3.12.56 \(\int \frac {(A+B x) (d+e x)^4}{a+c x^2} \, dx\)

Optimal. Leaf size=240 \[ -\frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (4 a B d e \left (c d^2-a e^2\right )-A \left (a^2 e^4-6 a c d^2 e^2+c^2 d^4\right )\right )}{\sqrt {a} c^{5/2}}+\frac {\log \left (a+c x^2\right ) \left (B \left (a^2 e^4-6 a c d^2 e^2+c^2 d^4\right )+4 A c d e \left (c d^2-a e^2\right )\right )}{2 c^3}+\frac {e^2 x^2 \left (-a B e^2+4 A c d e+6 B c d^2\right )}{2 c^2}+\frac {e x \left (-a A e^3-4 a B d e^2+6 A c d^2 e+4 B c d^3\right )}{c^2}+\frac {e^3 x^3 (A e+4 B d)}{3 c}+\frac {B e^4 x^4}{4 c} \]

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Rubi [A]  time = 0.26, antiderivative size = 240, 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 {\log \left (a+c x^2\right ) \left (B \left (a^2 e^4-6 a c d^2 e^2+c^2 d^4\right )+4 A c d e \left (c d^2-a e^2\right )\right )}{2 c^3}-\frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (4 a B d e \left (c d^2-a e^2\right )-A \left (a^2 e^4-6 a c d^2 e^2+c^2 d^4\right )\right )}{\sqrt {a} c^{5/2}}+\frac {e^2 x^2 \left (-a B e^2+4 A c d e+6 B c d^2\right )}{2 c^2}+\frac {e x \left (-a A e^3-4 a B d e^2+6 A c d^2 e+4 B c d^3\right )}{c^2}+\frac {e^3 x^3 (A e+4 B d)}{3 c}+\frac {B e^4 x^4}{4 c} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x)^4)/(a + c*x^2),x]

[Out]

