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

Optimal. Leaf size=220 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right )+4 a B d e \left (c d^2-3 a e^2\right )\right )}{2 a^{3/2} c^{5/2}}+\frac {e^2 \log \left (a+c x^2\right ) \left (-a B e^2+2 A c d e+3 B c d^2\right )}{c^3}-\frac {3 e^2 x \left (-a A e^2-4 a B d e+A c d^2\right )}{2 a c^2}-\frac {e^3 x^2 (A c d-2 a B e)}{2 a c^2}-\frac {(d+e x)^3 (a (A e+B d)-x (A c d-a B e))}{2 a c \left (a+c x^2\right )} \]

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Rubi [A]  time = 0.28, antiderivative size = 220, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 22, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {819, 801, 635, 205, 260} \begin {gather*} \frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right )+4 a B d e \left (c d^2-3 a e^2\right )\right )}{2 a^{3/2} c^{5/2}}+\frac {e^2 \log \left (a+c x^2\right ) \left (-a B e^2+2 A c d e+3 B c d^2\right )}{c^3}-\frac {3 e^2 x \left (-a A e^2-4 a B d e+A c d^2\right )}{2 a c^2}-\frac {e^3 x^2 (A c d-2 a B e)}{2 a c^2}-\frac {(d+e x)^3 (a (A e+B d)-x (A c d-a B e))}{2 a c \left (a+c x^2\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 819

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

Rubi steps

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

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Mathematica [A]  time = 0.20, size = 231, normalized size = 1.05 \begin {gather*} \frac {\frac {\sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (A \left (-3 a^2 e^4+6 a c d^2 e^2+c^2 d^4\right )+4 a B d e \left (c d^2-3 a e^2\right )\right )}{a^{3/2}}+\frac {-a^3 B e^4+a^2 c e^2 (A e (4 d+e x)+2 B d (3 d+2 e x))-a c^2 d^2 (2 A e (2 d+3 e x)+B d (d+4 e x))+A c^3 d^4 x}{a \left (a+c x^2\right )}+2 e^2 \log \left (a+c x^2\right ) \left (-a B e^2+2 A c d e+3 B c d^2\right )+2 c e^3 x (A e+4 B d)+B c e^4 x^2}{2 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*c*e^3*(4*B*d + A*e)*x + B*c*e^4*x^2 + (-(a^3*B*e^4) + A*c^3*d^4*x + a^2*c*e^2*(A*e*(4*d + e*x) + 2*B*d*(3*d
+ 2*e*x)) - a*c^2*d^2*(2*A*e*(2*d + 3*e*x) + B*d*(d + 4*e*x)))/(a*(a + c*x^2)) + (Sqrt[c]*(4*a*B*d*e*(c*d^2 -
3*a*e^2) + A*(c^2*d^4 + 6*a*c*d^2*e^2 - 3*a^2*e^4))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/a^(3/2) + 2*e^2*(3*B*c*d^2 +
2*A*c*d*e - a*B*e^2)*Log[a + c*x^2])/(2*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}{\left (a+c x^2\right )^2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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fricas [B]  time = 0.43, size = 849, normalized size = 3.86 \begin {gather*} \left [\frac {2 \, B a^{2} c^{2} e^{4} x^{4} + 2 \, B a^{3} c e^{4} x^{2} - 2 \, B a^{2} c^{2} d^{4} - 8 \, A a^{2} c^{2} d^{3} e + 12 \, B a^{3} c d^{2} e^{2} + 8 \, A a^{3} c d e^{3} - 2 \, B a^{4} e^{4} + 4 \, {\left (4 \, B a^{2} c^{2} d e^{3} + A a^{2} c^{2} e^{4}\right )} x^{3} + {\left (A a c^{2} d^{4} + 4 \, B a^{2} c d^{3} e + 6 \, A a^{2} c d^{2} e^{2} - 12 \, B a^{3} d e^{3} - 3 \, A a^{3} e^{4} + {\left (A c^{3} d^{4} + 4 \, B a c^{2} d^{3} e + 6 \, A a c^{2} d^{2} e^{2} - 12 \, B a^{2} c d e^{3} - 3 \, A a^{2} c e^{4}\right )} x^{2}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} + 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) + 2 \, {\left (A a c^{3} d^{4} - 4 \, B a^{2} c^{2} d^{3} e - 6 \, A a^{2} c^{2} d^{2} e^{2} + 12 \, B a^{3} c d e^{3} + 3 \, A a^{3} c e^{4}\right )} x + 4 \, {\left (3 \, B a^{3} c d^{2} e^{2} + 2 \, A a^{3} c d e^{3} - B a^{4} e^{4} + {\left (3 \, B a^{2} c^{2} d^{2} e^{2} + 2 \, A a^{2} c^{2} d e^{3} - B a^{3} c e^{4}\right )} x^{2}\right )} \log \left (c x^{2} + a\right )}{4 \, {\left (a^{2} c^{4} x^{2} + a^{3} c^{3}\right )}}, \frac {B a^{2} c^{2} e^{4} x^{4} + B a^{3} c e^{4} x^{2} - B a^{2} c^{2} d^{4} - 4 \, A a^{2} c^{2} d^{3} e + 6 \, B a^{3} c d^{2} e^{2} + 4 \, A a^{3} c d e^{3} - B a^{4} e^{4} + 2 \, {\left (4 \, B a^{2} c^{2} d e^{3} + A a^{2} c^{2} e^{4}\right )} x^{3} + {\left (A a c^{2} d^{4} + 4 \, B a^{2} c d^{3} e + 6 \, A a^{2} c d^{2} e^{2} - 12 \, B a^{3} d e^{3} - 3 \, A a^{3} e^{4} + {\left (A c^{3} d^{4} + 4 \, B a c^{2} d^{3} e + 6 \, A a c^{2} d^{2} e^{2} - 12 \, B a^{2} c d e^{3} - 3 \, A a^{2} c e^{4}\right )} x^{2}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) + {\left (A a c^{3} d^{4} - 4 \, B a^{2} c^{2} d^{3} e - 6 \, A a^{2} c^{2} d^{2} e^{2} + 12 \, B a^{3} c d e^{3} + 3 \, A a^{3} c e^{4}\right )} x + 2 \, {\left (3 \, B a^{3} c d^{2} e^{2} + 2 \, A a^{3} c d e^{3} - B a^{4} e^{4} + {\left (3 \, B a^{2} c^{2} d^{2} e^{2} + 2 \, A a^{2} c^{2} d e^{3} - B a^{3} c e^{4}\right )} x^{2}\right )} \log \left (c x^{2} + a\right )}{2 \, {\left (a^{2} c^{4} x^{2} + a^{3} c^{3}\right )}}\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)^2,x, algorithm="fricas")

