3.12.73 \(\int \frac {(A+B x) (d+e x)^4}{(a+c x^2)^3} \, dx\)

Optimal. Leaf size=216 \[ \frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (3 A \left (a e^2+c d^2\right )^2+4 a B d e \left (3 a e^2+c d^2\right )\right )}{8 a^{5/2} c^{5/2}}-\frac {(d+e x) \left (x \left (4 a^2 B e^3-c d \left (a e (3 A e+4 B d)+3 A c d^2\right )\right )+a e \left (3 A \left (a e^2+c d^2\right )+8 a B d e\right )\right )}{8 a^2 c^2 \left (a+c x^2\right )}-\frac {(d+e x)^3 (a (A e+B d)-x (A c d-a B e))}{4 a c \left (a+c x^2\right )^2}+\frac {B e^4 \log \left (a+c x^2\right )}{2 c^3} \]

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Rubi [A]  time = 0.21, antiderivative size = 216, 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 = {819, 635, 205, 260} \begin {gather*} -\frac {(d+e x) \left (x \left (4 a^2 B e^3-c d \left (a e (3 A e+4 B d)+3 A c d^2\right )\right )+a e \left (3 A \left (a e^2+c d^2\right )+8 a B d e\right )\right )}{8 a^2 c^2 \left (a+c x^2\right )}+\frac {\tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (3 A \left (a e^2+c d^2\right )^2+4 a B d e \left (3 a e^2+c d^2\right )\right )}{8 a^{5/2} c^{5/2}}-\frac {(d+e x)^3 (a (A e+B d)-x (A c d-a B e))}{4 a c \left (a+c x^2\right )^2}+\frac {B e^4 \log \left (a+c x^2\right )}{2 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[((A + B*x)*(d + e*x)^4)/(a + c*x^2)^3,x]

[Out]

-((d + e*x)^3*(a*(B*d + A*e) - (A*c*d - a*B*e)*x))/(4*a*c*(a + c*x^2)^2) - ((d + e*x)*(a*e*(8*a*B*d*e + 3*A*(c
*d^2 + a*e^2)) + (4*a^2*B*e^3 - c*d*(3*A*c*d^2 + a*e*(4*B*d + 3*A*e)))*x))/(8*a^2*c^2*(a + c*x^2)) + ((3*A*(c*
d^2 + a*e^2)^2 + 4*a*B*d*e*(c*d^2 + 3*a*e^2))*ArcTan[(Sqrt[c]*x)/Sqrt[a]])/(8*a^(5/2)*c^(5/2)) + (B*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 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 )^3} \, dx &=-\frac {(d+e x)^3 (a (B d+A e)-(A c d-a B e) x)}{4 a c \left (a+c x^2\right )^2}+\frac {\int \frac {(d+e x)^2 \left (3 A c d^2+a e (4 B d+3 A e)+4 a B e^2 x\right )}{\left (a+c x^2\right )^2} \, dx}{4 a c}\\ &=-\frac {(d+e x)^3 (a (B d+A e)-(A c d-a B e) x)}{4 a c \left (a+c x^2\right )^2}-\frac {(d+e x) \left (a e \left (8 a B d e+3 A \left (c d^2+a e^2\right )\right )+\left (4 a^2 B e^3-c d \left (3 A c d^2+a e (4 B d+3 A e)\right )\right ) x\right )}{8 a^2 c^2 \left (a+c x^2\right )}+\frac {\int \frac {3 A \left (c d^2+a e^2\right )^2+4 a B d e \left (c d^2+3 a e^2\right )+8 a^2 B e^4 x}{a+c x^2} \, dx}{8 a^2 c^2}\\ &=-\frac {(d+e x)^3 (a (B d+A e)-(A c d-a B e) x)}{4 a c \left (a+c x^2\right )^2}-\frac {(d+e x) \left (a e \left (8 a B d e+3 A \left (c d^2+a e^2\right )\right )+\left (4 a^2 B e^3-c d \left (3 A c d^2+a e (4 B d+3 A e)\right )\right ) x\right )}{8 a^2 c^2 \left (a+c x^2\right )}+\frac {\left (B e^4\right ) \int \frac {x}{a+c x^2} \, dx}{c^2}+\frac {\left (3 A \left (c d^2+a e^2\right )^2+4 a B d e \left (c d^2+3 a e^2\right )\right ) \int \frac {1}{a+c x^2} \, dx}{8 a^2 c^2}\\ &=-\frac {(d+e x)^3 (a (B d+A e)-(A c d-a B e) x)}{4 a c \left (a+c x^2\right )^2}-\frac {(d+e x) \left (a e \left (8 a B d e+3 A \left (c d^2+a e^2\right )\right )+\left (4 a^2 B e^3-c d \left (3 A c d^2+a e (4 B d+3 A e)\right )\right ) x\right )}{8 a^2 c^2 \left (a+c x^2\right )}+\frac {\left (3 A \left (c d^2+a e^2\right )^2+4 a B d e \left (c d^2+3 a e^2\right )\right ) \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right )}{8 a^{5/2} c^{5/2}}+\frac {B e^4 \log \left (a+c x^2\right )}{2 c^3}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 263, normalized size = 1.22 \begin {gather*} \frac {\frac {\sqrt {c} \tan ^{-1}\left (\frac {\sqrt {c} x}{\sqrt {a}}\right ) \left (3 A \left (a e^2+c d^2\right )^2+4 a B d e \left (3 a e^2+c d^2\right )\right )}{a^{5/2}}+\frac {-2 a^3 B e^4+2 a^2 c e^2 (A e (4 d+e x)+2 B d (3 d+2 e x))-2 a c^2 d^2 (2 A e (2 d+3 e x)+B d (d+4 e x))+2 A c^3 d^4 x}{a \left (a+c x^2\right )^2}+\frac {8 a^3 B e^4-a^2 c e^2 (A e (16 d+5 e x)+4 B d (6 d+5 e x))+2 a c^2 d^2 e x (3 A e+2 B d)+3 A c^3 d^4 x}{a^2 \left (a+c x^2\right )}+4 B e^4 \log \left (a+c x^2\right )}{8 c^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[((A + B*x)*(d + e*x)^4)/(a + c*x^2)^3,x]

