3.6.2 \(\int \frac {1}{(d+e x) (a+c x^2)^{5/2}} \, dx\)

Optimal. Leaf size=154 \[ \frac {3 a^2 e^3+c d x \left (5 a e^2+2 c d^2\right )}{3 a^2 \sqrt {a+c x^2} \left (a e^2+c d^2\right )^2}+\frac {a e+c d x}{3 a \left (a+c x^2\right )^{3/2} \left (a e^2+c d^2\right )}-\frac {e^4 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{5/2}} \]

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Rubi [A]  time = 0.13, antiderivative size = 154, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {741, 823, 12, 725, 206} \begin {gather*} \frac {3 a^2 e^3+c d x \left (5 a e^2+2 c d^2\right )}{3 a^2 \sqrt {a+c x^2} \left (a e^2+c d^2\right )^2}+\frac {a e+c d x}{3 a \left (a+c x^2\right )^{3/2} \left (a e^2+c d^2\right )}-\frac {e^4 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)*(a + c*x^2)^(5/2)),x]

[Out]

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 725

Int[1/(((d_) + (e_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> -Subst[Int[1/(c*d^2 + a*e^2 - x^2), x], x,
(a*e - c*d*x)/Sqrt[a + c*x^2]] /; FreeQ[{a, c, d, e}, x]

Rule 741

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*(p + 1)*(c*d^2 + a*e^2)), x] + Dist[1/(2*a*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*
Simp[c*d^2*(2*p + 3) + a*e^2*(m + 2*p + 3) + c*e*d*(m + 2*p + 4)*x, x]*(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a
, c, d, e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && IntQuadraticQ[a, 0, c, d, e, m, p, x]

Rule 823

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[((d + e*x)^(
m + 1)*(f*a*c*e - a*g*c*d + c*(c*d*f + a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), x] + Di
st[1/(2*a*c*(p + 1)*(c*d^2 + a*e^2)), Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Simp[f*(c^2*d^2*(2*p + 3) + a*c*e^2*
(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f + a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x]
&& NeQ[c*d^2 + a*e^2, 0] && LtQ[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])

Rubi steps

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

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Mathematica [A]  time = 0.14, size = 153, normalized size = 0.99 \begin {gather*} \frac {4 a^3 e^3+a^2 c e \left (d^2+6 d e x+3 e^2 x^2\right )+a c^2 d x \left (3 d^2+5 e^2 x^2\right )+2 c^3 d^3 x^3}{3 a^2 \left (a+c x^2\right )^{3/2} \left (a e^2+c d^2\right )^2}-\frac {e^4 \tanh ^{-1}\left (\frac {a e-c d x}{\sqrt {a+c x^2} \sqrt {a e^2+c d^2}}\right )}{\left (a e^2+c d^2\right )^{5/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)*(a + c*x^2)^(5/2)),x]

[Out]

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

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IntegrateAlgebraic [A]  time = 0.91, size = 224, normalized size = 1.45 \begin {gather*} \frac {4 a^3 e^3+a^2 c d^2 e+6 a^2 c d e^2 x+3 a^2 c e^3 x^2+3 a c^2 d^3 x+5 a c^2 d e^2 x^3+2 c^3 d^3 x^3}{3 a^2 \left (a+c x^2\right )^{3/2} \left (a e^2+c d^2\right )^2}+\frac {2 e^4 \sqrt {-a e^2-c d^2} \tan ^{-1}\left (-\frac {e \sqrt {a+c x^2}}{\sqrt {-a e^2-c d^2}}+\frac {\sqrt {c} e x}{\sqrt {-a e^2-c d^2}}+\frac {\sqrt {c} d}{\sqrt {-a e^2-c d^2}}\right )}{\left (a e^2+c d^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[1/((d + e*x)*(a + c*x^2)^(5/2)),x]

[Out]

