,a^3\,c^4\,d^3-3\,a^4\,c^3\,d\,e^2}{64\,a^6\,c^5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d + e*x)^(3/2)/(a - c*x^2)^2,x)

[Out]

2*atanh((2*c*e^6*(d + e*x)^(1/2)*(d^3/(16*a^3*c) - (3*d*e^2)/(64*a^2*c^2) - (e^3*(a^9*c^5)^(1/2))/(64*a^6*c^5)
)^(1/2))/((d*e^7)/(2*a) - (c*d^3*e^5)/(2*a^2) + (e^8*(a^9*c^5)^(1/2))/(4*a^5*c^3) - (d^2*e^6*(a^9*c^5)^(1/2))/
(4*a^6*c^2)) + (2*d*e^5*(a^9*c^5)^(1/2)*(d + e*x)^(1/2)*(d^3/(16*a^3*c) - (3*d*e^2)/(64*a^2*c^2) - (e^3*(a^9*c
^5)^(1/2))/(64*a^6*c^5))^(1/2))/((e^8*(a^9*c^5)^(1/2))/(4*c^2) - (a^3*c^2*d^3*e^5)/2 + (a^4*c*d*e^7)/2 - (d^2*
e^6*(a^9*c^5)^(1/2))/(4*a*c)))*(-(e^3*(a^9*c^5)^(1/2) - 4*a^3*c^4*d^3 + 3*a^4*c^3*d*e^2)/(64*a^6*c^5))^(1/2) -
(((a*e^3 - c*d^2*e)*(d + e*x)^(1/2))/(2*a*c) + (d*e*(d + e*x)^(3/2))/(2*a))/(c*(d + e*x)^2 - a*e^2 + c*d^2 -
2*c*d*(d + e*x)) + 2*atanh((2*c*e^6*(d + e*x)^(1/2)*(d^3/(16*a^3*c) - (3*d*e^2)/(64*a^2*c^2) + (e^3*(a^9*c^5)^
(1/2))/(64*a^6*c^5))^(1/2))/((d*e^7)/(2*a) - (c*d^3*e^5)/(2*a^2) - (e^8*(a^9*c^5)^(1/2))/(4*a^5*c^3) + (d^2*e^
6*(a^9*c^5)^(1/2))/(4*a^6*c^2)) + (2*d*e^5*(a^9*c^5)^(1/2)*(d + e*x)^(1/2)*(d^3/(16*a^3*c) - (3*d*e^2)/(64*a^2
*c^2) + (e^3*(a^9*c^5)^(1/2))/(64*a^6*c^5))^(1/2))/((e^8*(a^9*c^5)^(1/2))/(4*c^2) + (a^3*c^2*d^3*e^5)/2 - (a^4
*c*d*e^7)/2 - (d^2*e^6*(a^9*c^5)^(1/2))/(4*a*c)))*((e^3*(a^9*c^5)^(1/2) + 4*a^3*c^4*d^3 - 3*a^4*c^3*d*e^2)/(64
*a^6*c^5))^(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)**(3/2)/(-c*x**2+a)**2,x)

[Out]

Timed out

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