3.5.24 \(\int \frac {1}{(d+e x)^3 (a+c x^2)} \, dx\)

Optimal. Leaf size=176 \[ \frac {c^{3/2} d \left (c d^2-3 a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a} \left (a e^2+c d^2\right )^3}-\frac {c e \left (3 c d^2-a e^2\right ) \log \left (a+c x^2\right )}{2 \left (a e^2+c d^2\right )^3}-\frac {2 c d e}{(d+e x) \left (a e^2+c d^2\right )^2}-\frac {e}{2 (d+e x)^2 \left (a e^2+c d^2\right )}+\frac {c e \left (3 c d^2-a e^2\right ) \log (d+e x)}{\left (a e^2+c d^2\right )^3} \]

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Rubi [A]  time = 0.17, antiderivative size = 176, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {710, 801, 635, 205, 260} \begin {gather*} \frac {c^{3/2} d \left (c d^2-3 a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a} \left (a e^2+c d^2\right )^3}-\frac {c e \left (3 c d^2-a e^2\right ) \log \left (a+c x^2\right )}{2 \left (a e^2+c d^2\right )^3}-\frac {2 c d e}{(d+e x) \left (a e^2+c d^2\right )^2}-\frac {e}{2 (d+e x)^2 \left (a e^2+c d^2\right )}+\frac {c e \left (3 c d^2-a e^2\right ) \log (d+e x)}{\left (a e^2+c d^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((d + e*x)^3*(a + c*x^2)),x]

[Out]

-e/(2*(c*d^2 + a*e^2)*(d + e*x)^2) - (2*c*d*e)/((c*d^2 + a*e^2)^2*(d + e*x)) + (c^(3/2)*d*(c*d^2 - 3*a*e^2)*Ar
cTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*(c*d^2 + a*e^2)^3) + (c*e*(3*c*d^2 - a*e^2)*Log[d + e*x])/(c*d^2 + a*e^2)^
3 - (c*e*(3*c*d^2 - a*e^2)*Log[a + c*x^2])/(2*(c*d^2 + a*e^2)^3)

Rule 205

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

Rule 260

Int[(x_)^(m_.)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> Simp[Log[RemoveContent[a + b*x^n, x]]/(b*n), x] /; FreeQ
[{a, b, m, n}, x] && EqQ[m, n - 1]

Rule 635

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

Rule 710

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

Rule 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

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

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Mathematica [A]  time = 0.29, size = 140, normalized size = 0.80 \begin {gather*} \frac {\frac {2 c^{3/2} d \left (c d^2-3 a e^2\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a}}+e \left (c \left (a e^2-3 c d^2\right ) \log \left (a+c x^2\right )-\frac {\left (a e^2+c d^2\right ) \left (a e^2+c d (5 d+4 e x)\right )}{(d+e x)^2}+2 c \left (3 c d^2-a e^2\right ) \log (d+e x)\right )}{2 \left (a e^2+c d^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((d + e*x)^3*(a + c*x^2)),x]

[Out]

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

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IntegrateAlgebraic [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {1}{(d+e x)^3 \left (a+c x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[1/((d + e*x)^3*(a + c*x^2)),x]

[Out]

IntegrateAlgebraic[1/((d + e*x)^3*(a + c*x^2)), x]

