\(\int \frac {x^5}{(c+d x) (a+b x^2)^3} \, dx\) [243]

Optimal result

Integrand size = 20, antiderivative size = 222 \[ \int \frac {x^5}{(c+d x) \left (a+b x^2\right )^3} \, dx=-\frac {a^2 (c-d x)}{4 b^2 \left (b c^2+a d^2\right ) \left (a+b x^2\right )^2}+\frac {a \left (4 c \left (2 b c^2+a d^2\right )-d \left (9 b c^2+5 a d^2\right ) x\right )}{8 b^2 \left (b c^2+a d^2\right )^2 \left (a+b x^2\right )}+\frac {\sqrt {a} d \left (15 b^2 c^4+10 a b c^2 d^2+3 a^2 d^4\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 b^{5/2} \left (b c^2+a d^2\right )^3}-\frac {c^5 \log (c+d x)}{\left (b c^2+a d^2\right )^3}+\frac {c^5 \log \left (a+b x^2\right )}{2 \left (b c^2+a d^2\right )^3} \] Output:

-1/4*a^2*(-d*x+c)/b^2/(a*d^2+b*c^2)/(b*x^2+a)^2+1/8*a*(4*c*(a*d^2+2*b*c^2) 
-d*(5*a*d^2+9*b*c^2)*x)/b^2/(a*d^2+b*c^2)^2/(b*x^2+a)+1/8*a^(1/2)*d*(3*a^2 
*d^4+10*a*b*c^2*d^2+15*b^2*c^4)*arctan(b^(1/2)*x/a^(1/2))/b^(5/2)/(a*d^2+b 
*c^2)^3-c^5*ln(d*x+c)/(a*d^2+b*c^2)^3+1/2*c^5*ln(b*x^2+a)/(a*d^2+b*c^2)^3
 

Mathematica [A] (verified)

Time = 0.12 (sec) , antiderivative size = 182, normalized size of antiderivative = 0.82 \[ \int \frac {x^5}{(c+d x) \left (a+b x^2\right )^3} \, dx=\frac {-\frac {2 a^2 \left (b c^2+a d^2\right )^2 (c-d x)}{b^2 \left (a+b x^2\right )^2}+\frac {a \left (b c^2+a d^2\right ) \left (b c^2 (8 c-9 d x)+a d^2 (4 c-5 d x)\right )}{b^2 \left (a+b x^2\right )}+\frac {\sqrt {a} d \left (15 b^2 c^4+10 a b c^2 d^2+3 a^2 d^4\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{b^{5/2}}-8 c^5 \log (c+d x)+4 c^5 \log \left (a+b x^2\right )}{8 \left (b c^2+a d^2\right )^3} \] Input:

Integrate[x^5/((c + d*x)*(a + b*x^2)^3),x]
 

Output:

((-2*a^2*(b*c^2 + a*d^2)^2*(c - d*x))/(b^2*(a + b*x^2)^2) + (a*(b*c^2 + a* 
d^2)*(b*c^2*(8*c - 9*d*x) + a*d^2*(4*c - 5*d*x)))/(b^2*(a + b*x^2)) + (Sqr 
t[a]*d*(15*b^2*c^4 + 10*a*b*c^2*d^2 + 3*a^2*d^4)*ArcTan[(Sqrt[b]*x)/Sqrt[a 
]])/b^(5/2) - 8*c^5*Log[c + d*x] + 4*c^5*Log[a + b*x^2])/(8*(b*c^2 + a*d^2 
)^3)
 

Rubi [A] (verified)

Time = 1.17 (sec) , antiderivative size = 255, normalized size of antiderivative = 1.15, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {601, 2178, 27, 657, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

\(\displaystyle \int \frac {x^5}{\left (a+b x^2\right )^3 (c+d x)} \, dx\)