[Out]

[1/4*(2*B*a^2*c^2*e^4*x^4 + 2*B*a^3*c*e^4*x^2 - 2*B*a^2*c^2*d^4 - 8*A*a^2*c^2*d^3*e + 12*B*a^3*c*d^2*e^2 + 8*A
*a^3*c*d*e^3 - 2*B*a^4*e^4 + 4*(4*B*a^2*c^2*d*e^3 + A*a^2*c^2*e^4)*x^3 + (A*a*c^2*d^4 + 4*B*a^2*c*d^3*e + 6*A*
a^2*c*d^2*e^2 - 12*B*a^3*d*e^3 - 3*A*a^3*e^4 + (A*c^3*d^4 + 4*B*a*c^2*d^3*e + 6*A*a*c^2*d^2*e^2 - 12*B*a^2*c*d
*e^3 - 3*A*a^2*c*e^4)*x^2)*sqrt(-a*c)*log((c*x^2 + 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) + 2*(A*a*c^3*d^4 - 4*B*a^2
*c^2*d^3*e - 6*A*a^2*c^2*d^2*e^2 + 12*B*a^3*c*d*e^3 + 3*A*a^3*c*e^4)*x + 4*(3*B*a^3*c*d^2*e^2 + 2*A*a^3*c*d*e^
3 - B*a^4*e^4 + (3*B*a^2*c^2*d^2*e^2 + 2*A*a^2*c^2*d*e^3 - B*a^3*c*e^4)*x^2)*log(c*x^2 + a))/(a^2*c^4*x^2 + a^
3*c^3), 1/2*(B*a^2*c^2*e^4*x^4 + B*a^3*c*e^4*x^2 - B*a^2*c^2*d^4 - 4*A*a^2*c^2*d^3*e + 6*B*a^3*c*d^2*e^2 + 4*A
*a^3*c*d*e^3 - B*a^4*e^4 + 2*(4*B*a^2*c^2*d*e^3 + A*a^2*c^2*e^4)*x^3 + (A*a*c^2*d^4 + 4*B*a^2*c*d^3*e + 6*A*a^
2*c*d^2*e^2 - 12*B*a^3*d*e^3 - 3*A*a^3*e^4 + (A*c^3*d^4 + 4*B*a*c^2*d^3*e + 6*A*a*c^2*d^2*e^2 - 12*B*a^2*c*d*e
^3 - 3*A*a^2*c*e^4)*x^2)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) + (A*a*c^3*d^4 - 4*B*a^2*c^2*d^3*e - 6*A*a^2*c^2*d^2*
e^2 + 12*B*a^3*c*d*e^3 + 3*A*a^3*c*e^4)*x + 2*(3*B*a^3*c*d^2*e^2 + 2*A*a^3*c*d*e^3 - B*a^4*e^4 + (3*B*a^2*c^2*
d^2*e^2 + 2*A*a^2*c^2*d*e^3 - B*a^3*c*e^4)*x^2)*log(c*x^2 + a))/(a^2*c^4*x^2 + a^3*c^3)]