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fricas [B]  time = 1.36, size = 853, normalized size = 4.85 \begin {gather*} \left [-\frac {5 \, c^{2} d^{4} e + 6 \, a c d^{2} e^{3} + a^{2} e^{5} + {\left (c^{2} d^{5} - 3 \, a c d^{3} e^{2} + {\left (c^{2} d^{3} e^{2} - 3 \, a c d e^{4}\right )} x^{2} + 2 \, {\left (c^{2} d^{4} e - 3 \, a c d^{2} e^{3}\right )} x\right )} \sqrt {-\frac {c}{a}} \log \left (\frac {c x^{2} - 2 \, a x \sqrt {-\frac {c}{a}} - a}{c x^{2} + a}\right ) + 4 \, {\left (c^{2} d^{3} e^{2} + a c d e^{4}\right )} x + {\left (3 \, c^{2} d^{4} e - a c d^{2} e^{3} + {\left (3 \, c^{2} d^{2} e^{3} - a c e^{5}\right )} x^{2} + 2 \, {\left (3 \, c^{2} d^{3} e^{2} - a c d e^{4}\right )} x\right )} \log \left (c x^{2} + a\right ) - 2 \, {\left (3 \, c^{2} d^{4} e - a c d^{2} e^{3} + {\left (3 \, c^{2} d^{2} e^{3} - a c e^{5}\right )} x^{2} + 2 \, {\left (3 \, c^{2} d^{3} e^{2} - a c d e^{4}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (c^{3} d^{8} + 3 \, a c^{2} d^{6} e^{2} + 3 \, a^{2} c d^{4} e^{4} + a^{3} d^{2} e^{6} + {\left (c^{3} d^{6} e^{2} + 3 \, a c^{2} d^{4} e^{4} + 3 \, a^{2} c d^{2} e^{6} + a^{3} e^{8}\right )} x^{2} + 2 \, {\left (c^{3} d^{7} e + 3 \, a c^{2} d^{5} e^{3} + 3 \, a^{2} c d^{3} e^{5} + a^{3} d e^{7}\right )} x\right )}}, -\frac {5 \, c^{2} d^{4} e + 6 \, a c d^{2} e^{3} + a^{2} e^{5} - 2 \, {\left (c^{2} d^{5} - 3 \, a c d^{3} e^{2} + {\left (c^{2} d^{3} e^{2} - 3 \, a c d e^{4}\right )} x^{2} + 2 \, {\left (c^{2} d^{4} e - 3 \, a c d^{2} e^{3}\right )} x\right )} \sqrt {\frac {c}{a}} \arctan \left (x \sqrt {\frac {c}{a}}\right ) + 4 \, {\left (c^{2} d^{3} e^{2} + a c d e^{4}\right )} x + {\left (3 \, c^{2} d^{4} e - a c d^{2} e^{3} + {\left (3 \, c^{2} d^{2} e^{3} - a c e^{5}\right )} x^{2} + 2 \, {\left (3 \, c^{2} d^{3} e^{2} - a c d e^{4}\right )} x\right )} \log \left (c x^{2} + a\right ) - 2 \, {\left (3 \, c^{2} d^{4} e - a c d^{2} e^{3} + {\left (3 \, c^{2} d^{2} e^{3} - a c e^{5}\right )} x^{2} + 2 \, {\left (3 \, c^{2} d^{3} e^{2} - a c d e^{4}\right )} x\right )} \log \left (e x + d\right )}{2 \, {\left (c^{3} d^{8} + 3 \, a c^{2} d^{6} e^{2} + 3 \, a^{2} c d^{4} e^{4} + a^{3} d^{2} e^{6} + {\left (c^{3} d^{6} e^{2} + 3 \, a c^{2} d^{4} e^{4} + 3 \, a^{2} c d^{2} e^{6} + a^{3} e^{8}\right )} x^{2} + 2 \, {\left (c^{3} d^{7} e + 3 \, a c^{2} d^{5} e^{3} + 3 \, a^{2} c d^{3} e^{5} + a^{3} d e^{7}\right )} x\right )}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/(c*x^2+a),x, algorithm="fricas")

[Out]