\(\Big \downarrow \) 601

\(\displaystyle -\frac {\int \frac {\frac {c d a^3}{b^2 \left (b c^2+a d^2\right )}+\frac {\left (4 b c^2+a d^2\right ) x a^2}{b^2 \left (b c^2+a d^2\right )}-\frac {4 x^3 a}{b}}{(c+d x) \left (b x^2+a\right )^2}dx}{4 a}-\frac {a^2 (c-d x)}{4 b^2 \left (a+b x^2\right )^2 \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 2178

\(\displaystyle -\frac {-\frac {\int \frac {a^2 \left (a c d \left (7 b c^2+3 a d^2\right )+\left (8 b^2 c^4+7 a b d^2 c^2+3 a^2 d^4\right ) x\right )}{b \left (b c^2+a d^2\right )^2 (c+d x) \left (b x^2+a\right )}dx}{2 a b}-\frac {a^2 \left (4 c \left (a d^2+2 b c^2\right )-d x \left (5 a d^2+9 b c^2\right )\right )}{2 b^2 \left (a+b x^2\right ) \left (a d^2+b c^2\right )^2}}{4 a}-\frac {a^2 (c-d x)}{4 b^2 \left (a+b x^2\right )^2 \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {-\frac {a \int \frac {a c d \left (7 b c^2+3 a d^2\right )+\left (8 b^2 c^4+7 a b d^2 c^2+3 a^2 d^4\right ) x}{(c+d x) \left (b x^2+a\right )}dx}{2 b^2 \left (a d^2+b c^2\right )^2}-\frac {a^2 \left (4 c \left (a d^2+2 b c^2\right )-d x \left (5 a d^2+9 b c^2\right )\right )}{2 b^2 \left (a+b x^2\right ) \left (a d^2+b c^2\right )^2}}{4 a}-\frac {a^2 (c-d x)}{4 b^2 \left (a+b x^2\right )^2 \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 657

\(\displaystyle -\frac {-\frac {a \int \left (\frac {8 b^3 x c^5+15 a b^2 d c^4+10 a^2 b d^3 c^2+3 a^3 d^5}{\left (b c^2+a d^2\right ) \left (b x^2+a\right )}-\frac {8 b^2 c^5 d}{\left (b c^2+a d^2\right ) (c+d x)}\right )dx}{2 b^2 \left (a d^2+b c^2\right )^2}-\frac {a^2 \left (4 c \left (a d^2+2 b c^2\right )-d x \left (5 a d^2+9 b c^2\right )\right )}{2 b^2 \left (a+b x^2\right ) \left (a d^2+b c^2\right )^2}}{4 a}-\frac {a^2 (c-d x)}{4 b^2 \left (a+b x^2\right )^2 \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 2009

\(\displaystyle -\frac {-\frac {a \left (\frac {\sqrt {a} d \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (3 a^2 d^4+10 a b c^2 d^2+15 b^2 c^4\right )}{\sqrt {b} \left (a d^2+b c^2\right )}+\frac {4 b^2 c^5 \log \left (a+b x^2\right )}{a d^2+b c^2}-\frac {8 b^2 c^5 \log (c+d x)}{a d^2+b c^2}\right )}{2 b^2 \left (a d^2+b c^2\right )^2}-\frac {a^2 \left (4 c \left (a d^2+2 b c^2\right )-d x \left (5 a d^2+9 b c^2\right )\right )}{2 b^2 \left (a+b x^2\right ) \left (a d^2+b c^2\right )^2}}{4 a}-\frac {a^2 (c-d x)}{4 b^2 \left (a+b x^2\right )^2 \left (a d^2+b c^2\right )}\)

Input:

Int[x^5/((c + d*x)*(a + b*x^2)^3),x]
 

Output:

-1/4*(a^2*(c - d*x))/(b^2*(b*c^2 + a*d^2)*(a + b*x^2)^2) - (-1/2*(a^2*(4*c 
*(2*b*c^2 + a*d^2) - d*(9*b*c^2 + 5*a*d^2)*x))/(b^2*(b*c^2 + a*d^2)^2*(a + 
 b*x^2)) - (a*((Sqrt[a]*d*(15*b^2*c^4 + 10*a*b*c^2*d^2 + 3*a^2*d^4)*ArcTan 
[(Sqrt[b]*x)/Sqrt[a]])/(Sqrt[b]*(b*c^2 + a*d^2)) - (8*b^2*c^5*Log[c + d*x] 
)/(b*c^2 + a*d^2) + (4*b^2*c^5*Log[a + b*x^2])/(b*c^2 + a*d^2)))/(2*b^2*(b 
*c^2 + a*d^2)^2))/(4*a)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 601