(a^2*c*d^2*e + 4*a^3*e^3 + 3*a*c^2*d^3*x + 6*a^2*c*d*e^2*x + 3*a^2*c*e^3*x^2 + 2*c^3*d^3*x^3 + 5*a*c^2*d*e^2*x
^3)/(3*a^2*(c*d^2 + a*e^2)^2*(a + c*x^2)^(3/2)) + (2*e^4*Sqrt[-(c*d^2) - a*e^2]*ArcTan[(Sqrt[c]*d)/Sqrt[-(c*d^
2) - a*e^2] + (Sqrt[c]*e*x)/Sqrt[-(c*d^2) - a*e^2] - (e*Sqrt[a + c*x^2])/Sqrt[-(c*d^2) - a*e^2]])/(c*d^2 + a*e
^2)^3

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fricas [B]  time = 0.56, size = 858, normalized size = 5.57 \begin {gather*} \left [\frac {3 \, {\left (a^{2} c^{2} e^{4} x^{4} + 2 \, a^{3} c e^{4} x^{2} + a^{4} e^{4}\right )} \sqrt {c d^{2} + a e^{2}} \log \left (\frac {2 \, a c d e x - a c d^{2} - 2 \, a^{2} e^{2} - {\left (2 \, c^{2} d^{2} + a c e^{2}\right )} x^{2} - 2 \, \sqrt {c d^{2} + a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{e^{2} x^{2} + 2 \, d e x + d^{2}}\right ) + 2 \, {\left (a^{2} c^{2} d^{4} e + 5 \, a^{3} c d^{2} e^{3} + 4 \, a^{4} e^{5} + {\left (2 \, c^{4} d^{5} + 7 \, a c^{3} d^{3} e^{2} + 5 \, a^{2} c^{2} d e^{4}\right )} x^{3} + 3 \, {\left (a^{2} c^{2} d^{2} e^{3} + a^{3} c e^{5}\right )} x^{2} + 3 \, {\left (a c^{3} d^{5} + 3 \, a^{2} c^{2} d^{3} e^{2} + 2 \, a^{3} c d e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{6 \, {\left (a^{4} c^{3} d^{6} + 3 \, a^{5} c^{2} d^{4} e^{2} + 3 \, a^{6} c d^{2} e^{4} + a^{7} e^{6} + {\left (a^{2} c^{5} d^{6} + 3 \, a^{3} c^{4} d^{4} e^{2} + 3 \, a^{4} c^{3} d^{2} e^{4} + a^{5} c^{2} e^{6}\right )} x^{4} + 2 \, {\left (a^{3} c^{4} d^{6} + 3 \, a^{4} c^{3} d^{4} e^{2} + 3 \, a^{5} c^{2} d^{2} e^{4} + a^{6} c e^{6}\right )} x^{2}\right )}}, -\frac {3 \, {\left (a^{2} c^{2} e^{4} x^{4} + 2 \, a^{3} c e^{4} x^{2} + a^{4} e^{4}\right )} \sqrt {-c d^{2} - a e^{2}} \arctan \left (\frac {\sqrt {-c d^{2} - a e^{2}} {\left (c d x - a e\right )} \sqrt {c x^{2} + a}}{a c d^{2} + a^{2} e^{2} + {\left (c^{2} d^{2} + a c e^{2}\right )} x^{2}}\right ) - {\left (a^{2} c^{2} d^{4} e + 5 \, a^{3} c d^{2} e^{3} + 4 \, a^{4} e^{5} + {\left (2 \, c^{4} d^{5} + 7 \, a c^{3} d^{3} e^{2} + 5 \, a^{2} c^{2} d e^{4}\right )} x^{3} + 3 \, {\left (a^{2} c^{2} d^{2} e^{3} + a^{3} c e^{5}\right )} x^{2} + 3 \, {\left (a c^{3} d^{5} + 3 \, a^{2} c^{2} d^{3} e^{2} + 2 \, a^{3} c d e^{4}\right )} x\right )} \sqrt {c x^{2} + a}}{3 \, {\left (a^{4} c^{3} d^{6} + 3 \, a^{5} c^{2} d^{4} e^{2} + 3 \, a^{6} c d^{2} e^{4} + a^{7} e^{6} + {\left (a^{2} c^{5} d^{6} + 3 \, a^{3} c^{4} d^{4} e^{2} + 3 \, a^{4} c^{3} d^{2} e^{4} + a^{5} c^{2} e^{6}\right )} x^{4} + 2 \, {\left (a^{3} c^{4} d^{6} + 3 \, a^{4} c^{3} d^{4} e^{2} + 3 \, a^{5} c^{2} d^{2} e^{4} + a^{6} c e^{6}\right )} x^{2}\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*x^2+a)^(5/2),x, algorithm="fricas")