[Out]

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

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[((A + B*x)*(d + e*x)^4)/(a + c*x^2)^3,x]

[Out]

IntegrateAlgebraic[((A + B*x)*(d + e*x)^4)/(a + c*x^2)^3, x]

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fricas [B]  time = 0.45, size = 1055, normalized size = 4.88 \begin {gather*} \left [-\frac {4 \, B a^{3} c^{2} d^{4} + 16 \, A a^{3} c^{2} d^{3} e + 24 \, B a^{4} c d^{2} e^{2} + 16 \, A a^{4} c d e^{3} - 12 \, B a^{5} e^{4} - 2 \, {\left (3 \, A a c^{4} d^{4} + 4 \, B a^{2} c^{3} d^{3} e + 6 \, A a^{2} c^{3} d^{2} e^{2} - 20 \, B a^{3} c^{2} d e^{3} - 5 \, A a^{3} c^{2} e^{4}\right )} x^{3} + 16 \, {\left (3 \, B a^{3} c^{2} d^{2} e^{2} + 2 \, A a^{3} c^{2} d e^{3} - B a^{4} c e^{4}\right )} x^{2} + {\left (3 \, A a^{2} c^{2} d^{4} + 4 \, B a^{3} c d^{3} e + 6 \, A a^{3} c d^{2} e^{2} + 12 \, B a^{4} d e^{3} + 3 \, A a^{4} e^{4} + {\left (3 \, A c^{4} d^{4} + 4 \, B a c^{3} d^{3} e + 6 \, A a c^{3} d^{2} e^{2} + 12 \, B a^{2} c^{2} d e^{3} + 3 \, A a^{2} c^{2} e^{4}\right )} x^{4} + 2 \, {\left (3 \, 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}\right )} \sqrt {-a c} \log \left (\frac {c x^{2} - 2 \, \sqrt {-a c} x - a}{c x^{2} + a}\right ) - 2 \, {\left (5 \, A a^{2} c^{3} d^{4} - 4 \, B a^{3} c^{2} d^{3} e - 6 \, A a^{3} c^{2} d^{2} e^{2} - 12 \, B a^{4} c d e^{3} - 3 \, A a^{4} c e^{4}\right )} x - 8 \, {\left (B a^{3} c^{2} e^{4} x^{4} + 2 \, B a^{4} c e^{4} x^{2} + B a^{5} e^{4}\right )} \log \left (c x^{2} + a\right )}{16 \, {\left (a^{3} c^{5} x^{4} + 2 \, a^{4} c^{4} x^{2} + a^{5} c^{3}\right )}}, -\frac {2 \, B a^{3} c^{2} d^{4} + 8 \, A a^{3} c^{2} d^{3} e + 12 \, B a^{4} c d^{2} e^{2} + 8 \, A a^{4} c d e^{3} - 6 \, B a^{5} e^{4} - {\left (3 \, A a c^{4} d^{4} + 4 \, B a^{2} c^{3} d^{3} e + 6 \, A a^{2} c^{3} d^{2} e^{2} - 20 \, B a^{3} c^{2} d e^{3} - 5 \, A a^{3} c^{2} e^{4}\right )} x^{3} + 8 \, {\left (3 \, B a^{3} c^{2} d^{2} e^{2} + 2 \, A a^{3} c^{2} d e^{3} - B a^{4} c e^{4}\right )} x^{2} - {\left (3 \, A a^{2} c^{2} d^{4} + 4 \, B a^{3} c d^{3} e + 6 \, A a^{3} c d^{2} e^{2} + 12 \, B a^{4} d e^{3} + 3 \, A a^{4} e^{4} + {\left (3 \, A c^{4} d^{4} + 4 \, B a c^{3} d^{3} e + 6 \, A