3.12.63 \(\int \frac {A+B x}{(d+e x)^3 (a+c x^2)} \, dx\)

Optimal. Leaf size=251 \[ \frac {B d-A e}{2 (d+e x)^2 \left (a e^2+c d^2\right )}+\frac {-a B e^2-2 A c d e+B c d^2}{(d+e x) \left (a e^2+c d^2\right )^2}+\frac {\sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A c d \left (c d^2-3 a e^2\right )+a B e \left (3 c d^2-a e^2\right )\right )}{\sqrt {a} \left (a e^2+c d^2\right )^3}+\frac {c \log \left (a+c x^2\right ) \left (a A e^3-3 a B d e^2-3 A c d^2 e+B c d^3\right )}{2 \left (a e^2+c d^2\right )^3}-\frac {c \log (d+e x) \left (a A e^3-3 a B d e^2-3 A c d^2 e+B c d^3\right )}{\left (a e^2+c d^2\right )^3} \]

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Rubi [A]  time = 0.32, antiderivative size = 251, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.182, Rules used = {801, 635, 205, 260} \begin {gather*} \frac {c \log \left (a+c x^2\right ) \left (a A e^3-3 a B d e^2-3 A c d^2 e+B c d^3\right )}{2 \left (a e^2+c d^2\right )^3}+\frac {B d-A e}{2 (d+e x)^2 \left (a e^2+c d^2\right )}+\frac {-a B e^2-2 A c d e+B c d^2}{(d+e x) \left (a e^2+c d^2\right )^2}-\frac {c \log (d+e x) \left (a A e^3-3 a B d e^2-3 A c d^2 e+B c d^3\right )}{\left (a e^2+c d^2\right )^3}+\frac {\sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A c d \left (c d^2-3 a e^2\right )+a B e \left (3 c d^2-a e^2\right )\right )}{\sqrt {a} \left (a e^2+c d^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(A + B*x)/((d + e*x)^3*(a + c*x^2)),x]

[Out]