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giac [A]  time = 0.20, size = 261, normalized size = 1.19 \begin {gather*} \frac {{\left (3 \, B c d^{2} e^{2} + 2 \, A c d e^{3} - B a e^{4}\right )} \log \left (c x^{2} + a\right )}{c^{3}} + \frac {{\left (A c^{2} d^{4} + 4 \, B a c d^{3} e + 6 \, A a c d^{2} e^{2} - 12 \, B a^{2} d e^{3} - 3 \, A a^{2} e^{4}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} a c^{2}} + \frac {B c^{2} x^{2} e^{4} + 8 \, B c^{2} d x e^{3} + 2 \, A c^{2} x e^{4}}{2 \, c^{4}} - \frac {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} - {\left (A c^{3} d^{4} - 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}{2 \, {\left (c x^{2} + a\right )} a 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)^2,x, algorithm="giac")

[Out]

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

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maple [B]  time = 0.05, size = 414, normalized size = 1.88 \begin {gather*} \frac {A a \,e^{4} x}{2 \left (c \,x^{2}+a \right ) c^{2}}-\frac {3 A a \,e^{4} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \sqrt {a c}\, c^{2}}+\frac {A \,d^{4} x}{2 \left (c \,x^{2}+a \right ) a}+\frac {A \,d^{4} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \sqrt {a c}\, a}-\frac {3 A \,d^{2} e^{2} x}{\left (c \,x^{2}+a \right ) c}+\frac {3 A \,d^{2} e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}\, c}+\frac {2 B a d \,e^{3} x}{\left (c \,x^{2}+a \right ) c^{2}}-\frac {6 B a d \,e^{3} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}\, c^{2}}-\frac {2 B \,d^{3} e x}{\left (c \,x^{2}+a \right ) c}+\frac {2 B \,d^{3} e \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{\sqrt {a c}\, c}+\frac {B \,e^{4} x^{2}}{2 c^{2}}+\frac {2 A a d \,e^{3}}{\left (c \,x^{2}+a \right ) c^{2}}-\frac {2 A \,d^{3} e}{\left (c \,x^{2}+a \right ) c}+\frac {2 A d \,e^{3} \ln \left (c \,x^{2}+a \right )}{c^{2}}+\frac {A \,e^{4} x}{c^{2}}-\frac {B \,a^{2} e^{4}}{2 \left (c \,x^{2}+a \right ) c^{3}}+\frac {3 B a \,d^{2} e^{2}}{\left (c \,x^{2}+a \right ) c^{2}}-\frac {B a \,e^{4} \ln \left (c \,x^{2}+a \right )}{c^{3}}-\frac {B \,d^{4}}{2 \left (c \,x^{2}+a \right ) c}+\frac {3 B \,d^{2} e^{2} \ln \left (c \,x^{2}+a \right )}{c^{2}}+\frac {4 B d \,e^{3} x}{c^{2}} \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)^2,x)