(e*(4*B*c*d^3 + 6*A*c*d^2*e - 4*a*B*d*e^2 - a*A*e^3)*x)/c^2 + (e^2*(6*B*c*d^2 + 4*A*c*d*e - a*B*e^2)*x^2)/(2*c
^2) + (e^3*(4*B*d + A*e)*x^3)/(3*c) + (B*e^4*x^4)/(4*c) - ((4*a*B*d*e*(c*d^2 - a*e^2) - A*(c^2*d^4 - 6*a*c*d^2
*e^2 + a^2*e^4))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*c^(5/2)) + ((4*A*c*d*e*(c*d^2 - a*e^2) + B*(c^2*d^4 - 6
*a*c*d^2*e^2 + a^2*e^4))*Log[a + c*x^2])/(2*c^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)^4}{a+c x^2} \, dx &=\int \left (\frac {e \left (4 B c d^3+6 A c d^2 e-4 a B d e^2-a A e^3\right )}{c^2}+\frac {e^2 \left (6 B c d^2+4 A c d e-a B e^2\right ) x}{c^2}+\frac {e^3 (4 B d+A e) x^2}{c}+\frac {B e^4 x^3}{c}+\frac {-4 a B d e \left (c d^2-a e^2\right )+A \left (c^2 d^4-6 a c d^2 e^2+a^2 e^4\right )+\left (4 A c d e \left (c d^2-a e^2\right )+B \left (c^2 d^4-6 a c d^2 e^2+a^2 e^4\right )\right ) x}{c^2 \left (a+c x^2\right )}\right ) \, dx\\ &=\frac {e \left (4 B c d^3+6 A c d^2 e-4 a B d e^2-a A e^3\right ) x}{c^2}+\frac {e^2 \left (6 B c d^2+4 A c d e-a B e^2\right ) x^2}{2 c^2}+\frac {e^3 (4 B d+A e) x^3}{3 c}+\frac {B e^4 x^4}{4 c}+\frac {\int \frac {-4 a B d e \left (c d^2-a e^2\right )+A \left (c^2 d^4-6 a c d^2 e^2+a^2 e^4\right )+\left (4 A c d e \left (c d^2-a e^2\right )+B \left (c^2 d^4-6 a c d^2 e^2+a^2 e^4\right )\right ) x}{a+c x^2} \, dx}{c^2}\\ &=\frac {e \left (4 B c d^3+6 A c d^2 e-4 a B d e^2-a A e^3\right ) x}{c^2}+\frac {e^2 \left (6 B c d^2+4 A c d e-a B e^2\right ) x^2}{2 c^2}+\frac {e^3 (4 B d+A e) x^3}{3 c}+\frac {B e^4 x^4}{4 c}-\frac {\left (4 a B d e \left (c d^2-a e^2\right )-A \left (c^2 d^4-6 a c d^2 e^2+a^2 e^4\right )\right ) \int \frac {1}{a+c x^2} \, dx}{c^2}+\frac {\left (4 A c d e \left (c d^2-a e^2\right )+B \left (c^2 d^4-6 a c d^2 e^2+a^2 e^4\right )\right ) \int \frac {x}{a+c x^2} \, dx}{c^2}\\ &=\frac {e \left (4 B c d^3+6 A c d^2 e-4 a B d e^2-a A e^3\right ) x}{c^2}+\frac {e^2 \left (6 B c d^2+4 A c d e-a B e^2\right ) x^2}{2 c^2}+\frac {e^3 (4 B d+A e) x^3}{3 c}+\frac {B e^4 x^4}{4 c}-\frac {\left (4 a B d e \left (c d^2-a e^2\right )-A \left (c^2 d^4-6 a c d^2 e^2+a^2 e^4\right )\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{\sqrt {a} c^{5/2}}+\frac {\left (4 A c d e \left (c d^2-a e^2\right )+B \left (c^2 d^4-6 a c d^2 e^2+a^2 e^4\right )\right ) \log \left (a+c x^2\right )}{2 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.18, size = 217, normalized size = 0.90 \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A \left (a^2 e^4-6 a c d^2 e^2+c^2 d^4\right )+4 a B d e \left (a e^2-c d^2\right )\right )}{\sqrt {a} c^{5/2}}+\frac {6 \log \left (a+c x^2\right ) \left (B \left (a^2 e^4-6 a c d^2 e^2+c^2 d^4\right )+4 A c d e \left (c d^2-a e^2\right )\right )+c e x \left (-12 a A e^3-6 a B e^2 (8 d+e x)+4 A c e \left (18 d^2+6 d e x+e^2 x^2\right )+B c \left (48 d^3+36 d^2 e x+16 d e^2 x^2+3 e^3 x^3\right )\right )}{12 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(d + e*x)^4)/(a + c*x^2),x]

[Out]

((4*a*B*d*e*(-(c*d^2) + a*e^2) + A*(c^2*d^4 - 6*a*c*d^2*e^2 + a^2*e^4))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(Sqrt[a]*
c^(5/2)) + (c*e*x*(-12*a*A*e^3 - 6*a*B*e^2*(8*d + e*x) + 4*A*c*e*(18*d^2 + 6*d*e*x + e^2*x^2) + B*c*(48*d^3 +
36*d^2*e*x + 16*d*e^2*x^2 + 3*e^3*x^3)) + 6*(4*A*c*d*e*(c*d^2 - a*e^2) + B*(c^2*d^4 - 6*a*c*d^2*e^2 + a^2*e^4)
)*Log[a + c*x^2])/(12*c^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)^4}{a+c x^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((A + B*x)*(d + e*x)^4)/(a + c*x^2),x]

[Out]

IntegrateAlgebraic[((A + B*x)*(d + e*x)^4)/(a + c*x^2), x]