a c^{3} d^{2} e^{2} + 12 \, B a^{2} c^{2} d e^{3} + 3 \, A a^{2} c^{2} e^{4}\right )} x^{4} + 2 \, {\left (3 \, 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}\right )} \sqrt {a c} \arctan \left (\frac {\sqrt {a c} x}{a}\right ) - {\left (5 \, A a^{2} c^{3} d^{4} - 4 \, B a^{3} c^{2} d^{3} e - 6 \, A a^{3} c^{2} d^{2} e^{2} - 12 \, B a^{4} c d e^{3} - 3 \, A a^{4} c e^{4}\right )} x - 4 \, {\left (B a^{3} c^{2} e^{4} x^{4} + 2 \, B a^{4} c e^{4} x^{2} + B a^{5} e^{4}\right )} \log \left (c x^{2} + a\right )}{8 \, {\left (a^{3} c^{5} x^{4} + 2 \, a^{4} c^{4} x^{2} + a^{5} 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)^3,x, algorithm="fricas")

[Out]

[-1/16*(4*B*a^3*c^2*d^4 + 16*A*a^3*c^2*d^3*e + 24*B*a^4*c*d^2*e^2 + 16*A*a^4*c*d*e^3 - 12*B*a^5*e^4 - 2*(3*A*a
*c^4*d^4 + 4*B*a^2*c^3*d^3*e + 6*A*a^2*c^3*d^2*e^2 - 20*B*a^3*c^2*d*e^3 - 5*A*a^3*c^2*e^4)*x^3 + 16*(3*B*a^3*c
^2*d^2*e^2 + 2*A*a^3*c^2*d*e^3 - B*a^4*c*e^4)*x^2 + (3*A*a^2*c^2*d^4 + 4*B*a^3*c*d^3*e + 6*A*a^3*c*d^2*e^2 + 1
2*B*a^4*d*e^3 + 3*A*a^4*e^4 + (3*A*c^4*d^4 + 4*B*a*c^3*d^3*e + 6*A*a*c^3*d^2*e^2 + 12*B*a^2*c^2*d*e^3 + 3*A*a^
2*c^2*e^4)*x^4 + 2*(3*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)*sqrt(-a*c)*log((c*x^2 - 2*sqrt(-a*c)*x - a)/(c*x^2 + a)) - 2*(5*A*a^2*c^3*d^4 - 4*B*a^3*c^2*d^3*e - 6*A
*a^3*c^2*d^2*e^2 - 12*B*a^4*c*d*e^3 - 3*A*a^4*c*e^4)*x - 8*(B*a^3*c^2*e^4*x^4 + 2*B*a^4*c*e^4*x^2 + B*a^5*e^4)
*log(c*x^2 + a))/(a^3*c^5*x^4 + 2*a^4*c^4*x^2 + a^5*c^3), -1/8*(2*B*a^3*c^2*d^4 + 8*A*a^3*c^2*d^3*e + 12*B*a^4
*c*d^2*e^2 + 8*A*a^4*c*d*e^3 - 6*B*a^5*e^4 - (3*A*a*c^4*d^4 + 4*B*a^2*c^3*d^3*e + 6*A*a^2*c^3*d^2*e^2 - 20*B*a
^3*c^2*d*e^3 - 5*A*a^3*c^2*e^4)*x^3 + 8*(3*B*a^3*c^2*d^2*e^2 + 2*A*a^3*c^2*d*e^3 - B*a^4*c*e^4)*x^2 - (3*A*a^2
*c^2*d^4 + 4*B*a^3*c*d^3*e + 6*A*a^3*c*d^2*e^2 + 12*B*a^4*d*e^3 + 3*A*a^4*e^4 + (3*A*c^4*d^4 + 4*B*a*c^3*d^3*e
+ 6*A*a*c^3*d^2*e^2 + 12*B*a^2*c^2*d*e^3 + 3*A*a^2*c^2*e^4)*x^4 + 2*(3*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)*sqrt(a*c)*arctan(sqrt(a*c)*x/a) - (5*A*a^2*c^3*d^4 -
4*B*a^3*c^2*d^3*e - 6*A*a^3*c^2*d^2*e^2 - 12*B*a^4*c*d*e^3 - 3*A*a^4*c*e^4)*x - 4*(B*a^3*c^2*e^4*x^4 + 2*B*a^4
*c*e^4*x^2 + B*a^5*e^4)*log(c*x^2 + a))/(a^3*c^5*x^4 + 2*a^4*c^4*x^2 + a^5*c^3)]