[-1/2*(5*c^2*d^4*e + 6*a*c*d^2*e^3 + a^2*e^5 + (c^2*d^5 - 3*a*c*d^3*e^2 + (c^2*d^3*e^2 - 3*a*c*d*e^4)*x^2 + 2*
(c^2*d^4*e - 3*a*c*d^2*e^3)*x)*sqrt(-c/a)*log((c*x^2 - 2*a*x*sqrt(-c/a) - a)/(c*x^2 + a)) + 4*(c^2*d^3*e^2 + a
*c*d*e^4)*x + (3*c^2*d^4*e - a*c*d^2*e^3 + (3*c^2*d^2*e^3 - a*c*e^5)*x^2 + 2*(3*c^2*d^3*e^2 - a*c*d*e^4)*x)*lo
g(c*x^2 + a) - 2*(3*c^2*d^4*e - a*c*d^2*e^3 + (3*c^2*d^2*e^3 - a*c*e^5)*x^2 + 2*(3*c^2*d^3*e^2 - a*c*d*e^4)*x)
*log(e*x + d))/(c^3*d^8 + 3*a*c^2*d^6*e^2 + 3*a^2*c*d^4*e^4 + a^3*d^2*e^6 + (c^3*d^6*e^2 + 3*a*c^2*d^4*e^4 + 3
*a^2*c*d^2*e^6 + a^3*e^8)*x^2 + 2*(c^3*d^7*e + 3*a*c^2*d^5*e^3 + 3*a^2*c*d^3*e^5 + a^3*d*e^7)*x), -1/2*(5*c^2*
d^4*e + 6*a*c*d^2*e^3 + a^2*e^5 - 2*(c^2*d^5 - 3*a*c*d^3*e^2 + (c^2*d^3*e^2 - 3*a*c*d*e^4)*x^2 + 2*(c^2*d^4*e
- 3*a*c*d^2*e^3)*x)*sqrt(c/a)*arctan(x*sqrt(c/a)) + 4*(c^2*d^3*e^2 + a*c*d*e^4)*x + (3*c^2*d^4*e - a*c*d^2*e^3
+ (3*c^2*d^2*e^3 - a*c*e^5)*x^2 + 2*(3*c^2*d^3*e^2 - a*c*d*e^4)*x)*log(c*x^2 + a) - 2*(3*c^2*d^4*e - a*c*d^2*
e^3 + (3*c^2*d^2*e^3 - a*c*e^5)*x^2 + 2*(3*c^2*d^3*e^2 - a*c*d*e^4)*x)*log(e*x + d))/(c^3*d^8 + 3*a*c^2*d^6*e^
2 + 3*a^2*c*d^4*e^4 + a^3*d^2*e^6 + (c^3*d^6*e^2 + 3*a*c^2*d^4*e^4 + 3*a^2*c*d^2*e^6 + a^3*e^8)*x^2 + 2*(c^3*d
^7*e + 3*a*c^2*d^5*e^3 + 3*a^2*c*d^3*e^5 + a^3*d*e^7)*x)]

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giac [A]  time = 0.20, size = 269, normalized size = 1.53 \begin {gather*} -\frac {{\left (3 \, c^{2} d^{2} e - a c e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )}} + \frac {{\left (3 \, c^{2} d^{2} e^{2} - a c e^{4}\right )} \log \left ({\left | x e + d \right |}\right )}{c^{3} d^{6} e + 3 \, a c^{2} d^{4} e^{3} + 3 \, a^{2} c d^{2} e^{5} + a^{3} e^{7}} + \frac {{\left (c^{3} d^{3} - 3 \, a c^{2} d e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{{\left (c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )} \sqrt {a c}} - \frac {5 \, c^{2} d^{4} e + 6 \, a c d^{2} e^{3} + a^{2} e^{5} + 4 \, {\left (c^{2} d^{3} e^{2} + a c d e^{4}\right )} x}{2 \, {\left (c d^{2} + a e^{2}\right )}^{3} {\left (x e + d\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/(c*x^2+a),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.05, size = 233, normalized size = 1.32 \begin {gather*} -\frac {3 a \,c^{2} d \,e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3} \sqrt {a c}}+\frac {c^{3} d^{3} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3} \sqrt {a c}}+\frac {a c \,e^{3} \ln \left (c \,x^{2}+a \right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{3}}-\frac {a c \,e^{3} \ln \left (e x +d \right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3}}-\frac {3 c^{2} d^{2} e \ln \left (c \,x^{2}+a \right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{3}}+\frac {3 c^{2} d^{2} e \ln \left (e x +d \right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3}}-\frac {2 c d e}{\left (a \,e^{2}+c \,d^{2}\right )^{2} \left (e x +d \right )}-\frac {e}{2 \left (a \,e^{2}+c \,d^{2}\right ) \left (e x +d \right )^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*x+d)^3/(c*x^2+a),x)