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fricas [A]  time = 0.44, size = 550, normalized size = 2.29 \begin {gather*} \left [\frac {3 \, B a c^{2} e^{4} x^{4} + 4 \, {\left (4 \, B a c^{2} d e^{3} + A a c^{2} e^{4}\right )} x^{3} + 6 \, {\left (6 \, B a c^{2} d^{2} e^{2} + 4 \, A a c^{2} d e^{3} - B a^{2} c e^{4}\right )} x^{2} - 6 \, {\left (A c^{2} d^{4} - 4 \, B a c d^{3} e - 6 \, A a c d^{2} e^{2} + 4 \, B a^{2} d e^{3} + A a^{2} e^{4}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) + 12 \, {\left (4 \, B a c^{2} d^{3} e + 6 \, A a c^{2} d^{2} e^{2} - 4 \, B a^{2} c d e^{3} - A a^{2} c e^{4}\right )} x + 6 \, {\left (B a c^{2} d^{4} + 4 \, A a c^{2} d^{3} e - 6 \, B a^{2} c d^{2} e^{2} - 4 \, A a^{2} c d e^{3} + B a^{3} e^{4}\right )} \log \left (c x^{2} + a\right )}{12 \, a c^{3}}, \frac {3 \, B a c^{2} e^{4} x^{4} + 4 \, {\left (4 \, B a c^{2} d e^{3} + A a c^{2} e^{4}\right )} x^{3} + 6 \, {\left (6 \, B a c^{2} d^{2} e^{2} + 4 \, A a c^{2} d e^{3} - B a^{2} c e^{4}\right )} x^{2} + 12 \, {\left (A c^{2} d^{4} - 4 \, B a c d^{3} e - 6 \, A a c d^{2} e^{2} + 4 \, B a^{2} d e^{3} + A a^{2} e^{4}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) + 12 \, {\left (4 \, B a c^{2} d^{3} e + 6 \, A a c^{2} d^{2} e^{2} - 4 \, B a^{2} c d e^{3} - A a^{2} c e^{4}\right )} x + 6 \, {\left (B a c^{2} d^{4} + 4 \, A a c^{2} d^{3} e - 6 \, B a^{2} c d^{2} e^{2} - 4 \, A a^{2} c d e^{3} + B a^{3} e^{4}\right )} \log \left (c x^{2} + a\right )}{12 \, a c^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^4/(c*x^2+a),x, algorithm="fricas")

[Out]

[1/12*(3*B*a*c^2*e^4*x^4 + 4*(4*B*a*c^2*d*e^3 + A*a*c^2*e^4)*x^3 + 6*(6*B*a*c^2*d^2*e^2 + 4*A*a*c^2*d*e^3 - B*
a^2*c*e^4)*x^2 - 6*(A*c^2*d^4 - 4*B*a*c*d^3*e - 6*A*a*c*d^2*e^2 + 4*B*a^2*d*e^3 + A*a^2*e^4)*sqrt(-a*c)*log((c
*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) + 12*(4*B*a*c^2*d^3*e + 6*A*a*c^2*d^2*e^2 - 4*B*a^2*c*d*e^3 - A*a^2*c*
e^4)*x + 6*(B*a*c^2*d^4 + 4*A*a*c^2*d^3*e - 6*B*a^2*c*d^2*e^2 - 4*A*a^2*c*d*e^3 + B*a^3*e^4)*log(c*x^2 + a))/(
a*c^3), 1/12*(3*B*a*c^2*e^4*x^4 + 4*(4*B*a*c^2*d*e^3 + A*a*c^2*e^4)*x^3 + 6*(6*B*a*c^2*d^2*e^2 + 4*A*a*c^2*d*e
^3 - B*a^2*c*e^4)*x^2 + 12*(A*c^2*d^4 - 4*B*a*c*d^3*e - 6*A*a*c*d^2*e^2 + 4*B*a^2*d*e^3 + A*a^2*e^4)*sqrt(a*c)
*arctan(sqrt(a*c)*x/a) + 12*(4*B*a*c^2*d^3*e + 6*A*a*c^2*d^2*e^2 - 4*B*a^2*c*d*e^3 - A*a^2*c*e^4)*x + 6*(B*a*c
^2*d^4 + 4*A*a*c^2*d^3*e - 6*B*a^2*c*d^2*e^2 - 4*A*a^2*c*d*e^3 + B*a^3*e^4)*log(c*x^2 + a))/(a*c^3)]