(B*d - A*e)/(2*(c*d^2 + a*e^2)*(d + e*x)^2) + (B*c*d^2 - 2*A*c*d*e - a*B*e^2)/((c*d^2 + a*e^2)^2*(d + e*x)) +
(Sqrt[c]*(A*c*d*(c*d^2 - 3*a*e^2) + a*B*e*(3*c*d^2 - a*e^2))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*(c*d^2 + a*
e^2)^3) - (c*(B*c*d^3 - 3*A*c*d^2*e - 3*a*B*d*e^2 + a*A*e^3)*Log[d + e*x])/(c*d^2 + a*e^2)^3 + (c*(B*c*d^3 - 3
*A*c*d^2*e - 3*a*B*d*e^2 + a*A*e^3)*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 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 {A+B x}{(d+e x)^3 \left (a+c x^2\right )} \, dx &=\int \left (\frac {e (-B d+A e)}{\left (c d^2+a e^2\right ) (d+e x)^3}+\frac {e \left (-B c d^2+2 A c d e+a B e^2\right )}{\left (c d^2+a e^2\right )^2 (d+e x)^2}+\frac {c e \left (-B c d^3+3 A c d^2 e+3 a B d e^2-a A e^3\right )}{\left (c d^2+a e^2\right )^3 (d+e x)}+\frac {c \left (A c d \left (c d^2-3 a e^2\right )+a B e \left (3 c d^2-a e^2\right )+c \left (B c d^3-3 A c d^2 e-3 a B d e^2+a A e^3\right ) x\right )}{\left (c d^2+a e^2\right )^3 \left (a+c x^2\right )}\right ) \, dx\\ &=\frac {B d-A e}{2 \left (c d^2+a e^2\right ) (d+e x)^2}+\frac {B c d^2-2 A c d e-a B e^2}{\left (c d^2+a e^2\right )^2 (d+e x)}-\frac {c \left (B c d^3-3 A c d^2 e-3 a B d e^2+a A e^3\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^3}+\frac {c \int \frac {A c d \left (c d^2-3 a e^2\right )+a B e \left (3 c d^2-a e^2\right )+c \left (B c d^3-3 A c d^2 e-3 a B d e^2+a A e^3\right ) x}{a+c x^2} \, dx}{\left (c d^2+a e^2\right )^3}\\ &=\frac {B d-A e}{2 \left (c d^2+a e^2\right ) (d+e x)^2}+\frac {B c d^2-2 A c d e-a B e^2}{\left (c d^2+a e^2\right )^2 (d+e x)}-\frac {c \left (B c d^3-3 A c d^2 e-3 a B d e^2+a A e^3\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^3}+\frac {\left (c^2 \left (B c d^3-3 A c d^2 e-3 a B d e^2+a A e^3\right )\right ) \int \frac {x}{a+c x^2} \, dx}{\left (c d^2+a e^2\right )^3}+\frac {\left (c \left (A c d \left (c d^2-3 a e^2\right )+a B e \left (3 c d^2-a e^2\right )\right )\right ) \int \frac {1}{a+c x^2} \, dx}{\left (c d^2+a e^2\right )^3}\\ &=\frac {B d-A e}{2 \left (c d^2+a e^2\right ) (d+e x)^2}+\frac {B c d^2-2 A c d e-a B e^2}{\left (c d^2+a e^2\right )^2 (d+e x)}+\frac {\sqrt {c} \left (A c d \left (c d^2-3 a e^2\right )+a B e \left (3 c d^2-a e^2\right )\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a} \left (c d^2+a e^2\right )^3}-\frac {c \left (B c d^3-3 A c d^2 e-3 a B d e^2+a A e^3\right ) \log (d+e x)}{\left (c d^2+a e^2\right )^3}+\frac {c \left (B c d^3-3 A c d^2 e-3 a B d e^2+a A e^3\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 = 0.36, size = 223, normalized size = 0.89 \begin {gather*} \frac {\frac {\left (a e^2+c d^2\right ) \left (B \left (c d^2 (3 d+2 e x)-a e^2 (d+2 e x)\right )-A e \left (a e^2+c d (5 d+4 e x)\right )\right )}{(d+e x)^2}+\frac {2 \sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A c d \left (c d^2-3 a e^2\right )+a B e \left (3 c d^2-a e^2\right )\right )}{\sqrt {a}}+c \log \left (a+c x^2\right ) \left (a A e^3-3 a B d e^2-3 A c d^2 e+B c d^3\right )-2 c \log (d+e x) \left (a A e^3-3 a B d e^2-3 A c d^2 e+B c d^3\right )}{2 \left (a e^2+c d^2\right )^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((c*d^2 + a*e^2)*(B*(-(a*e^2*(d + 2*e*x)) + c*d^2*(3*d + 2*e*x)) - A*e*(a*e^2 + c*d*(5*d + 4*e*x))))/(d + e*x
)^2 + (2*Sqrt[c]*(A*c*d*(c*d^2 - 3*a*e^2) + a*B*e*(3*c*d^2 - a*e^2))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/Sqrt[a] - 2*
c*(B*c*d^3 - 3*A*c*d^2*e - 3*a*B*d*e^2 + a*A*e^3)*Log[d + e*x] + c*(B*c*d^3 - 3*A*c*d^2*e - 3*a*B*d*e^2 + a*A*
e^3)*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 {A+B x}{(d+e x)^3 \left (a+c x^2\right )} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 35.12, size = 1350, normalized size = 5.38