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giac [A]  time = 0.16, size = 312, normalized size = 1.44 \begin {gather*} \frac {B e^{4} \log \left (c x^{2} + a\right )}{2 \, c^{3}} + \frac {{\left (3 \, 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 )}{8 \, \sqrt {a c} a^{2} c^{2}} + \frac {{\left (3 \, A c^{3} d^{4} + 4 \, B a c^{2} d^{3} e + 6 \, A a c^{2} d^{2} e^{2} - 20 \, B a^{2} c d e^{3} - 5 \, A a^{2} c e^{4}\right )} x^{3} - 8 \, {\left (3 \, B a^{2} c d^{2} e^{2} + 2 \, A a^{2} c d e^{3} - B a^{3} e^{4}\right )} x^{2} + {\left (5 \, 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}\right )} x - \frac {2 \, {\left (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} - 3 \, B a^{4} e^{4}\right )}}{c}}{8 \, {\left (c x^{2} + a\right )}^{2} a^{2} 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)^3,x, algorithm="giac")

[Out]

1/2*B*e^4*log(c*x^2 + a)/c^3 + 1/8*(3*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^2*c^2) + 1/8*((3*A*c^3*d^4 + 4*B*a*c^2*d^3*e + 6*A*a*c^2*d^2*e^2 - 20*B
*a^2*c*d*e^3 - 5*A*a^2*c*e^4)*x^3 - 8*(3*B*a^2*c*d^2*e^2 + 2*A*a^2*c*d*e^3 - B*a^3*e^4)*x^2 + (5*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)*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 - 3*B*a^4*e^4)/c)/((c*x^2 + a)^2*a^2*c^2)

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maple [A]  time = 0.05, size = 359, normalized size = 1.66 \begin {gather*} \frac {3 A \,d^{2} e^{2} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{4 \sqrt {a c}\, a c}+\frac {3 A \,d^{4} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \sqrt {a c}\, a^{2}}+\frac {3 A \,e^{4} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{8 \sqrt {a c}\, c^{2}}+\frac {B \,d^{3} e \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \sqrt {a c}\, a c}+\frac {3 B d \,e^{3} \arctan \left (\frac {c x}{\sqrt {a c}}\right )}{2 \sqrt {a c}\, c^{2}}+\frac {B \,e^{4} \ln \left (c \,x^{2}+a \right )}{2 c^{3}}+\frac {-\frac {\left (2 A c d e -B a \,e^{2}+3 B c \,d^{2}\right ) e^{2} x^{2}}{c^{2}}-\frac {\left (5 A \,a^{2} e^{4}-6 A a c \,d^{2} e^{2}-3 A \,c^{2} d^{4}+20 B \,a^{2} d \,e^{3}-4 B a c \,d^{3} e \right ) x^{3}}{8 a^{2} c}-\frac {\left (3 A \,a^{2} e^{4}+6 A a c \,d^{2} e^{2}-5 A \,c^{2} d^{4}+12 B \,a^{2} d \,e^{3}+4 B a c \,d^{3} e \right ) x}{8 a \,c^{2}}-\frac {4 A d a c \,e^{3}+4 A \,c^{2} d^{3} e -3 B \,a^{2} e^{4}+6 a \,d^{2} B \,e^{2} c +B \,c^{2} d^{4}}{4 c^{3}}}{\left (c \,x^{2}+a \right )^{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)^3,x)