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giac [A]  time = 0.16, size = 245, normalized size = 1.02 \begin {gather*} \frac {{\left (A c^{2} d^{4} - 4 \, B a c d^{3} e - 6 \, A a c d^{2} e^{2} + 4 \, B a^{2} d e^{3} + A a^{2} e^{4}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c} c^{2}} + \frac {{\left (B c^{2} d^{4} + 4 \, A c^{2} d^{3} e - 6 \, B a c d^{2} e^{2} - 4 \, A a c d e^{3} + B a^{2} e^{4}\right )} \log \left (c x^{2} + a\right )}{2 \, c^{3}} + \frac {3 \, B c^{3} x^{4} e^{4} + 16 \, B c^{3} d x^{3} e^{3} + 36 \, B c^{3} d^{2} x^{2} e^{2} + 48 \, B c^{3} d^{3} x e + 4 \, A c^{3} x^{3} e^{4} + 24 \, A c^{3} d x^{2} e^{3} + 72 \, A c^{3} d^{2} x e^{2} - 6 \, B a c^{2} x^{2} e^{4} - 48 \, B a c^{2} d x e^{3} - 12 \, A a c^{2} x e^{4}}{12 \, c^{4}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^4/(c*x^2+a),x, algorithm="giac")

[Out]

(A*c^2*d^4 - 4*B*a*c*d^3*e - 6*A*a*c*d^2*e^2 + 4*B*a^2*d*e^3 + A*a^2*e^4)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*c^2
) + 1/2*(B*c^2*d^4 + 4*A*c^2*d^3*e - 6*B*a*c*d^2*e^2 - 4*A*a*c*d*e^3 + B*a^2*e^4)*log(c*x^2 + a)/c^3 + 1/12*(3
*B*c^3*x^4*e^4 + 16*B*c^3*d*x^3*e^3 + 36*B*c^3*d^2*x^2*e^2 + 48*B*c^3*d^3*x*e + 4*A*c^3*x^3*e^4 + 24*A*c^3*d*x
^2*e^3 + 72*A*c^3*d^2*x*e^2 - 6*B*a*c^2*x^2*e^4 - 48*B*a*c^2*d*x*e^3 - 12*A*a*c^2*x*e^4)/c^4

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maple [A]  time = 0.05, size = 345, normalized size = 1.44 \begin {gather*} \frac {B \,e^{4} x^{4}}{4 c}+\frac {A \,e^{4} x^{3}}{3 c}+\frac {4 B d \,e^{3} x^{3}}{3 c}+\frac {A \,a^{2} e^{4} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}\, c^{2}}-\frac {6 A a \,d^{2} e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}\, c}+\frac {2 A d \,e^{3} x^{2}}{c}+\frac {A \,d^{4} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}}+\frac {4 B \,a^{2} d \,e^{3} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}\, c^{2}}-\frac {4 B a \,d^{3} e \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}\, c}-\frac {B a \,e^{4} x^{2}}{2 c^{2}}+\frac {3 B \,d^{2} e^{2} x^{2}}{c}-\frac {2 A a d \,e^{3} \ln \left (c \,x^{2}+a \right )}{c^{2}}-\frac {A a \,e^{4} x}{c^{2}}+\frac {2 A \,d^{3} e \ln \left (c \,x^{2}+a \right )}{c}+\frac {6 A \,d^{2} e^{2} x}{c}+\frac {B \,a^{2} e^{4} \ln \left (c \,x^{2}+a \right )}{2 c^{3}}-\frac {3 B a \,d^{2} e^{2} \ln \left (c \,x^{2}+a \right )}{c^{2}}-\frac {4 B a d \,e^{3} x}{c^{2}}+\frac {B \,d^{4} \ln \left (c \,x^{2}+a \right )}{2 c}+\frac {4 B \,d^{3} e x}{c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((B*x+A)*(e*x+d)^4/(c*x^2+a),x)