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/2*(3*B*c^2*d^5 - 5*A*c^2*d^4*e + 2*B*a*c*d^3*e^2 - 6*A*a*c*d^2*e^3 - B*a^2*d*e^4 - A*a^2*e^5 - (A*c^2*d^5 +
3*B*a*c*d^4*e - 3*A*a*c*d^3*e^2 - B*a^2*d^2*e^3 + (A*c^2*d^3*e^2 + 3*B*a*c*d^2*e^3 - 3*A*a*c*d*e^4 - B*a^2*e^
5)*x^2 + 2*(A*c^2*d^4*e + 3*B*a*c*d^3*e^2 - 3*A*a*c*d^2*e^3 - B*a^2*d*e^4)*x)*sqrt(-c/a)*log((c*x^2 - 2*a*x*sq
rt(-c/a) - a)/(c*x^2 + a)) + 2*(B*c^2*d^4*e - 2*A*c^2*d^3*e^2 - 2*A*a*c*d*e^4 - B*a^2*e^5)*x + (B*c^2*d^5 - 3*
A*c^2*d^4*e - 3*B*a*c*d^3*e^2 + A*a*c*d^2*e^3 + (B*c^2*d^3*e^2 - 3*A*c^2*d^2*e^3 - 3*B*a*c*d*e^4 + A*a*c*e^5)*
x^2 + 2*(B*c^2*d^4*e - 3*A*c^2*d^3*e^2 - 3*B*a*c*d^2*e^3 + A*a*c*d*e^4)*x)*log(c*x^2 + a) - 2*(B*c^2*d^5 - 3*A
*c^2*d^4*e - 3*B*a*c*d^3*e^2 + A*a*c*d^2*e^3 + (B*c^2*d^3*e^2 - 3*A*c^2*d^2*e^3 - 3*B*a*c*d*e^4 + A*a*c*e^5)*x
^2 + 2*(B*c^2*d^4*e - 3*A*c^2*d^3*e^2 - 3*B*a*c*d^2*e^3 + A*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*(3*B*c^2*d^5 - 5*A*c^2*d^4*e + 2*B*a*c*d^3*e^
2 - 6*A*a*c*d^2*e^3 - B*a^2*d*e^4 - A*a^2*e^5 + 2*(A*c^2*d^5 + 3*B*a*c*d^4*e - 3*A*a*c*d^3*e^2 - B*a^2*d^2*e^3
+ (A*c^2*d^3*e^2 + 3*B*a*c*d^2*e^3 - 3*A*a*c*d*e^4 - B*a^2*e^5)*x^2 + 2*(A*c^2*d^4*e + 3*B*a*c*d^3*e^2 - 3*A*
a*c*d^2*e^3 - B*a^2*d*e^4)*x)*sqrt(c/a)*arctan(x*sqrt(c/a)) + 2*(B*c^2*d^4*e - 2*A*c^2*d^3*e^2 - 2*A*a*c*d*e^4
- B*a^2*e^5)*x + (B*c^2*d^5 - 3*A*c^2*d^4*e - 3*B*a*c*d^3*e^2 + A*a*c*d^2*e^3 + (B*c^2*d^3*e^2 - 3*A*c^2*d^2*
e^3 - 3*B*a*c*d*e^4 + A*a*c*e^5)*x^2 + 2*(B*c^2*d^4*e - 3*A*c^2*d^3*e^2 - 3*B*a*c*d^2*e^3 + A*a*c*d*e^4)*x)*lo
g(c*x^2 + a) - 2*(B*c^2*d^5 - 3*A*c^2*d^4*e - 3*B*a*c*d^3*e^2 + A*a*c*d^2*e^3 + (B*c^2*d^3*e^2 - 3*A*c^2*d^2*e
^3 - 3*B*a*c*d*e^4 + A*a*c*e^5)*x^2 + 2*(B*c^2*d^4*e - 3*A*c^2*d^3*e^2 - 3*B*a*c*d^2*e^3 + A*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 = 0.17, size = 383, normalized size = 1.53 \begin {gather*} \frac {{\left (B c^{2} d^{3} - 3 \, A c^{2} d^{2} e - 3 \, B a c d e^{2} + A 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 (B c^{2} d^{3} e - 3 \, A c^{2} d^{2} e^{2} - 3 \, B a c d e^{3} + A 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 (A c^{3} d^{3} + 3 \, B a c^{2} d^{2} e - 3 \, A a c^{2} d e^{2} - B a^{2} c e^{3}\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 {3 \, B c^{2} d^{5} - 5 \, A c^{2} d^{4} e + 2 \, B a c d^{3} e^{2} - 6 \, A a c d^{2} e^{3} - B a^{2} d e^{4} - A a^{2} e^{5} + 2 \, {\left (B c^{2} d^{4} e - 2 \, A c^{2} d^{3} e^{2} - 2 \, A a c d e^{4} - B a^{2} e^{5}\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((B*x+A)/(e*x+d)^3/(c*x^2+a),x, algorithm="giac")