[Out]

[1/6*(3*(a^2*c^2*e^4*x^4 + 2*a^3*c*e^4*x^2 + a^4*e^4)*sqrt(c*d^2 + a*e^2)*log((2*a*c*d*e*x - a*c*d^2 - 2*a^2*e
^2 - (2*c^2*d^2 + a*c*e^2)*x^2 - 2*sqrt(c*d^2 + a*e^2)*(c*d*x - a*e)*sqrt(c*x^2 + a))/(e^2*x^2 + 2*d*e*x + d^2
)) + 2*(a^2*c^2*d^4*e + 5*a^3*c*d^2*e^3 + 4*a^4*e^5 + (2*c^4*d^5 + 7*a*c^3*d^3*e^2 + 5*a^2*c^2*d*e^4)*x^3 + 3*
(a^2*c^2*d^2*e^3 + a^3*c*e^5)*x^2 + 3*(a*c^3*d^5 + 3*a^2*c^2*d^3*e^2 + 2*a^3*c*d*e^4)*x)*sqrt(c*x^2 + a))/(a^4
*c^3*d^6 + 3*a^5*c^2*d^4*e^2 + 3*a^6*c*d^2*e^4 + a^7*e^6 + (a^2*c^5*d^6 + 3*a^3*c^4*d^4*e^2 + 3*a^4*c^3*d^2*e^
4 + a^5*c^2*e^6)*x^4 + 2*(a^3*c^4*d^6 + 3*a^4*c^3*d^4*e^2 + 3*a^5*c^2*d^2*e^4 + a^6*c*e^6)*x^2), -1/3*(3*(a^2*
c^2*e^4*x^4 + 2*a^3*c*e^4*x^2 + a^4*e^4)*sqrt(-c*d^2 - a*e^2)*arctan(sqrt(-c*d^2 - a*e^2)*(c*d*x - a*e)*sqrt(c
*x^2 + a)/(a*c*d^2 + a^2*e^2 + (c^2*d^2 + a*c*e^2)*x^2)) - (a^2*c^2*d^4*e + 5*a^3*c*d^2*e^3 + 4*a^4*e^5 + (2*c
^4*d^5 + 7*a*c^3*d^3*e^2 + 5*a^2*c^2*d*e^4)*x^3 + 3*(a^2*c^2*d^2*e^3 + a^3*c*e^5)*x^2 + 3*(a*c^3*d^5 + 3*a^2*c
^2*d^3*e^2 + 2*a^3*c*d*e^4)*x)*sqrt(c*x^2 + a))/(a^4*c^3*d^6 + 3*a^5*c^2*d^4*e^2 + 3*a^6*c*d^2*e^4 + a^7*e^6 +
(a^2*c^5*d^6 + 3*a^3*c^4*d^4*e^2 + 3*a^4*c^3*d^2*e^4 + a^5*c^2*e^6)*x^4 + 2*(a^3*c^4*d^6 + 3*a^4*c^3*d^4*e^2
+ 3*a^5*c^2*d^2*e^4 + a^6*c*e^6)*x^2)]