[Out]

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

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maxima [A]  time = 3.03, size = 323, normalized size = 1.84 \begin {gather*} -\frac {{\left (3 \, c^{2} d^{2} e - a c e^{3}\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )}} + \frac {{\left (3 \, c^{2} d^{2} e - a c e^{3}\right )} \log \left (e x + d\right )}{c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}} + \frac {{\left (c^{3} d^{3} - 3 \, a c^{2} d e^{2}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{{\left (c^{3} d^{6} + 3 \, a c^{2} d^{4} e^{2} + 3 \, a^{2} c d^{2} e^{4} + a^{3} e^{6}\right )} \sqrt {a c}} - \frac {4 \, c d e^{2} x + 5 \, c d^{2} e + a e^{3}}{2 \, {\left (c^{2} d^{6} + 2 \, a c d^{4} e^{2} + a^{2} d^{2} e^{4} + {\left (c^{2} d^{4} e^{2} + 2 \, a c d^{2} e^{4} + a^{2} e^{6}\right )} x^{2} + 2 \, {\left (c^{2} d^{5} e + 2 \, a c d^{3} e^{3} + a^{2} d e^{5}\right )} x\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*x+d)^3/(c*x^2+a),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.14, size = 745, normalized size = 4.23 \begin {gather*} \frac {\ln \left (d+e\,x\right )\,\left (3\,c^2\,d^2\,e-a\,c\,e^3\right )}{a^3\,e^6+3\,a^2\,c\,d^2\,e^4+3\,a\,c^2\,d^4\,e^2+c^3\,d^6}-\frac {\ln \left (c^2\,d^{10}\,{\left (-a\,c^3\right )}^{3/2}-9\,a^6\,e^{10}\,\sqrt {-a\,c^3}+9\,a^6\,c^2\,e^{10}\,x+106\,a^2\,d^6\,e^4\,{\left (-a\,c^3\right )}^{3/2}+a\,c^7\,d^{10}\,x+6\,a^4\,c^2\,d^4\,e^6\,\sqrt {-a\,c^3}+77\,a\,c\,d^8\,e^2\,{\left (-a\,c^3\right )}^{3/2}+77\,a^2\,c^6\,d^8\,e^2\,x+106\,a^3\,c^5\,d^6\,e^4\,x-6\,a^4\,c^4\,d^4\,e^6\,x-27\,a^5\,c^3\,d^2\,e^8\,x+27\,a^5\,c\,d^2\,e^8\,\sqrt {-a\,c^3}\right )\,\left (c\,\left (\frac {d^3\,\sqrt {-a\,c^3}}{2}-\frac {a^2\,e^3}{2}\right )+\frac {3\,a\,c^2\,d^2\,e}{2}-\frac {3\,a\,d\,e^2\,\sqrt {-a\,c^3}}{2}\right )}{a^4\,e^6+3\,a^3\,c\,d^2\,e^4+3\,a^2\,c^2\,d^4\,e^2+a\,c^3\,d^6}-\frac {\ln \left (c^2\,d^{10}\,{\left (-a\,c^3\right )}^{3/2}-9\,a^6\,e^{10}\,\sqrt {-a\,c^3}-9\,a^6\,c^2\,e^{10}\,x+106\,a^2\,d^6\,e^4\,{\left (-a\,c^3\right )}^{3/2}-a\,c^7\,d^{10}\,x+6\,a^4\,c^2\,d^4\,e^6\,\sqrt {-a\,c^3}+77\,a\,c\,d^8\,e^2\,{\left (-a\,c^3\right )}^{3/2}-77\,a^2\,c^6\,d^8\,e^2\,x-106\,a^3\,c^5\,d^6\,e^4\,x+6\,a^4\,c^4\,d^4\,e^6\,x+27\,a^5\,c^3\,d^2\,e^8\,x+27\,a^5\,c\,d^2\,e^8\,\sqrt {-a\,c^3}\right )\,\left (\frac {3\,a\,c^2\,d^2\,e}{2}-c\,\left (\frac {d^3\,\sqrt {-a\,c^3}}{2}+\frac {a^2\,e^3}{2}\right )+\frac {3\,a\,d\,e^2\,\sqrt {-a\,c^3}}{2}\right )}{a^4\,e^6+3\,a^3\,c\,d^2\,e^4+3\,a^2\,c^2\,d^4\,e^2+a\,c^3\,d^6}-\frac {\frac {5\,c\,d^2\,e+a\,e^3}{2\,\left (a^2\,e^4+2\,a\,c\,d^2\,e^2+c^2\,d^4\right )}+\frac {2\,c\,d\,e^2\,x}{a^2\,e^4+2\,a\,c\,d^2\,e^2+c^2\,d^4}}{d^2+2\,d\,e\,x+e^2\,x^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a + c*x^2)*(d + e*x)^3),x)