Int[(x_)^(m_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbo 
l] :> With[{Qx = PolynomialQuotient[x^m*(c + d*x)^n, a + b*x^2, x], e = Coe 
ff[PolynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 0], f = Coeff[Pol 
ynomialRemainder[x^m*(c + d*x)^n, a + b*x^2, x], x, 1]}, Simp[(a*f - b*e*x) 
*((a + b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[(c 
+ d*x)^n*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*(p + 1)*Qx)/(c + d*x)^n + (e* 
(2*p + 3))/(c + d*x)^n, x], x], x]] /; FreeQ[{a, b, c, d}, x] && IGtQ[m, 1] 
 && LtQ[p, -1] && ILtQ[n, 0] && NeQ[b*c^2 + a*d^2, 0]
 

rule 657

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

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 2178

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{Qx = PolynomialQuotient[(d + e*x)^m*Pq, a + b*x^2, x], R = Coeff[Po 
lynomialRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 0], S = Coeff[Polynomia 
lRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*S - b*R*x)*((a + 
b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1))   Int[(d + e*x 
)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*b*(p + 1)*Qx)/(d + e*x)^m + (b*R*( 
2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x 
] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]
 

Maple [A] (verified)

Time = 0.36 (sec) , antiderivative size = 265, normalized size of antiderivative = 1.19

method	result	size
default	\(\frac {\frac {-\frac {a d \left (5 a^{2} d^{4}+14 b \,c^{2} d^{2} a +9 b^{2} c^{4}\right ) x^{3}}{8 b}+\frac {a c \left (a^{2} d^{4}+3 b \,c^{2} d^{2} a +2 b^{2} c^{4}\right ) x^{2}}{2 b}-\frac {a^{2} d \left (3 a^{2} d^{4}+10 b \,c^{2} d^{2} a +7 b^{2} c^{4}\right ) x}{8 b^{2}}+\frac {a^{2} c \left (a^{2} d^{4}+4 b \,c^{2} d^{2} a +3 b^{2} c^{4}\right )}{4 b^{2}}}{\left (b \,x^{2}+a \right )^{2}}+\frac {4 b^{2} c^{5} \ln \left (b \,x^{2}+a \right )+\frac {\left (3 a^{3} d^{5}+10 a^{2} b \,c^{2} d^{3}+15 a \,b^{2} c^{4} d \right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}}{8 b^{2}}}{\left (a \,d^{2}+b \,c^{2}\right )^{3}}-\frac {c^{5} \ln \left (d x +c \right )}{\left (a \,d^{2}+b \,c^{2}\right )^{3}}\)	\(265\)
risch	\(\text {Expression too large to display}\)	\(2722\)

Input:

int(x^5/(d*x+c)/(b*x^2+a)^3,x,method=_RETURNVERBOSE)
 

Output:

1/(a*d^2+b*c^2)^3*((-1/8*a*d*(5*a^2*d^4+14*a*b*c^2*d^2+9*b^2*c^4)/b*x^3+1/ 
2*a*c*(a^2*d^4+3*a*b*c^2*d^2+2*b^2*c^4)/b*x^2-1/8*a^2*d*(3*a^2*d^4+10*a*b* 
c^2*d^2+7*b^2*c^4)/b^2*x+1/4*a^2*c*(a^2*d^4+4*a*b*c^2*d^2+3*b^2*c^4)/b^2)/ 
(b*x^2+a)^2+1/8/b^2*(4*b^2*c^5*ln(b*x^2+a)+(3*a^3*d^5+10*a^2*b*c^2*d^3+15* 
a*b^2*c^4*d)/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))))-c^5*ln(d*x+c)/(a*d^2+b* 
c^2)^3
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 514 vs. \(2 (206) = 412\).