[Out]

1/2*(B*c^2*d^3 - 3*A*c^2*d^2*e - 3*B*a*c*d*e^2 + A*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) - (B*c^2*d^3*e - 3*A*c^2*d^2*e^2 - 3*B*a*c*d*e^3 + A*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) + (A*c^3*d^3 + 3*B*a*c^2*d^2*e - 3*A*a*c^2*d*e^2 - B*a^2*c*e^
3)*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*(3*B*c^2*d^
5 - 5*A*c^2*d^4*e + 2*B*a*c*d^3*e^2 - 6*A*a*c*d^2*e^3 - B*a^2*d*e^4 - A*a^2*e^5 + 2*(B*c^2*d^4*e - 2*A*c^2*d^3
*e^2 - 2*A*a*c*d*e^4 - B*a^2*e^5)*x)/((c*d^2 + a*e^2)^3*(x*e + d)^2)

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maple [B]  time = 0.06, size = 509, normalized size = 2.03 \begin {gather*} -\frac {3 A 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 {A \,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 {B \,a^{2} c \,e^{3} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3} \sqrt {a c}}+\frac {3 B a \,c^{2} d^{2} e \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3} \sqrt {a c}}+\frac {A a c \,e^{3} \ln \left (c \,x^{2}+a \right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{3}}-\frac {A a c \,e^{3} \ln \left (e x +d \right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3}}-\frac {3 A \,c^{2} d^{2} e \ln \left (c \,x^{2}+a \right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{3}}+\frac {3 A \,c^{2} d^{2} e \ln \left (e x +d \right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3}}-\frac {3 B a c d \,e^{2} \ln \left (c \,x^{2}+a \right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{3}}+\frac {3 B a c d \,e^{2} \ln \left (e x +d \right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3}}+\frac {B \,c^{2} d^{3} \ln \left (c \,x^{2}+a \right )}{2 \left (a \,e^{2}+c \,d^{2}\right )^{3}}-\frac {B \,c^{2} d^{3} \ln \left (e x +d \right )}{\left (a \,e^{2}+c \,d^{2}\right )^{3}}-\frac {2 A c d e}{\left (a \,e^{2}+c \,d^{2}\right )^{2} \left (e x +d \right )}-\frac {B a \,e^{2}}{\left (a \,e^{2}+c \,d^{2}\right )^{2} \left (e x +d \right )}+\frac {B c \,d^{2}}{\left (a \,e^{2}+c \,d^{2}\right )^{2} \left (e x +d \right )}-\frac {A e}{2 \left (a \,e^{2}+c \,d^{2}\right ) \left (e x +d \right )^{2}}+\frac {B d}{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((B*x+A)/(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)*a*A*e^3-3/2*c^2/(a*e^2+c*d^2)^3*ln(c*x^2+a)*A*d^2*e-3/2*c/(a*e^2+c*d^2)^3*ln
(c*x^2+a)*a*B*d*e^2+1/2*c^2/(a*e^2+c*d^2)^3*ln(c*x^2+a)*B*d^3-3*c^2/(a*e^2+c*d^2)^3/(a*c)^(1/2)*arctan(1/(a*c)
^(1/2)*c*x)*A*a*d*e^2+c^3/(a*e^2+c*d^2)^3/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*A*d^3-c/(a*e^2+c*d^2)^3/(a*c)^
(1/2)*arctan(1/(a*c)^(1/2)*c*x)*B*a^2*e^3+3*c^2/(a*e^2+c*d^2)^3/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*B*a*d^2*
e-1/2/(a*e^2+c*d^2)/(e*x+d)^2*A*e+1/2/(a*e^2+c*d^2)/(e*x+d)^2*B*d-2/(a*e^2+c*d^2)^2/(e*x+d)*A*c*d*e-1/(a*e^2+c
*d^2)^2/(e*x+d)*B*a*e^2+1/(a*e^2+c*d^2)^2/(e*x+d)*B*c*d^2-c/(a*e^2+c*d^2)^3*ln(e*x+d)*a*A*e^3+3*c^2/(a*e^2+c*d
^2)^3*ln(e*x+d)*A*d^2*e+3*c/(a*e^2+c*d^2)^3*ln(e*x+d)*a*B*d*e^2-c^2/(a*e^2+c*d^2)^3*ln(e*x+d)*B*d^3