[Out]

(-1/8*(5*A*a^2*e^4-6*A*a*c*d^2*e^2-3*A*c^2*d^4+20*B*a^2*d*e^3-4*B*a*c*d^3*e)/a^2/c*x^3-e^2*(2*A*c*d*e-B*a*e^2+
3*B*c*d^2)/c^2*x^2-1/8*(3*A*a^2*e^4+6*A*a*c*d^2*e^2-5*A*c^2*d^4+12*B*a^2*d*e^3+4*B*a*c*d^3*e)/a/c^2*x-1/4*(4*A
*a*c*d*e^3+4*A*c^2*d^3*e-3*B*a^2*e^4+6*B*a*c*d^2*e^2+B*c^2*d^4)/c^3)/(c*x^2+a)^2+1/2*B*e^4*ln(c*x^2+a)/c^3+3/8
/c^2/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*A*e^4+3/4/a/c/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*A*d^2*e^2+3/8/a
^2/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*A*d^4+3/2/c^2/(a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*B*d*e^3+1/2/a/c/(
a*c)^(1/2)*arctan(1/(a*c)^(1/2)*c*x)*B*d^3*e

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maxima [A]  time = 1.34, size = 352, normalized size = 1.63 \begin {gather*} \frac {B e^{4} \log \left (c x^{2} + a\right )}{2 \, c^{3}} - \frac {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} - 6 \, B a^{4} e^{4} - {\left (3 \, A c^{4} d^{4} + 4 \, B a c^{3} d^{3} e + 6 \, A a c^{3} d^{2} e^{2} - 20 \, B a^{2} c^{2} d e^{3} - 5 \, A a^{2} c^{2} e^{4}\right )} x^{3} + 8 \, {\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} - {\left (5 \, 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}{8 \, {\left (a^{2} c^{5} x^{4} + 2 \, a^{3} c^{4} x^{2} + a^{4} c^{3}\right )}} + \frac {{\left (3 \, 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 )}{8 \, \sqrt {a c} a^{2} 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)^3,x, algorithm="maxima")

[Out]

1/2*B*e^4*log(c*x^2 + a)/c^3 - 1/8*(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
- 6*B*a^4*e^4 - (3*A*c^4*d^4 + 4*B*a*c^3*d^3*e + 6*A*a*c^3*d^2*e^2 - 20*B*a^2*c^2*d*e^3 - 5*A*a^2*c^2*e^4)*x^
3 + 8*(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 - (5*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)/(a^2*c^5*x^4 + 2*a^3*c^4*x^2 + a^4*c^3) + 1/8*(3*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^2*c^2)