[Out]

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

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maxima [A]  time = 1.51, size = 268, normalized size = 1.22 \begin {gather*} -\frac {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} - {\left (A c^{3} d^{4} - 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}{2 \, {\left (a c^{4} x^{2} + a^{2} c^{3}\right )}} + \frac {B e^{4} x^{2} + 2 \, {\left (4 \, B d e^{3} + A e^{4}\right )} x}{2 \, c^{2}} + \frac {{\left (3 \, B c d^{2} e^{2} + 2 \, A c d e^{3} - B a e^{4}\right )} \log \left (c x^{2} + a\right )}{c^{3}} + \frac {{\left (A c^{2} d^{4} + 4 \, B a c d^{3} e + 6 \, A a c d^{2} e^{2} - 12 \, B a^{2} d e^{3} - 3 \, A a^{2} e^{4}\right )} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \, \sqrt {a c} a c^{2}} \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)^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.91, size = 276, normalized size = 1.25 \begin {gather*} \frac {x\,\left (A\,e^4+4\,B\,d\,e^3\right )}{c^2}-\frac {\frac {B\,a^2\,e^4-6\,B\,a\,c\,d^2\,e^2-4\,A\,a\,c\,d\,e^3+B\,c^2\,d^4+4\,A\,c^2\,d^3\,e}{2\,c}-\frac {x\,\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 )}{2\,a}}{c^3\,x^2+a\,c^2}+\frac {\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )\,\left (-12\,B\,a^2\,d\,e^3-3\,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 )}{2\,a^{3/2}\,c^{5/2}}+\frac {\ln \left (c\,x^2+a\right )\,\left (-32\,B\,a^4\,c^3\,e^4+96\,B\,a^3\,c^4\,d^2\,e^2+64\,A\,a^3\,c^4\,d\,e^3\right )}{32\,a^3\,c^6}+\frac {B\,e^4\,x^2}{2\,c^2} \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)^2,x)

[Out]

(x*(A*e^4 + 4*B*d*e^3))/c^2 - ((B*a^2*e^4 + B*c^2*d^4 + 4*A*c^2*d^3*e - 4*A*a*c*d*e^3 - 6*B*a*c*d^2*e^2)/(2*c)
- (x*(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))/(2*a))/(a*c^2 + c^3*x^2) + (a
tan((c^(1/2)*x)/a^(1/2))*(A*c^2*d^4 - 3*A*a^2*e^4 - 12*B*a^2*d*e^3 + 4*B*a*c*d^3*e + 6*A*a*c*d^2*e^2))/(2*a^(3
/2)*c^(5/2)) + (log(a + c*x^2)*(64*A*a^3*c^4*d*e^3 - 32*B*a^4*c^3*e^4 + 96*B*a^3*c^4*d^2*e^2))/(32*a^3*c^6) +
(B*e^4*x^2)/(2*c^2)