[Out]

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

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maxima [A]  time = 1.17, size = 230, normalized size = 0.96 \begin {gather*} \frac {{\left (A c^{2} d^{4} - 4 \, B a c d^{3} e - 6 \, A a c d^{2} e^{2} + 4 \, B a^{2} d e^{3} + A a^{2} e^{4}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c} c^{2}} + \frac {3 \, B c e^{4} x^{4} + 4 \, {\left (4 \, B c d e^{3} + A c e^{4}\right )} x^{3} + 6 \, {\left (6 \, B c d^{2} e^{2} + 4 \, A c d e^{3} - B a e^{4}\right )} x^{2} + 12 \, {\left (4 \, B c d^{3} e + 6 \, A c d^{2} e^{2} - 4 \, B a d e^{3} - A a e^{4}\right )} x}{12 \, c^{2}} + \frac {{\left (B c^{2} d^{4} + 4 \, A c^{2} d^{3} e - 6 \, B a c d^{2} e^{2} - 4 \, A a c d e^{3} + B a^{2} e^{4}\right )} \log \left (c x^{2} + a\right )}{2 \, c^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)^4/(c*x^2+a),x, algorithm="maxima")

[Out]

(A*c^2*d^4 - 4*B*a*c*d^3*e - 6*A*a*c*d^2*e^2 + 4*B*a^2*d*e^3 + A*a^2*e^4)*arctan(c*x/sqrt(a*c))/(sqrt(a*c)*c^2
) + 1/12*(3*B*c*e^4*x^4 + 4*(4*B*c*d*e^3 + A*c*e^4)*x^3 + 6*(6*B*c*d^2*e^2 + 4*A*c*d*e^3 - B*a*e^4)*x^2 + 12*(
4*B*c*d^3*e + 6*A*c*d^2*e^2 - 4*B*a*d*e^3 - A*a*e^4)*x)/c^2 + 1/2*(B*c^2*d^4 + 4*A*c^2*d^3*e - 6*B*a*c*d^2*e^2
- 4*A*a*c*d*e^3 + B*a^2*e^4)*log(c*x^2 + a)/c^3

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mupad [B]  time = 0.24, size = 249, normalized size = 1.04 \begin {gather*} \frac {x^3\,\left (A\,e^4+4\,B\,d\,e^3\right )}{3\,c}-x\,\left (\frac {a\,\left (A\,e^4+4\,B\,d\,e^3\right )}{c^2}-\frac {2\,d^2\,e\,\left (3\,A\,e+2\,B\,d\right )}{c}\right )-x^2\,\left (\frac {B\,a\,e^4}{2\,c^2}-\frac {d\,e^2\,\left (2\,A\,e+3\,B\,d\right )}{c}\right )+\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (4\,B\,a^2\,d\,e^3+A\,a^2\,e^4-4\,B\,a\,c\,d^3\,e-6\,A\,a\,c\,d^2\,e^2+A\,c^2\,d^4\right )}{\sqrt {a}\,c^{5/2}}+\frac {\ln \left (c\,x^2+a\right )\,\left (4\,B\,a^3\,c^3\,e^4-24\,B\,a^2\,c^4\,d^2\,e^2-16\,A\,a^2\,c^4\,d\,e^3+4\,B\,a\,c^5\,d^4+16\,A\,a\,c^5\,d^3\,e\right )}{8\,a\,c^6}+\frac {B\,e^4\,x^4}{4\,c} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*x)*(d + e*x)^4)/(a + c*x^2),x)