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mupad [B]  time = 2.56, size = 763, normalized size = 3.53 \begin {gather*} \frac {5\,A\,d^4\,x}{8\,\left (a^3+2\,a^2\,c\,x^2+a\,c^2\,x^4\right )}-\frac {B\,d^4}{4\,\left (a^2\,c+2\,a\,c^2\,x^2+c^3\,x^4\right )}+\frac {3\,B\,a^2\,e^4}{4\,\left (a^2\,c^3+2\,a\,c^4\,x^2+c^5\,x^4\right )}-\frac {A\,d^3\,e}{a^2\,c+2\,a\,c^2\,x^2+c^3\,x^4}-\frac {5\,A\,e^4\,x^3}{8\,\left (a^2\,c+2\,a\,c^2\,x^2+c^3\,x^4\right )}+\frac {B\,e^4\,\ln \left (c\,x^2+a\right )}{2\,c^3}-\frac {A\,a\,d\,e^3}{a^2\,c^2+2\,a\,c^3\,x^2+c^4\,x^4}+\frac {3\,A\,c\,d^4\,x^3}{8\,\left (a^4+2\,a^3\,c\,x^2+a^2\,c^2\,x^4\right )}-\frac {3\,A\,a\,e^4\,x}{8\,\left (a^2\,c^2+2\,a\,c^3\,x^2+c^4\,x^4\right )}+\frac {3\,A\,d^2\,e^2\,x^3}{4\,\left (a^3+2\,a^2\,c\,x^2+a\,c^2\,x^4\right )}-\frac {3\,A\,d^2\,e^2\,x}{4\,\left (a^2\,c+2\,a\,c^2\,x^2+c^3\,x^4\right )}-\frac {2\,A\,d\,e^3\,x^2}{a^2\,c+2\,a\,c^2\,x^2+c^3\,x^4}-\frac {5\,B\,d\,e^3\,x^3}{2\,\left (a^2\,c+2\,a\,c^2\,x^2+c^3\,x^4\right )}-\frac {3\,B\,a\,d^2\,e^2}{2\,\left (a^2\,c^2+2\,a\,c^3\,x^2+c^4\,x^4\right )}+\frac {B\,a\,e^4\,x^2}{a^2\,c^2+2\,a\,c^3\,x^2+c^4\,x^4}-\frac {3\,B\,d^2\,e^2\,x^2}{a^2\,c+2\,a\,c^2\,x^2+c^3\,x^4}+\frac {3\,A\,d^4\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )}{8\,a^{5/2}\,\sqrt {c}}+\frac {3\,A\,e^4\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )}{8\,\sqrt {a}\,c^{5/2}}+\frac {B\,d^3\,e\,x^3}{2\,\left (a^3+2\,a^2\,c\,x^2+a\,c^2\,x^4\right )}-\frac {B\,d^3\,e\,x}{2\,\left (a^2\,c+2\,a\,c^2\,x^2+c^3\,x^4\right )}+\frac {3\,B\,d\,e^3\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )}{2\,\sqrt {a}\,c^{5/2}}+\frac {B\,d^3\,e\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )}{2\,a^{3/2}\,c^{3/2}}+\frac {3\,A\,d^2\,e^2\,\mathrm {atan}\left (\frac {\sqrt {c}\,x}{\sqrt {a}}\right )}{4\,a^{3/2}\,c^{3/2}}-\frac {3\,B\,a\,d\,e^3\,x}{2\,\left (a^2\,c^2+2\,a\,c^3\,x^2+c^4\,x^4\right )} \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)^3,x)

[Out]

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

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sympy [B]  time = 46.14, size = 816, normalized size = 3.78 \begin {gather*} \left (\frac {B e^{4}}{2 c^{3}} - \frac {\sqrt {- a^{5} c^{7}} \left (3 A a^{2} e^{4} + 6 A a c d^{2} e^{2} + 3 A c^{2} d^{4} + 12 B a^{2} d e^{3} + 4 B a c d^{3} e\right )}{16 a^{5} c^{6}}\right ) \log {\left (x + \frac {- 8 B a^{3} e^{4} + 16 a^{3} c^{3} \left (\frac {B e^{4}}{2 c^{3}} - \frac {\sqrt {- a^{5} c^{7}} \left (3 A a^{2} e^{4} + 6 A a c d^{2} e^{2} + 3 A c^{2} d^{4} + 12 B a^{2} d e^{3} + 4 B a c d^{3} e\right )}{16 a^{5} c^{6}}\right )}{3 A a^{2} c e^{4} + 6 A a c^{2} d^{2} e^{2} + 3 A c^{3} d^{4} + 12 B a^{2} c d e^{3} + 4 B a c^{2} d^{3} e} \right )} + \left (\frac {B e^{4}}{2 c^{3}} + \frac {\sqrt {- a^{5} c^{7}} \left (3 A a^{2} e^{4} + 6 A a c d^{2} e^{2} + 3 A c^{2} d^{4} + 12 B a^{2} d e^{3} + 4 B a c d^{3} e\right )}{16 a^{5} c^{6}}\right ) \log {\left (x + \frac {- 8 B a^{3} e^{4} + 16 a^{3} c^{3} \left (\frac {B e^{4}}{2 c^{3}} + \frac {\sqrt {- a^{5} c^{7}} \left (3 A a^{2} e^{4} + 6 A a c d^{2} e^{2} + 3 A c^{2} d^{4} + 12 B a^{2} d e^{3} + 4 B a c d^{3} e\right )}{16 a^{5} c^{6}}\right )}{3 A a^{2} c e^{4} + 6 A a c^{2} d^{2} e^{2} + 3 A c^{3} d^{4} + 12 B a^{2} c d e^{3} + 4 B a c^{2} d^{3} e} \right )} + \frac {- 8 A a^{3} c d e^{3} - 8 A a^{2} c^{2} d^{3} e + 6 B a^{4} e^{4} - 12 B a^{3} c d^{2} e^{2} - 2 B a^{2} c^{2} d^{4} + x^{3} \left (- 5 A a^{2} c^{2} e^{4} + 6 A a c^{3} d^{2} e^{2} + 3 A c^{4} d^{4} - 20 B a^{2} c^{2} d e^{3} + 4 B a c^{3} d^{3} e\right ) + x^{2} \left (- 16 A a^{2} c^{2} d e^{3} + 8 B a^{3} c e^{4} - 24 B a^{2} c^{2} d^{2} e^{2}\right ) + x \left (- 3 A a^{3} c e^{4} - 6 A a^{2} c^{2} d^{2} e^{2} + 5 A a c^{3} d^{4} - 12 B a^{3} c d e^{3} - 4 B a^{2} c^{2} d^{3} e\right )}{8 a^{4} c^{3} + 16 a^{3} c^{4} x^{2} + 8 a^{2} c^{5} x^{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)**3,x)