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giac [B]  time = 0.35, size = 935, normalized size = 6.07 \begin {gather*} -\frac {2 \, \arctan \left (\frac {{\left (\sqrt {c} x - \sqrt {c x^{2} + a}\right )} e + \sqrt {c} d}{\sqrt {-c d^{2} - a e^{2}}}\right ) e^{4}}{{\left (c^{2} d^{4} + 2 \, a c d^{2} e^{2} + a^{2} e^{4}\right )} \sqrt {-c d^{2} - a e^{2}}} + \frac {{\left ({\left (\frac {{\left (2 \, c^{10} d^{15} + 17 \, a c^{9} d^{13} e^{2} + 60 \, a^{2} c^{8} d^{11} e^{4} + 115 \, a^{3} c^{7} d^{9} e^{6} + 130 \, a^{4} c^{6} d^{7} e^{8} + 87 \, a^{5} c^{5} d^{5} e^{10} + 32 \, a^{6} c^{4} d^{3} e^{12} + 5 \, a^{7} c^{3} d e^{14}\right )} x}{a^{2} c^{9} d^{16} + 8 \, a^{3} c^{8} d^{14} e^{2} + 28 \, a^{4} c^{7} d^{12} e^{4} + 56 \, a^{5} c^{6} d^{10} e^{6} + 70 \, a^{6} c^{5} d^{8} e^{8} + 56 \, a^{7} c^{4} d^{6} e^{10} + 28 \, a^{8} c^{3} d^{4} e^{12} + 8 \, a^{9} c^{2} d^{2} e^{14} + a^{10} c e^{16}} + \frac {3 \, {\left (a^{2} c^{8} d^{12} e^{3} + 6 \, a^{3} c^{7} d^{10} e^{5} + 15 \, a^{4} c^{6} d^{8} e^{7} + 20 \, a^{5} c^{5} d^{6} e^{9} + 15 \, a^{6} c^{4} d^{4} e^{11} + 6 \, a^{7} c^{3} d^{2} e^{13} + a^{8} c^{2} e^{15}\right )}}{a^{2} c^{9} d^{16} + 8 \, a^{3} c^{8} d^{14} e^{2} + 28 \, a^{4} c^{7} d^{12} e^{4} + 56 \, a^{5} c^{6} d^{10} e^{6} + 70 \, a^{6} c^{5} d^{8} e^{8} + 56 \, a^{7} c^{4} d^{6} e^{10} + 28 \, a^{8} c^{3} d^{4} e^{12} + 8 \, a^{9} c^{2} d^{2} e^{14} + a^{10} c e^{16}}\right )} x + \frac {3 \, {\left (a c^{9} d^{15} + 8 \, a^{2} c^{8} d^{13} e^{2} + 27 \, a^{3} c^{7} d^{11} e^{4} + 50 \, a^{4} c^{6} d^{9} e^{6} + 55 \, a^{5} c^{5} d^{7} e^{8} + 36 \, a^{6} c^{4} d^{5} e^{10} + 13 \, a^{7} c^{3} d^{3} e^{12} + 2 \, a^{8} c^{2} d e^{14}\right )}}{a^{2} c^{9} d^{16} + 8 \, a^{3} c^{8} d^{14} e^{2} + 28 \, a^{4} c^{7} d^{12} e^{4} + 56 \, a^{5} c^{6} d^{10} e^{6} + 70 \, a^{6} c^{5} d^{8} e^{8} + 56 \, a^{7} c^{4} d^{6} e^{10} + 28 \, a^{8} c^{3} d^{4} e^{12} + 8 \, a^{9} c^{2} d^{2} e^{14} + a^{10} c e^{16}}\right )} x + \frac {a^{2} c^{8} d^{14} e + 10 \, a^{3} c^{7} d^{12} e^{3} + 39 \, a^{4} c^{6} d^{10} e^{5} + 80 \, a^{5} c^{5} d^{8} e^{7} + 95 \, a^{6} c^{4} d^{6} e^{9} + 66 \, a^{7} c^{3} d^{4} e^{11} + 25 \, a^{8} c^{2} d^{2} e^{13} + 4 \, a^{9} c e^{15}}{a^{2} c^{9} d^{16} + 8 \, a^{3} c^{8} d^{14} e^{2} + 28 \, a^{4} c^{7} d^{12} e^{4} + 56 \, a^{5} c^{6} d^{10} e^{6} + 70 \, a^{6} c^{5} d^{8} e^{8} + 56 \, a^{7} c^{4} d^{6} e^{10} + 28 \, a^{8} c^{3} d^{4} e^{12} + 8 \, a^{9} c^{2} d^{2} e^{14} + a^{10} c e^{16}}}{3 \, {\left (c x^{2} + a\right )}^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*x^2+a)^(5/2),x, algorithm="giac")