[Out]

(x^3*(A*e^4 + 4*B*d*e^3))/(3*c) - x*((a*(A*e^4 + 4*B*d*e^3))/c^2 - (2*d^2*e*(3*A*e + 2*B*d))/c) - x^2*((B*a*e^
4)/(2*c^2) - (d*e^2*(2*A*e + 3*B*d))/c) + (atan((c^(1/2)*x)/a^(1/2))*(A*a^2*e^4 + A*c^2*d^4 + 4*B*a^2*d*e^3 -
4*B*a*c*d^3*e - 6*A*a*c*d^2*e^2))/(a^(1/2)*c^(5/2)) + (log(a + c*x^2)*(4*B*a*c^5*d^4 + 4*B*a^3*c^3*e^4 - 16*A*
a^2*c^4*d*e^3 - 24*B*a^2*c^4*d^2*e^2 + 16*A*a*c^5*d^3*e))/(8*a*c^6) + (B*e^4*x^4)/(4*c)

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sympy [B]  time = 3.50, size = 908, normalized size = 3.78 \begin {gather*} \frac {B e^{4} x^{4}}{4 c} + x^{3} \left (\frac {A e^{4}}{3 c} + \frac {4 B d e^{3}}{3 c}\right ) + x^{2} \left (\frac {2 A d e^{3}}{c} - \frac {B a e^{4}}{2 c^{2}} + \frac {3 B d^{2} e^{2}}{c}\right ) + x \left (- \frac {A a e^{4}}{c^{2}} + \frac {6 A d^{2} e^{2}}{c} - \frac {4 B a d e^{3}}{c^{2}} + \frac {4 B d^{3} e}{c}\right ) + \left (\frac {- 4 A a c d e^{3} + 4 A c^{2} d^{3} e + B a^{2} e^{4} - 6 B a c d^{2} e^{2} + B c^{2} d^{4}}{2 c^{3}} - \frac {\sqrt {- a c^{7}} \left (A a^{2} e^{4} - 6 A a c d^{2} e^{2} + A c^{2} d^{4} + 4 B a^{2} d e^{3} - 4 B a c d^{3} e\right )}{2 a c^{6}}\right ) \log {\left (x + \frac {4 A a^{2} c d e^{3} - 4 A a c^{2} d^{3} e - B a^{3} e^{4} + 6 B a^{2} c d^{2} e^{2} - B a c^{2} d^{4} + 2 a c^{3} \left (\frac {- 4 A a c d e^{3} + 4 A c^{2} d^{3} e + B a^{2} e^{4} - 6 B a c d^{2} e^{2} + B c^{2} d^{4}}{2 c^{3}} - \frac {\sqrt {- a c^{7}} \left (A a^{2} e^{4} - 6 A a c d^{2} e^{2} + A c^{2} d^{4} + 4 B a^{2} d e^{3} - 4 B a c d^{3} e\right )}{2 a c^{6}}\right )}{A a^{2} c e^{4} - 6 A a c^{2} d^{2} e^{2} + A c^{3} d^{4} + 4 B a^{2} c d e^{3} - 4 B a c^{2} d^{3} e} \right )} + \left (\frac {- 4 A a c d e^{3} + 4 A c^{2} d^{3} e + B a^{2} e^{4} - 6 B a c d^{2} e^{2} + B c^{2} d^{4}}{2 c^{3}} + \frac {\sqrt {- a c^{7}} \left (A a^{2} e^{4} - 6 A a c d^{2} e^{2} + A c^{2} d^{4} + 4 B a^{2} d e^{3} - 4 B a c d^{3} e\right )}{2 a c^{6}}\right ) \log {\left (x + \frac {4 A a^{2} c d e^{3} - 4 A a c^{2} d^{3} e - B a^{3} e^{4} + 6 B a^{2} c d^{2} e^{2} - B a c^{2} d^{4} + 2 a c^{3} \left (\frac {- 4 A a c d e^{3} + 4 A c^{2} d^{3} e + B a^{2} e^{4} - 6 B a c d^{2} e^{2} + B c^{2} d^{4}}{2 c^{3}} + \frac {\sqrt {- a c^{7}} \left (A a^{2} e^{4} - 6 A a c d^{2} e^{2} + A c^{2} d^{4} + 4 B a^{2} d e^{3} - 4 B a c d^{3} e\right )}{2 a c^{6}}\right )}{A a^{2} c e^{4} - 6 A a c^{2} d^{2} e^{2} + A c^{3} d^{4} + 4 B a^{2} c d e^{3} - 4 B a c^{2} d^{3} e} \right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((B*x+A)*(e*x+d)**4/(c*x**2+a),x)