[Out]

(B*e**4/(2*c**3) - sqrt(-a**5*c**7)*(3*A*a**2*e**4 + 6*A*a*c*d**2*e**2 + 3*A*c**2*d**4 + 12*B*a**2*d*e**3 + 4*
B*a*c*d**3*e)/(16*a**5*c**6))*log(x + (-8*B*a**3*e**4 + 16*a**3*c**3*(B*e**4/(2*c**3) - sqrt(-a**5*c**7)*(3*A*
a**2*e**4 + 6*A*a*c*d**2*e**2 + 3*A*c**2*d**4 + 12*B*a**2*d*e**3 + 4*B*a*c*d**3*e)/(16*a**5*c**6)))/(3*A*a**2*
c*e**4 + 6*A*a*c**2*d**2*e**2 + 3*A*c**3*d**4 + 12*B*a**2*c*d*e**3 + 4*B*a*c**2*d**3*e)) + (B*e**4/(2*c**3) +
sqrt(-a**5*c**7)*(3*A*a**2*e**4 + 6*A*a*c*d**2*e**2 + 3*A*c**2*d**4 + 12*B*a**2*d*e**3 + 4*B*a*c*d**3*e)/(16*a
**5*c**6))*log(x + (-8*B*a**3*e**4 + 16*a**3*c**3*(B*e**4/(2*c**3) + sqrt(-a**5*c**7)*(3*A*a**2*e**4 + 6*A*a*c
*d**2*e**2 + 3*A*c**2*d**4 + 12*B*a**2*d*e**3 + 4*B*a*c*d**3*e)/(16*a**5*c**6)))/(3*A*a**2*c*e**4 + 6*A*a*c**2
*d**2*e**2 + 3*A*c**3*d**4 + 12*B*a**2*c*d*e**3 + 4*B*a*c**2*d**3*e)) + (-8*A*a**3*c*d*e**3 - 8*A*a**2*c**2*d*
*3*e + 6*B*a**4*e**4 - 12*B*a**3*c*d**2*e**2 - 2*B*a**2*c**2*d**4 + x**3*(-5*A*a**2*c**2*e**4 + 6*A*a*c**3*d**
2*e**2 + 3*A*c**4*d**4 - 20*B*a**2*c**2*d*e**3 + 4*B*a*c**3*d**3*e) + x**2*(-16*A*a**2*c**2*d*e**3 + 8*B*a**3*
c*e**4 - 24*B*a**2*c**2*d**2*e**2) + x*(-3*A*a**3*c*e**4 - 6*A*a**2*c**2*d**2*e**2 + 5*A*a*c**3*d**4 - 12*B*a*
*3*c*d*e**3 - 4*B*a**2*c**2*d**3*e))/(8*a**4*c**3 + 16*a**3*c**4*x**2 + 8*a**2*c**5*x**4)

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