[Out]

-2*arctan(((sqrt(c)*x - sqrt(c*x^2 + a))*e + sqrt(c)*d)/sqrt(-c*d^2 - a*e^2))*e^4/((c^2*d^4 + 2*a*c*d^2*e^2 +
a^2*e^4)*sqrt(-c*d^2 - a*e^2)) + 1/3*((((2*c^10*d^15 + 17*a*c^9*d^13*e^2 + 60*a^2*c^8*d^11*e^4 + 115*a^3*c^7*d
^9*e^6 + 130*a^4*c^6*d^7*e^8 + 87*a^5*c^5*d^5*e^10 + 32*a^6*c^4*d^3*e^12 + 5*a^7*c^3*d*e^14)*x/(a^2*c^9*d^16 +
8*a^3*c^8*d^14*e^2 + 28*a^4*c^7*d^12*e^4 + 56*a^5*c^6*d^10*e^6 + 70*a^6*c^5*d^8*e^8 + 56*a^7*c^4*d^6*e^10 + 2
8*a^8*c^3*d^4*e^12 + 8*a^9*c^2*d^2*e^14 + a^10*c*e^16) + 3*(a^2*c^8*d^12*e^3 + 6*a^3*c^7*d^10*e^5 + 15*a^4*c^6
*d^8*e^7 + 20*a^5*c^5*d^6*e^9 + 15*a^6*c^4*d^4*e^11 + 6*a^7*c^3*d^2*e^13 + a^8*c^2*e^15)/(a^2*c^9*d^16 + 8*a^3
*c^8*d^14*e^2 + 28*a^4*c^7*d^12*e^4 + 56*a^5*c^6*d^10*e^6 + 70*a^6*c^5*d^8*e^8 + 56*a^7*c^4*d^6*e^10 + 28*a^8*
c^3*d^4*e^12 + 8*a^9*c^2*d^2*e^14 + a^10*c*e^16))*x + 3*(a*c^9*d^15 + 8*a^2*c^8*d^13*e^2 + 27*a^3*c^7*d^11*e^4
+ 50*a^4*c^6*d^9*e^6 + 55*a^5*c^5*d^7*e^8 + 36*a^6*c^4*d^5*e^10 + 13*a^7*c^3*d^3*e^12 + 2*a^8*c^2*d*e^14)/(a^
2*c^9*d^16 + 8*a^3*c^8*d^14*e^2 + 28*a^4*c^7*d^12*e^4 + 56*a^5*c^6*d^10*e^6 + 70*a^6*c^5*d^8*e^8 + 56*a^7*c^4*
d^6*e^10 + 28*a^8*c^3*d^4*e^12 + 8*a^9*c^2*d^2*e^14 + a^10*c*e^16))*x + (a^2*c^8*d^14*e + 10*a^3*c^7*d^12*e^3
+ 39*a^4*c^6*d^10*e^5 + 80*a^5*c^5*d^8*e^7 + 95*a^6*c^4*d^6*e^9 + 66*a^7*c^3*d^4*e^11 + 25*a^8*c^2*d^2*e^13 +
4*a^9*c*e^15)/(a^2*c^9*d^16 + 8*a^3*c^8*d^14*e^2 + 28*a^4*c^7*d^12*e^4 + 56*a^5*c^6*d^10*e^6 + 70*a^6*c^5*d^8*
e^8 + 56*a^7*c^4*d^6*e^10 + 28*a^8*c^3*d^4*e^12 + 8*a^9*c^2*d^2*e^14 + a^10*c*e^16))/(c*x^2 + a)^(3/2)