Time = 1.86 (sec) , antiderivative size = 1048, normalized size of antiderivative = 4.72 \[ \int \frac {x^5}{(c+d x) \left (a+b x^2\right )^3} \, dx =\text {Too large to display} \] Input:

integrate(x^5/(d*x+c)/(b*x^2+a)^3,x, algorithm="fricas")
 

Output:

[1/16*(12*a^2*b^2*c^5 + 16*a^3*b*c^3*d^2 + 4*a^4*c*d^4 - 2*(9*a*b^3*c^4*d 
+ 14*a^2*b^2*c^2*d^3 + 5*a^3*b*d^5)*x^3 + 8*(2*a*b^3*c^5 + 3*a^2*b^2*c^3*d 
^2 + a^3*b*c*d^4)*x^2 + (15*a^2*b^2*c^4*d + 10*a^3*b*c^2*d^3 + 3*a^4*d^5 + 
 (15*b^4*c^4*d + 10*a*b^3*c^2*d^3 + 3*a^2*b^2*d^5)*x^4 + 2*(15*a*b^3*c^4*d 
 + 10*a^2*b^2*c^2*d^3 + 3*a^3*b*d^5)*x^2)*sqrt(-a/b)*log((b*x^2 + 2*b*x*sq 
rt(-a/b) - a)/(b*x^2 + a)) - 2*(7*a^2*b^2*c^4*d + 10*a^3*b*c^2*d^3 + 3*a^4 
*d^5)*x + 8*(b^4*c^5*x^4 + 2*a*b^3*c^5*x^2 + a^2*b^2*c^5)*log(b*x^2 + a) - 
 16*(b^4*c^5*x^4 + 2*a*b^3*c^5*x^2 + a^2*b^2*c^5)*log(d*x + c))/(a^2*b^5*c 
^6 + 3*a^3*b^4*c^4*d^2 + 3*a^4*b^3*c^2*d^4 + a^5*b^2*d^6 + (b^7*c^6 + 3*a* 
b^6*c^4*d^2 + 3*a^2*b^5*c^2*d^4 + a^3*b^4*d^6)*x^4 + 2*(a*b^6*c^6 + 3*a^2* 
b^5*c^4*d^2 + 3*a^3*b^4*c^2*d^4 + a^4*b^3*d^6)*x^2), 1/8*(6*a^2*b^2*c^5 + 
8*a^3*b*c^3*d^2 + 2*a^4*c*d^4 - (9*a*b^3*c^4*d + 14*a^2*b^2*c^2*d^3 + 5*a^ 
3*b*d^5)*x^3 + 4*(2*a*b^3*c^5 + 3*a^2*b^2*c^3*d^2 + a^3*b*c*d^4)*x^2 + (15 
*a^2*b^2*c^4*d + 10*a^3*b*c^2*d^3 + 3*a^4*d^5 + (15*b^4*c^4*d + 10*a*b^3*c 
^2*d^3 + 3*a^2*b^2*d^5)*x^4 + 2*(15*a*b^3*c^4*d + 10*a^2*b^2*c^2*d^3 + 3*a 
^3*b*d^5)*x^2)*sqrt(a/b)*arctan(b*x*sqrt(a/b)/a) - (7*a^2*b^2*c^4*d + 10*a 
^3*b*c^2*d^3 + 3*a^4*d^5)*x + 4*(b^4*c^5*x^4 + 2*a*b^3*c^5*x^2 + a^2*b^2*c 
^5)*log(b*x^2 + a) - 8*(b^4*c^5*x^4 + 2*a*b^3*c^5*x^2 + a^2*b^2*c^5)*log(d 
*x + c))/(a^2*b^5*c^6 + 3*a^3*b^4*c^4*d^2 + 3*a^4*b^3*c^2*d^4 + a^5*b^2*d^ 
6 + (b^7*c^6 + 3*a*b^6*c^4*d^2 + 3*a^2*b^5*c^2*d^4 + a^3*b^4*d^6)*x^4 +...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^5}{(c+d x) \left (a+b x^2\right )^3} \, dx=\text {Timed out} \] Input:

integrate(x**5/(d*x+c)/(b*x**2+a)**3,x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 401, normalized size of antiderivative = 1.81 \[ \int \frac {x^5}{(c+d x) \left (a+b x^2\right )^3} \, dx=\frac {c^{5} \log \left (b x^{2} + a\right )}{2 \, {\left (b^{3} c^{6} + 3 \, a b^{2} c^{4} d^{2} + 3 \, a^{2} b c^{2} d^{4} + a^{3} d^{6}\right )}} - \frac {c^{5} \log \left (d x + c\right )}{b^{3} c^{6} + 3 \, a b^{2} c^{4} d^{2} + 3 \, a^{2} b c^{2} d^{4} + a^{3} d^{6}} + \frac {{\left (15 \, a b^{2} c^{4} d + 10 \, a^{2} b c^{2} d^{3} + 3 \, a^{3} d^{5}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, {\left (b^{5} c^{6} + 3 \, a b^{4} c^{4} d^{2} + 3 \, a^{2} b^{3} c^{2} d^{4} + a^{3} b^{2} d^{6}\right )} \sqrt {a b}} + \frac {6 \, a^{2} b c^{3} + 2 \, a^{3} c d^{2} - {\left (9 \, a b^{2} c^{2} d + 5 \, a^{2} b d^{3}\right )} x^{3} + 4 \, {\left (2 \, a b^{2} c^{3} + a^{2} b c d^{2}\right )} x^{2} - {\left (7 \, a^{2} b c^{2} d + 3 \, a^{3} d^{3}\right )} x}{8 \, {\left (a^{2} b^{4} c^{4} + 2 \, a^{3} b^{3} c^{2} d^{2} + a^{4} b^{2} d^{4} + {\left (b^{6} c^{4} + 2 \, a b^{5} c^{2} d^{2} + a^{2} b^{4} d^{4}\right )} x^{4} + 2 \, {\left (a b^{5} c^{4} + 2 \, a^{2} b^{4} c^{2} d^{2} + a^{3} b^{3} d^{4}\right )} x^{2}\right )}} \] Input:

integrate(x^5/(d*x+c)/(b*x^2+a)^3,x, algorithm="maxima")
 

Output:

1/2*c^5*log(b*x^2 + a)/(b^3*c^6 + 3*a*b^2*c^4*d^2 + 3*a^2*b*c^2*d^4 + a^3* 
d^6) - c^5*log(d*x + c)/(b^3*c^6 + 3*a*b^2*c^4*d^2 + 3*a^2*b*c^2*d^4 + a^3 
*d^6) + 1/8*(15*a*b^2*c^4*d + 10*a^2*b*c^2*d^3 + 3*a^3*d^5)*arctan(b*x/sqr 
t(a*b))/((b^5*c^6 + 3*a*b^4*c^4*d^2 + 3*a^2*b^3*c^2*d^4 + a^3*b^2*d^6)*sqr 
t(a*b)) + 1/8*(6*a^2*b*c^3 + 2*a^3*c*d^2 - (9*a*b^2*c^2*d + 5*a^2*b*d^3)*x 
^3 + 4*(2*a*b^2*c^3 + a^2*b*c*d^2)*x^2 - (7*a^2*b*c^2*d + 3*a^3*d^3)*x)/(a 
^2*b^4*c^4 + 2*a^3*b^3*c^2*d^2 + a^4*b^2*d^4 + (b^6*c^4 + 2*a*b^5*c^2*d^2 
+ a^2*b^4*d^4)*x^4 + 2*(a*b^5*c^4 + 2*a^2*b^4*c^2*d^2 + a^3*b^3*d^4)*x^2)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 376, normalized size of antiderivative = 1.69 \[ \int \frac {x^5}{(c+d x) \left (a+b x^2\right )^3} \, dx=-\frac {c^{5} d \log \left ({\left | d x + c \right |}\right )}{b^{3} c^{6} d + 3 \, a b^{2} c^{4} d^{3} + 3 \, a^{2} b c^{2} d^{5} + a^{3} d^{7}} + \frac {c^{5} \log \left (b x^{2} + a\right )}{2 \, {\left (b^{3} c^{6} + 3 \, a b^{2} c^{4} d^{2} + 3 \, a^{2} b c^{2} d^{4} + a^{3} d^{6}\right )}} + \frac {{\left (15 \, a b^{2} c^{4} d + 10 \, a^{2} b c^{2} d^{3} + 3 \, a^{3} d^{5}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, {\left (b^{5} c^{6} + 3 \, a b^{4} c^{4} d^{2} + 3 \, a^{2} b^{3} c^{2} d^{4} + a^{3} b^{2} d^{6}\right )} \sqrt {a b}} + \frac {6 \, a^{2} b^{2} c^{5} + 8 \, a^{3} b c^{3} d^{2} + 2 \, a^{4} c d^{4} - {\left (9 \, a b^{3} c^{4} d + 14 \, a^{2} b^{2} c^{2} d^{3} + 5 \, a^{3} b d^{5}\right )} x^{3} + 4 \, {\left (2 \, a b^{3} c^{5} + 3 \, a^{2} b^{2} c^{3} d^{2} + a^{3} b c d^{4}\right )} x^{2} - {\left (7 \, a^{2} b^{2} c^{4} d + 10 \, a^{3} b c^{2} d^{3} + 3 \, a^{4} d^{5}\right )} x}{8 \, {\left (b c^{2} + a d^{2}\right )}^{3} {\left (b x^{2} + a\right )}^{2} b^{2}} \] Input:

integrate(x^5/(d*x+c)/(b*x^2+a)^3,x, algorithm="giac")
 

Output:

-c^5*d*log(abs(d*x + c))/(b^3*c^6*d + 3*a*b^2*c^4*d^3 + 3*a^2*b*c^2*d^5 + 
a^3*d^7) + 1/2*c^5*log(b*x^2 + a)/(b^3*c^6 + 3*a*b^2*c^4*d^2 + 3*a^2*b*c^2 
*d^4 + a^3*d^6) + 1/8*(15*a*b^2*c^4*d + 10*a^2*b*c^2*d^3 + 3*a^3*d^5)*arct 
an(b*x/sqrt(a*b))/((b^5*c^6 + 3*a*b^4*c^4*d^2 + 3*a^2*b^3*c^2*d^4 + a^3*b^ 
2*d^6)*sqrt(a*b)) + 1/8*(6*a^2*b^2*c^5 + 8*a^3*b*c^3*d^2 + 2*a^4*c*d^4 - ( 
9*a*b^3*c^4*d + 14*a^2*b^2*c^2*d^3 + 5*a^3*b*d^5)*x^3 + 4*(2*a*b^3*c^5 + 3 
*a^2*b^2*c^3*d^2 + a^3*b*c*d^4)*x^2 - (7*a^2*b^2*c^4*d + 10*a^3*b*c^2*d^3 
+ 3*a^4*d^5)*x)/((b*c^2 + a*d^2)^3*(b*x^2 + a)^2*b^2)
 

Mupad [B] (verification not implemented)

Time = 7.82 (sec) , antiderivative size = 978, normalized size of antiderivative = 4.41 \[ \int \frac {x^5}{(c+d x) \left (a+b x^2\right )^3} \, dx =\text {Too large to display} \] Input:

int(x^5/((a + b*x^2)^3*(c + d*x)),x)
 

Output:

(log(576*b^12*c^14*x + 576*b^9*c^14*(-a*b^5)^(1/2) + 9*a^7*b^5*d^14*x + 9* 
a^7*b^2*d^14*(-a*b^5)^(1/2) - 740*a^3*c^6*d^8*(-a*b^5)^(3/2) - 1377*b^3*c^ 
12*d^2*(-a*b^5)^(3/2) + 319*a^5*b^4*c^4*d^10*(-a*b^5)^(1/2) + 78*a^6*b^3*c 
^2*d^12*(-a*b^5)^(1/2) + 1377*a*b^11*c^12*d^2*x + 1326*a^2*b^10*c^10*d^4*x 
 + 1015*a^3*b^9*c^8*d^6*x + 740*a^4*b^8*c^6*d^8*x + 319*a^5*b^7*c^4*d^10*x 
 + 78*a^6*b^6*c^2*d^12*x - 1326*a*b^2*c^10*d^4*(-a*b^5)^(3/2) - 1015*a^2*b 
*c^8*d^6*(-a*b^5)^(3/2))*(8*b^5*c^5 + 3*a^2*d^5*(-a*b^5)^(1/2) + 15*b^2*c^ 
4*d*(-a*b^5)^(1/2) + 10*a*b*c^2*d^3*(-a*b^5)^(1/2)))/(16*(b^8*c^6 + a^3*b^ 
5*d^6 + 3*a*b^7*c^4*d^2 + 3*a^2*b^6*c^2*d^4)) - (c^5*log(c + d*x))/(a*d^2 
+ b*c^2)^3 - (log(576*b^12*c^14*x - 576*b^9*c^14*(-a*b^5)^(1/2) + 9*a^7*b^ 
5*d^14*x - 9*a^7*b^2*d^14*(-a*b^5)^(1/2) + 740*a^3*c^6*d^8*(-a*b^5)^(3/2) 
+ 1377*b^3*c^12*d^2*(-a*b^5)^(3/2) - 319*a^5*b^4*c^4*d^10*(-a*b^5)^(1/2) - 
 78*a^6*b^3*c^2*d^12*(-a*b^5)^(1/2) + 1377*a*b^11*c^12*d^2*x + 1326*a^2*b^ 
10*c^10*d^4*x + 1015*a^3*b^9*c^8*d^6*x + 740*a^4*b^8*c^6*d^8*x + 319*a^5*b 
^7*c^4*d^10*x + 78*a^6*b^6*c^2*d^12*x + 1326*a*b^2*c^10*d^4*(-a*b^5)^(3/2) 
 + 1015*a^2*b*c^8*d^6*(-a*b^5)^(3/2))*(3*a^2*d^5*(-a*b^5)^(1/2) - 8*b^5*c^ 
5 + 15*b^2*c^4*d*(-a*b^5)^(1/2) + 10*a*b*c^2*d^3*(-a*b^5)^(1/2)))/(16*(b^8 
*c^6 + a^3*b^5*d^6 + 3*a*b^7*c^4*d^2 + 3*a^2*b^6*c^2*d^4)) - ((x^3*(5*a^2* 
d^3 + 9*a*b*c^2*d))/(8*(b^3*c^4 + a^2*b*d^4 + 2*a*b^2*c^2*d^2)) - (a^2*(3* 
b*c^3 + a*c*d^2))/(4*b^2*(a*d^2 + b*c^2)^2) - (a*x^2*(2*b*c^3 + a*c*d^2...
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 682, normalized size of antiderivative = 3.07 \[ \int \frac {x^5}{(c+d x) \left (a+b x^2\right )^3} \, dx=\frac {2 a^{2} b^{3} c^{5}-4 b^{5} c^{5} x^{4}+20 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} c^{2} d^{3} x^{2}+30 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{3} c^{4} d \,x^{2}+10 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{3} c^{2} d^{3} x^{4}+10 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} b \,c^{2} d^{3}+6 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} b \,d^{5} x^{2}+15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} c^{4} d +3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} d^{5} x^{4}+15 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{4} c^{4} d \,x^{4}+3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{4} d^{5}+8 \,\mathrm {log}\left (b \,x^{2}+a \right ) a \,b^{4} c^{5} x^{2}-16 \,\mathrm {log}\left (d x +c \right ) a \,b^{4} c^{5} x^{2}-10 a^{3} b^{2} c^{2} d^{3} x -7 a^{2} b^{3} c^{4} d x -14 a^{2} b^{3} c^{2} d^{3} x^{3}-2 a^{2} b^{3} c \,d^{4} x^{4}-9 a \,b^{4} c^{4} d \,x^{3}-6 a \,b^{4} c^{3} d^{2} x^{4}+4 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} b^{3} c^{5}+4 \,\mathrm {log}\left (b \,x^{2}+a \right ) b^{5} c^{5} x^{4}-8 \,\mathrm {log}\left (d x +c \right ) a^{2} b^{3} c^{5}-8 \,\mathrm {log}\left (d x +c \right ) b^{5} c^{5} x^{4}-3 a^{4} b \,d^{5} x +2 a^{3} b^{2} c^{3} d^{2}-5 a^{3} b^{2} d^{5} x^{3}}{8 b^{3} \left (a^{3} b^{2} d^{6} x^{4}+3 a^{2} b^{3} c^{2} d^{4} x^{4}+3 a \,b^{4} c^{4} d^{2} x^{4}+b^{5} c^{6} x^{4}+2 a^{4} b \,d^{6} x^{2}+6 a^{3} b^{2} c^{2} d^{4} x^{2}+6 a^{2} b^{3} c^{4} d^{2} x^{2}+2 a \,b^{4} c^{6} x^{2}+a^{5} d^{6}+3 a^{4} b \,c^{2} d^{4}+3 a^{3} b^{2} c^{4} d^{2}+a^{2} b^{3} c^{6}\right )} \] Input:

int(x^5/(d*x+c)/(b*x^2+a)^3,x)
 

Output:

(3*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**4*d**5 + 10*sqrt(b)*sq 
rt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b*c**2*d**3 + 6*sqrt(b)*sqrt(a)*a 
tan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b*d**5*x**2 + 15*sqrt(b)*sqrt(a)*atan((b 
*x)/(sqrt(b)*sqrt(a)))*a**2*b**2*c**4*d + 20*sqrt(b)*sqrt(a)*atan((b*x)/(s 
qrt(b)*sqrt(a)))*a**2*b**2*c**2*d**3*x**2 + 3*sqrt(b)*sqrt(a)*atan((b*x)/( 
sqrt(b)*sqrt(a)))*a**2*b**2*d**5*x**4 + 30*sqrt(b)*sqrt(a)*atan((b*x)/(sqr 
t(b)*sqrt(a)))*a*b**3*c**4*d*x**2 + 10*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b) 
*sqrt(a)))*a*b**3*c**2*d**3*x**4 + 15*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)* 
sqrt(a)))*b**4*c**4*d*x**4 + 4*log(a + b*x**2)*a**2*b**3*c**5 + 8*log(a + 
b*x**2)*a*b**4*c**5*x**2 + 4*log(a + b*x**2)*b**5*c**5*x**4 - 8*log(c + d* 
x)*a**2*b**3*c**5 - 16*log(c + d*x)*a*b**4*c**5*x**2 - 8*log(c + d*x)*b**5 
*c**5*x**4 - 3*a**4*b*d**5*x + 2*a**3*b**2*c**3*d**2 - 10*a**3*b**2*c**2*d 
**3*x - 5*a**3*b**2*d**5*x**3 + 2*a**2*b**3*c**5 - 7*a**2*b**3*c**4*d*x - 
14*a**2*b**3*c**2*d**3*x**3 - 2*a**2*b**3*c*d**4*x**4 - 9*a*b**4*c**4*d*x* 
*3 - 6*a*b**4*c**3*d**2*x**4 - 4*b**5*c**5*x**4)/(8*b**3*(a**5*d**6 + 3*a* 
*4*b*c**2*d**4 + 2*a**4*b*d**6*x**2 + 3*a**3*b**2*c**4*d**2 + 6*a**3*b**2* 
c**2*d**4*x**2 + a**3*b**2*d**6*x**4 + a**2*b**3*c**6 + 6*a**2*b**3*c**4*d 
**2*x**2 + 3*a**2*b**3*c**2*d**4*x**4 + 2*a*b**4*c**6*x**2 + 3*a*b**4*c**4 
*d**2*x**4 + b**5*c**6*x**4))