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maxima [A]  time = 1.35, size = 419, normalized size = 1.67 \begin {gather*} \frac {{\left (B c^{2} d^{3} - 3 \, A c^{2} d^{2} e - 3 \, B a c d e^{2} + A 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 (B c^{2} d^{3} - 3 \, A c^{2} d^{2} e - 3 \, B a c d e^{2} + A 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 (A c^{3} d^{3} + 3 \, B a c^{2} d^{2} e - 3 \, A a c^{2} d e^{2} - B a^{2} c e^{3}\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 {3 \, B c d^{3} - 5 \, A c d^{2} e - B a d e^{2} - A a e^{3} + 2 \, {\left (B c d^{2} e - 2 \, A c d e^{2} - B a e^{3}\right )} x}{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((B*x+A)/(e*x+d)^3/(c*x^2+a),x, algorithm="maxima")

[Out]

1/2*(B*c^2*d^3 - 3*A*c^2*d^2*e - 3*B*a*c*d*e^2 + A*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) - (B*c^2*d^3 - 3*A*c^2*d^2*e - 3*B*a*c*d*e^2 + A*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) + (A*c^3*d^3 + 3*B*a*c^2*d^2*e - 3*A*a*c^2*d*e^2 - B*a^2*c*e^3)*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*(3*B*c*d^3 - 5*A*c*d^2
*e - B*a*d*e^2 - A*a*e^3 + 2*(B*c*d^2*e - 2*A*c*d*e^2 - B*a*e^3)*x)/(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 = 3.64, size = 1680, normalized size = 6.69