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maple [B]  time = 0.05, size = 454, normalized size = 2.95 \begin {gather*} \frac {c d \,e^{2} x}{\left (a \,e^{2}+c \,d^{2}\right )^{2} \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, a}-\frac {e^{3} \ln \left (\frac {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\frac {2 a \,e^{2}+2 c \,d^{2}}{e^{2}}+2 \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}{x +\frac {d}{e}}\right )}{\left (a \,e^{2}+c \,d^{2}\right )^{2} \sqrt {\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}+\frac {e^{3}}{\left (a \,e^{2}+c \,d^{2}\right )^{2} \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}}+\frac {c d x}{3 \left (a \,e^{2}+c \,d^{2}\right ) \left (-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}\right )^{\frac {3}{2}} a}+\frac {2 c d x}{3 \left (a \,e^{2}+c \,d^{2}\right ) \sqrt {-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}}\, a^{2}}+\frac {e}{3 \left (a \,e^{2}+c \,d^{2}\right ) \left (-\frac {2 \left (x +\frac {d}{e}\right ) c d}{e}+\left (x +\frac {d}{e}\right )^{2} c +\frac {a \,e^{2}+c \,d^{2}}{e^{2}}\right )^{\frac {3}{2}}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)/(c*x^2+a)^(5/2),x)

[Out]

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

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maxima [B]  time = 1.79, size = 283, normalized size = 1.84 \begin {gather*} \frac {c d x}{3 \, {\left ({\left (c x^{2} + a\right )}^{\frac {3}{2}} a c d^{2} + {\left (c x^{2} + a\right )}^{\frac {3}{2}} a^{2} e^{2}\right )}} + \frac {c d x}{2 \, \sqrt {c x^{2} + a} a^{2} c d^{2} + \frac {\sqrt {c x^{2} + a} a c^{2} d^{4}}{e^{2}} + \sqrt {c x^{2} + a} a^{3} e^{2}} + \frac {2 \, c d x}{3 \, {\left (\sqrt {c x^{2} + a} a^{2} c d^{2} + \sqrt {c x^{2} + a} a^{3} e^{2}\right )}} + \frac {1}{\frac {\sqrt {c x^{2} + a} c^{2} d^{4}}{e^{3}} + \frac {2 \, \sqrt {c x^{2} + a} a c d^{2}}{e} + \sqrt {c x^{2} + a} a^{2} e} + \frac {1}{3 \, {\left (\frac {{\left (c x^{2} + a\right )}^{\frac {3}{2}} c d^{2}}{e} + {\left (c x^{2} + a\right )}^{\frac {3}{2}} a e\right )}} + \frac {\operatorname {arsinh}\left (\frac {c d x}{\sqrt {a c} {\left | e x + d \right |}} - \frac {a e}{\sqrt {a c} {\left | e x + d \right |}}\right )}{{\left (a + \frac {c d^{2}}{e^{2}}\right )}^{\frac {5}{2}} e} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*x^2+a)^(5/2),x, algorithm="maxima")

[Out]

1/3*c*d*x/((c*x^2 + a)^(3/2)*a*c*d^2 + (c*x^2 + a)^(3/2)*a^2*e^2) + c*d*x/(2*sqrt(c*x^2 + a)*a^2*c*d^2 + sqrt(
c*x^2 + a)*a*c^2*d^4/e^2 + sqrt(c*x^2 + a)*a^3*e^2) + 2/3*c*d*x/(sqrt(c*x^2 + a)*a^2*c*d^2 + sqrt(c*x^2 + a)*a
^3*e^2) + 1/(sqrt(c*x^2 + a)*c^2*d^4/e^3 + 2*sqrt(c*x^2 + a)*a*c*d^2/e + sqrt(c*x^2 + a)*a^2*e) + 1/3/((c*x^2
+ a)^(3/2)*c*d^2/e + (c*x^2 + a)^(3/2)*a*e) + arcsinh(c*d*x/(sqrt(a*c)*abs(e*x + d)) - a*e/(sqrt(a*c)*abs(e*x
+ d)))/((a + c*d^2/e^2)^(5/2)*e)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {1}{{\left (c\,x^2+a\right )}^{5/2}\,\left (d+e\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + c*x^2)^(5/2)*(d + e*x)),x)

[Out]

int(1/((a + c*x^2)^(5/2)*(d + e*x)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{\left (a + c x^{2}\right )^{\frac {5}{2}} \left (d + e x\right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)/(c*x**2+a)**(5/2),x)

[Out]

Integral(1/((a + c*x**2)**(5/2)*(d + e*x)), x)

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