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sympy [B]  time = 10.54, size = 836, normalized size = 3.80 \begin {gather*} \frac {B e^{4} x^{2}}{2 c^{2}} + x \left (\frac {A e^{4}}{c^{2}} + \frac {4 B d e^{3}}{c^{2}}\right ) + \left (- \frac {e^{2} \left (- 2 A c d e + B a e^{2} - 3 B c d^{2}\right )}{c^{3}} - \frac {\sqrt {- a^{3} c^{7}} \left (3 A a^{2} e^{4} - 6 A a c d^{2} e^{2} - A c^{2} d^{4} + 12 B a^{2} d e^{3} - 4 B a c d^{3} e\right )}{4 a^{3} c^{6}}\right ) \log {\left (x + \frac {8 A a^{2} c d e^{3} - 4 B a^{3} e^{4} + 12 B a^{2} c d^{2} e^{2} - 4 a^{2} c^{3} \left (- \frac {e^{2} \left (- 2 A c d e + B a e^{2} - 3 B c d^{2}\right )}{c^{3}} - \frac {\sqrt {- a^{3} c^{7}} \left (3 A a^{2} e^{4} - 6 A a c d^{2} e^{2} - A c^{2} d^{4} + 12 B a^{2} d e^{3} - 4 B a c d^{3} e\right )}{4 a^{3} c^{6}}\right )}{3 A a^{2} c e^{4} - 6 A a c^{2} d^{2} e^{2} - A c^{3} d^{4} + 12 B a^{2} c d e^{3} - 4 B a c^{2} d^{3} e} \right )} + \left (- \frac {e^{2} \left (- 2 A c d e + B a e^{2} - 3 B c d^{2}\right )}{c^{3}} + \frac {\sqrt {- a^{3} c^{7}} \left (3 A a^{2} e^{4} - 6 A a c d^{2} e^{2} - A c^{2} d^{4} + 12 B a^{2} d e^{3} - 4 B a c d^{3} e\right )}{4 a^{3} c^{6}}\right ) \log {\left (x + \frac {8 A a^{2} c d e^{3} - 4 B a^{3} e^{4} + 12 B a^{2} c d^{2} e^{2} - 4 a^{2} c^{3} \left (- \frac {e^{2} \left (- 2 A c d e + B a e^{2} - 3 B c d^{2}\right )}{c^{3}} + \frac {\sqrt {- a^{3} c^{7}} \left (3 A a^{2} e^{4} - 6 A a c d^{2} e^{2} - A c^{2} d^{4} + 12 B a^{2} d e^{3} - 4 B a c d^{3} e\right )}{4 a^{3} c^{6}}\right )}{3 A a^{2} c e^{4} - 6 A a c^{2} d^{2} e^{2} - A c^{3} d^{4} + 12 B a^{2} c d e^{3} - 4 B a c^{2} d^{3} e} \right )} + \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} + x \left (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 )}{2 a^{2} c^{3} + 2 a c^{4} x^{2}} \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)**2,x)

[Out]

B*e**4*x**2/(2*c**2) + x*(A*e**4/c**2 + 4*B*d*e**3/c**2) + (-e**2*(-2*A*c*d*e + B*a*e**2 - 3*B*c*d**2)/c**3 -
sqrt(-a**3*c**7)*(3*A*a**2*e**4 - 6*A*a*c*d**2*e**2 - A*c**2*d**4 + 12*B*a**2*d*e**3 - 4*B*a*c*d**3*e)/(4*a**3
*c**6))*log(x + (8*A*a**2*c*d*e**3 - 4*B*a**3*e**4 + 12*B*a**2*c*d**2*e**2 - 4*a**2*c**3*(-e**2*(-2*A*c*d*e +
B*a*e**2 - 3*B*c*d**2)/c**3 - sqrt(-a**3*c**7)*(3*A*a**2*e**4 - 6*A*a*c*d**2*e**2 - A*c**2*d**4 + 12*B*a**2*d*
e**3 - 4*B*a*c*d**3*e)/(4*a**3*c**6)))/(3*A*a**2*c*e**4 - 6*A*a*c**2*d**2*e**2 - A*c**3*d**4 + 12*B*a**2*c*d*e
**3 - 4*B*a*c**2*d**3*e)) + (-e**2*(-2*A*c*d*e + B*a*e**2 - 3*B*c*d**2)/c**3 + sqrt(-a**3*c**7)*(3*A*a**2*e**4
- 6*A*a*c*d**2*e**2 - A*c**2*d**4 + 12*B*a**2*d*e**3 - 4*B*a*c*d**3*e)/(4*a**3*c**6))*log(x + (8*A*a**2*c*d*e
**3 - 4*B*a**3*e**4 + 12*B*a**2*c*d**2*e**2 - 4*a**2*c**3*(-e**2*(-2*A*c*d*e + B*a*e**2 - 3*B*c*d**2)/c**3 + s
qrt(-a**3*c**7)*(3*A*a**2*e**4 - 6*A*a*c*d**2*e**2 - A*c**2*d**4 + 12*B*a**2*d*e**3 - 4*B*a*c*d**3*e)/(4*a**3*
c**6)))/(3*A*a**2*c*e**4 - 6*A*a*c**2*d**2*e**2 - A*c**3*d**4 + 12*B*a**2*c*d*e**3 - 4*B*a*c**2*d**3*e)) + (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 + x*(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))/(2*a**2*c**3 + 2*a*c**4*x**2)

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