[Out]

(log(d + e*x)*(3*c^2*d^2*e - a*c*e^3))/(a^3*e^6 + c^3*d^6 + 3*a*c^2*d^4*e^2 + 3*a^2*c*d^2*e^4) - (log(c^2*d^10
*(-a*c^3)^(3/2) - 9*a^6*e^10*(-a*c^3)^(1/2) + 9*a^6*c^2*e^10*x + 106*a^2*d^6*e^4*(-a*c^3)^(3/2) + a*c^7*d^10*x
+ 6*a^4*c^2*d^4*e^6*(-a*c^3)^(1/2) + 77*a*c*d^8*e^2*(-a*c^3)^(3/2) + 77*a^2*c^6*d^8*e^2*x + 106*a^3*c^5*d^6*e
^4*x - 6*a^4*c^4*d^4*e^6*x - 27*a^5*c^3*d^2*e^8*x + 27*a^5*c*d^2*e^8*(-a*c^3)^(1/2))*(c*((d^3*(-a*c^3)^(1/2))/
2 - (a^2*e^3)/2) + (3*a*c^2*d^2*e)/2 - (3*a*d*e^2*(-a*c^3)^(1/2))/2))/(a^4*e^6 + a*c^3*d^6 + 3*a^3*c*d^2*e^4 +
3*a^2*c^2*d^4*e^2) - (log(c^2*d^10*(-a*c^3)^(3/2) - 9*a^6*e^10*(-a*c^3)^(1/2) - 9*a^6*c^2*e^10*x + 106*a^2*d^
6*e^4*(-a*c^3)^(3/2) - a*c^7*d^10*x + 6*a^4*c^2*d^4*e^6*(-a*c^3)^(1/2) + 77*a*c*d^8*e^2*(-a*c^3)^(3/2) - 77*a^
2*c^6*d^8*e^2*x - 106*a^3*c^5*d^6*e^4*x + 6*a^4*c^4*d^4*e^6*x + 27*a^5*c^3*d^2*e^8*x + 27*a^5*c*d^2*e^8*(-a*c^
3)^(1/2))*((3*a*c^2*d^2*e)/2 - c*((d^3*(-a*c^3)^(1/2))/2 + (a^2*e^3)/2) + (3*a*d*e^2*(-a*c^3)^(1/2))/2))/(a^4*
e^6 + a*c^3*d^6 + 3*a^3*c*d^2*e^4 + 3*a^2*c^2*d^4*e^2) - ((a*e^3 + 5*c*d^2*e)/(2*(a^2*e^4 + c^2*d^4 + 2*a*c*d^
2*e^2)) + (2*c*d*e^2*x)/(a^2*e^4 + c^2*d^4 + 2*a*c*d^2*e^2))/(d^2 + e^2*x^2 + 2*d*e*x)

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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(1/(e*x+d)**3/(c*x**2+a),x)

[Out]

Timed out

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