[Out]

B*e**4*x**4/(4*c) + x**3*(A*e**4/(3*c) + 4*B*d*e**3/(3*c)) + x**2*(2*A*d*e**3/c - B*a*e**4/(2*c**2) + 3*B*d**2
*e**2/c) + x*(-A*a*e**4/c**2 + 6*A*d**2*e**2/c - 4*B*a*d*e**3/c**2 + 4*B*d**3*e/c) + ((-4*A*a*c*d*e**3 + 4*A*c
**2*d**3*e + B*a**2*e**4 - 6*B*a*c*d**2*e**2 + B*c**2*d**4)/(2*c**3) - sqrt(-a*c**7)*(A*a**2*e**4 - 6*A*a*c*d*
*2*e**2 + A*c**2*d**4 + 4*B*a**2*d*e**3 - 4*B*a*c*d**3*e)/(2*a*c**6))*log(x + (4*A*a**2*c*d*e**3 - 4*A*a*c**2*
d**3*e - B*a**3*e**4 + 6*B*a**2*c*d**2*e**2 - B*a*c**2*d**4 + 2*a*c**3*((-4*A*a*c*d*e**3 + 4*A*c**2*d**3*e + B
*a**2*e**4 - 6*B*a*c*d**2*e**2 + B*c**2*d**4)/(2*c**3) - sqrt(-a*c**7)*(A*a**2*e**4 - 6*A*a*c*d**2*e**2 + A*c*
*2*d**4 + 4*B*a**2*d*e**3 - 4*B*a*c*d**3*e)/(2*a*c**6)))/(A*a**2*c*e**4 - 6*A*a*c**2*d**2*e**2 + A*c**3*d**4 +
4*B*a**2*c*d*e**3 - 4*B*a*c**2*d**3*e)) + ((-4*A*a*c*d*e**3 + 4*A*c**2*d**3*e + B*a**2*e**4 - 6*B*a*c*d**2*e*
*2 + B*c**2*d**4)/(2*c**3) + sqrt(-a*c**7)*(A*a**2*e**4 - 6*A*a*c*d**2*e**2 + A*c**2*d**4 + 4*B*a**2*d*e**3 -
4*B*a*c*d**3*e)/(2*a*c**6))*log(x + (4*A*a**2*c*d*e**3 - 4*A*a*c**2*d**3*e - B*a**3*e**4 + 6*B*a**2*c*d**2*e**
2 - B*a*c**2*d**4 + 2*a*c**3*((-4*A*a*c*d*e**3 + 4*A*c**2*d**3*e + B*a**2*e**4 - 6*B*a*c*d**2*e**2 + B*c**2*d*
*4)/(2*c**3) + sqrt(-a*c**7)*(A*a**2*e**4 - 6*A*a*c*d**2*e**2 + A*c**2*d**4 + 4*B*a**2*d*e**3 - 4*B*a*c*d**3*e
)/(2*a*c**6)))/(A*a**2*c*e**4 - 6*A*a*c**2*d**2*e**2 + A*c**3*d**4 + 4*B*a**2*c*d*e**3 - 4*B*a*c**2*d**3*e))

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