result too large to display

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(log(B^2*a^7*e^10*(-a*c)^(1/2) - A^2*c^5*d^10*(-a*c)^(3/2) - 9*A^2*a^5*e^10*(-a*c)^(3/2) + 9*B^2*c^3*d^10*(-a*
c)^(5/2) + A^2*a*c^7*d^10*x + B^2*a^7*c*e^10*x + 6*A^2*a*d^4*e^6*(-a*c)^(7/2) + 6*B^2*a*d^6*e^4*(-a*c)^(7/2) -
106*A^2*c*d^6*e^4*(-a*c)^(7/2) + 27*B^2*c*d^8*e^2*(-a*c)^(7/2) + 9*A^2*a^6*c^2*e^10*x + 9*B^2*a^2*c^6*d^10*x
- 27*A^2*a^3*d^2*e^8*(-a*c)^(5/2) + 106*B^2*a^3*d^4*e^6*(-a*c)^(5/2) - 77*B^2*a^5*d^2*e^8*(-a*c)^(3/2) + 77*A^
2*c^3*d^8*e^2*(-a*c)^(5/2) - 224*A*B*a*d^5*e^5*(-a*c)^(7/2) + 48*A*B*a^5*d*e^9*(-a*c)^(3/2) - 64*A*B*c*d^7*e^3
*(-a*c)^(7/2) - 48*A*B*c^3*d^9*e*(-a*c)^(5/2) + 77*A^2*a^2*c^6*d^8*e^2*x + 106*A^2*a^3*c^5*d^6*e^4*x - 6*A^2*a
^4*c^4*d^4*e^6*x - 27*A^2*a^5*c^3*d^2*e^8*x - 27*B^2*a^3*c^5*d^8*e^2*x - 6*B^2*a^4*c^4*d^6*e^4*x + 106*B^2*a^5
*c^3*d^4*e^6*x + 77*B^2*a^6*c^2*d^2*e^8*x + 64*A*B*a^3*d^3*e^7*(-a*c)^(5/2) - 48*A*B*a^2*c^6*d^9*e*x - 48*A*B*
a^6*c^2*d*e^9*x + 64*A*B*a^3*c^5*d^7*e^3*x + 224*A*B*a^4*c^4*d^5*e^5*x + 64*A*B*a^5*c^3*d^3*e^7*x)*(a^2*e^3*((
A*c)/2 - (B*(-a*c)^(1/2))/2) - e^2*((3*B*a^2*c*d)/2 + (3*A*a*c*d*(-a*c)^(1/2))/2) - a*e*((3*A*c^2*d^2)/2 - (3*
B*c*d^2*(-a*c)^(1/2))/2) + (B*a*c^2*d^3)/2 + (A*c^2*d^3*(-a*c)^(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(d + e*x)*(c^2*(B*d^3 - 3*A*d^2*e) + a*c*(A*e^3 - 3*B*d*e^2)))/(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(9*A^2*a^5*e^10*(-a*c)^(3/2) + A^2*c^5*d^10*(-a*c)^(3/2) - B^2*a^
7*e^10*(-a*c)^(1/2) - 9*B^2*c^3*d^10*(-a*c)^(5/2) + A^2*a*c^7*d^10*x + B^2*a^7*c*e^10*x - 6*A^2*a*d^4*e^6*(-a*
c)^(7/2) - 6*B^2*a*d^6*e^4*(-a*c)^(7/2) + 106*A^2*c*d^6*e^4*(-a*c)^(7/2) - 27*B^2*c*d^8*e^2*(-a*c)^(7/2) + 9*A
^2*a^6*c^2*e^10*x + 9*B^2*a^2*c^6*d^10*x + 27*A^2*a^3*d^2*e^8*(-a*c)^(5/2) - 106*B^2*a^3*d^4*e^6*(-a*c)^(5/2)
+ 77*B^2*a^5*d^2*e^8*(-a*c)^(3/2) - 77*A^2*c^3*d^8*e^2*(-a*c)^(5/2) + 224*A*B*a*d^5*e^5*(-a*c)^(7/2) - 48*A*B*
a^5*d*e^9*(-a*c)^(3/2) + 64*A*B*c*d^7*e^3*(-a*c)^(7/2) + 48*A*B*c^3*d^9*e*(-a*c)^(5/2) + 77*A^2*a^2*c^6*d^8*e^
2*x + 106*A^2*a^3*c^5*d^6*e^4*x - 6*A^2*a^4*c^4*d^4*e^6*x - 27*A^2*a^5*c^3*d^2*e^8*x - 27*B^2*a^3*c^5*d^8*e^2*
x - 6*B^2*a^4*c^4*d^6*e^4*x + 106*B^2*a^5*c^3*d^4*e^6*x + 77*B^2*a^6*c^2*d^2*e^8*x - 64*A*B*a^3*d^3*e^7*(-a*c)
^(5/2) - 48*A*B*a^2*c^6*d^9*e*x - 48*A*B*a^6*c^2*d*e^9*x + 64*A*B*a^3*c^5*d^7*e^3*x + 224*A*B*a^4*c^4*d^5*e^5*
x + 64*A*B*a^5*c^3*d^3*e^7*x)*(e^2*((3*B*a^2*c*d)/2 - (3*A*a*c*d*(-a*c)^(1/2))/2) - a^2*e^3*((A*c)/2 + (B*(-a*
c)^(1/2))/2) + a*e*((3*A*c^2*d^2)/2 + (3*B*c*d^2*(-a*c)^(1/2))/2) - (B*a*c^2*d^3)/2 + (A*c^2*d^3*(-a*c)^(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*a*e^3 - 3*B*c*d^3 + B*a*d*e^2 + 5*A*c*d
^2*e)/(2*(a^2*e^4 + c^2*d^4 + 2*a*c*d^2*e^2)) + (x*(B*a*e^3 + 2*A*c*d*e^2 - B*c*d^2*e))/(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((B*x+A)/(e*x+d)**3/(c*x**2+a),x)

[Out]

Timed out

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