\(\int \frac {x^3}{(c+d x) (a+b x^2)^3} \, dx\) [245]

Optimal result

Integrand size = 20, antiderivative size = 215 \[ \int \frac {x^3}{(c+d x) \left (a+b x^2\right )^3} \, dx=\frac {a (c-d x)}{4 b \left (b c^2+a d^2\right ) \left (a+b x^2\right )^2}-\frac {4 b c^3-d \left (5 b c^2+a d^2\right ) x}{8 b \left (b c^2+a d^2\right )^2 \left (a+b x^2\right )}-\frac {d \left (3 b^2 c^4-6 a b c^2 d^2-a^2 d^4\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{8 \sqrt {a} b^{3/2} \left (b c^2+a d^2\right )^3}-\frac {c^3 d^2 \log (c+d x)}{\left (b c^2+a d^2\right )^3}+\frac {c^3 d^2 \log \left (a+b x^2\right )}{2 \left (b c^2+a d^2\right )^3} \] Output:

1/4*a*(-d*x+c)/b/(a*d^2+b*c^2)/(b*x^2+a)^2-1/8*(4*b*c^3-d*(a*d^2+5*b*c^2)* 
x)/b/(a*d^2+b*c^2)^2/(b*x^2+a)-1/8*d*(-a^2*d^4-6*a*b*c^2*d^2+3*b^2*c^4)*ar 
ctan(b^(1/2)*x/a^(1/2))/a^(1/2)/b^(3/2)/(a*d^2+b*c^2)^3-c^3*d^2*ln(d*x+c)/ 
(a*d^2+b*c^2)^3+1/2*c^3*d^2*ln(b*x^2+a)/(a*d^2+b*c^2)^3
 

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 177, normalized size of antiderivative = 0.82 \[ \int \frac {x^3}{(c+d x) \left (a+b x^2\right )^3} \, dx=\frac {\frac {2 a \left (b c^2+a d^2\right )^2 (c-d x)}{b \left (a+b x^2\right )^2}+\frac {\left (b c^2+a d^2\right ) \left (a d^3 x+b c^2 (-4 c+5 d x)\right )}{b \left (a+b x^2\right )}+\frac {d \left (-3 b^2 c^4+6 a b c^2 d^2+a^2 d^4\right ) \arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right )}{\sqrt {a} b^{3/2}}-8 c^3 d^2 \log (c+d x)+4 c^3 d^2 \log \left (a+b x^2\right )}{8 \left (b c^2+a d^2\right )^3} \] Input:

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

Output:

((2*a*(b*c^2 + a*d^2)^2*(c - d*x))/(b*(a + b*x^2)^2) + ((b*c^2 + a*d^2)*(a 
*d^3*x + b*c^2*(-4*c + 5*d*x)))/(b*(a + b*x^2)) + (d*(-3*b^2*c^4 + 6*a*b*c 
^2*d^2 + a^2*d^4)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(Sqrt[a]*b^(3/2)) - 8*c^3*d 
^2*Log[c + d*x] + 4*c^3*d^2*Log[a + b*x^2])/(8*(b*c^2 + a*d^2)^3)
 

Rubi [A] (verified)

Time = 0.78 (sec) , antiderivative size = 244, normalized size of antiderivative = 1.13, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.350, Rules used = {601, 25, 27, 686, 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.

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

\(\Big \downarrow \) 601

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 686

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 657

\(\displaystyle \frac {-\frac {d \int \left (\frac {8 b d^2 c^3}{\left (b c^2+a d^2\right ) (c+d x)}+\frac {3 b^2 c^4-8 b^2 d x c^3-6 a b d^2 c^2-a^2 d^4}{\left (b c^2+a d^2\right ) \left (b x^2+a\right )}\right )dx}{2 \left (a d^2+b c^2\right )}-\frac {4 b c^3-d x \left (a d^2+5 b c^2\right )}{2 \left (a+b x^2\right ) \left (a d^2+b c^2\right )}}{4 b \left (a d^2+b c^2\right )}+\frac {a (c-d x)}{4 b \left (a+b x^2\right )^2 \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 2009

\(\displaystyle \frac {-\frac {d \left (\frac {\arctan \left (\frac {\sqrt {b} x}{\sqrt {a}}\right ) \left (-a^2 d^4-6 a b c^2 d^2+3 b^2 c^4\right )}{\sqrt {a} \sqrt {b} \left (a d^2+b c^2\right )}-\frac {4 b c^3 d \log \left (a+b x^2\right )}{a d^2+b c^2}+\frac {8 b c^3 d \log (c+d x)}{a d^2+b c^2}\right )}{2 \left (a d^2+b c^2\right )}-\frac {4 b c^3-d x \left (a d^2+5 b c^2\right )}{2 \left (a+b x^2\right ) \left (a d^2+b c^2\right )}}{4 b \left (a d^2+b c^2\right )}+\frac {a (c-d x)}{4 b \left (a+b x^2\right )^2 \left (a d^2+b c^2\right )}\)

Input:

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

Output:

(a*(c - d*x))/(4*b*(b*c^2 + a*d^2)*(a + b*x^2)^2) + (-1/2*(4*b*c^3 - d*(5* 
b*c^2 + a*d^2)*x)/((b*c^2 + a*d^2)*(a + b*x^2)) - (d*(((3*b^2*c^4 - 6*a*b* 
c^2*d^2 - a^2*d^4)*ArcTan[(Sqrt[b]*x)/Sqrt[a]])/(Sqrt[a]*Sqrt[b]*(b*c^2 + 
a*d^2)) + (8*b*c^3*d*Log[c + d*x])/(b*c^2 + a*d^2) - (4*b*c^3*d*Log[a + b* 
x^2])/(b*c^2 + a*d^2)))/(2*(b*c^2 + a*d^2)))/(4*b*(b*c^2 + a*d^2))
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 686

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_), x_Symbol] :> Simp[(-(d + e*x)^(m + 1))*(f*a*c*e - a*g*c*d + c*(c*d*f + 
a*e*g)*x)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[ 
1/(2*a*c*(p + 1)*(c*d^2 + a*e^2))   Int[(d + e*x)^m*(a + c*x^2)^(p + 1)*Sim 
p[f*(c^2*d^2*(2*p + 3) + a*c*e^2*(m + 2*p + 3)) - a*c*d*e*g*m + c*e*(c*d*f 
+ a*e*g)*(m + 2*p + 4)*x, x], x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ 
[p, -1] && (IntegerQ[m] || IntegerQ[p] || IntegersQ[2*m, 2*p])
 

rule 2009

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

Maple [A] (verified)

Time = 0.32 (sec) , antiderivative size = 230, normalized size of antiderivative = 1.07

method	result	size
default	\(\frac {\frac {\left (\frac {1}{8} a^{2} d^{5}+\frac {3}{4} d^{3} a \,c^{2} b +\frac {5}{8} b^{2} c^{4} d \right ) x^{3}+\left (-\frac {1}{2} a \,c^{3} d^{2} b -\frac {1}{2} c^{5} b^{2}\right ) x^{2}-\frac {a d \left (a^{2} d^{4}-2 b \,c^{2} d^{2} a -3 b^{2} c^{4}\right ) x}{8 b}+\frac {a c \left (a^{2} d^{4}-b^{2} c^{4}\right )}{4 b}}{\left (b \,x^{2}+a \right )^{2}}+\frac {d \left (4 b \,c^{3} d \ln \left (b \,x^{2}+a \right )+\frac {\left (a^{2} d^{4}+6 b \,c^{2} d^{2} a -3 b^{2} c^{4}\right ) \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{\sqrt {a b}}\right )}{8 b}}{\left (a \,d^{2}+b \,c^{2}\right )^{3}}-\frac {c^{3} d^{2} \ln \left (d x +c \right )}{\left (a \,d^{2}+b \,c^{2}\right )^{3}}\)	\(230\)
risch	\(\frac {\frac {\left (a \,d^{2}+5 b \,c^{2}\right ) d \,x^{3}}{8 a^{2} d^{4}+16 b \,c^{2} d^{2} a +8 b^{2} c^{4}}-\frac {b \,c^{3} x^{2}}{2 \left (a^{2} d^{4}+2 b \,c^{2} d^{2} a +b^{2} c^{4}\right )}-\frac {a d \left (a \,d^{2}-3 b \,c^{2}\right ) x}{8 \left (a^{2} d^{4}+2 b \,c^{2} d^{2} a +b^{2} c^{4}\right ) b}+\frac {a c \left (a \,d^{2}-b \,c^{2}\right )}{4 \left (a^{2} d^{4}+2 b \,c^{2} d^{2} a +b^{2} c^{4}\right ) b}}{\left (b \,x^{2}+a \right )^{2}}-\frac {c^{3} d^{2} \ln \left (d x +c \right )}{a^{3} d^{6}+3 a^{2} b \,c^{2} d^{4}+3 a \,b^{2} c^{4} d^{2}+b^{3} c^{6}}+\frac {\left (\munderset {\textit {\_R} =\operatorname {RootOf}\left (\left (b^{3} a^{4} d^{6}+3 a^{3} b^{4} c^{2} d^{4}+3 a^{2} b^{5} c^{4} d^{2}+a \,b^{6} c^{6}\right ) \textit {\_Z}^{2}-16 a \,b^{3} c^{3} d^{2} \textit {\_Z} +a \,d^{4}+9 b \,c^{2} d^{2}\right )}{\sum }\textit {\_R} \ln \left (\left (\left (3 a^{5} b^{3} d^{10}+11 a^{4} b^{4} c^{2} d^{8}+14 a^{3} b^{5} c^{4} d^{6}+6 a^{2} b^{6} c^{6} d^{4}-a \,b^{7} c^{8} d^{2}-b^{8} c^{10}\right ) \textit {\_R}^{2}+\left (2 a^{3} b^{2} c \,d^{8}-10 a^{2} b^{3} c^{3} d^{6}-26 a \,b^{4} c^{5} d^{4}-14 b^{5} c^{7} d^{2}\right ) \textit {\_R} +2 a^{2} d^{8}+20 a b \,c^{2} d^{6}+50 b^{2} c^{4} d^{4}\right ) x +\left (4 a^{5} b^{3} c \,d^{9}+16 a^{4} b^{4} c^{3} d^{7}+24 a^{3} b^{5} c^{5} d^{5}+16 a^{2} b^{6} c^{7} d^{3}+4 a \,b^{7} c^{9} d \right ) \textit {\_R}^{2}+\left (-a^{4} b \,d^{9}-6 a^{3} b^{2} c^{2} d^{7}-12 a^{2} b^{3} c^{4} d^{5}-10 a \,b^{4} c^{6} d^{3}-3 b^{5} c^{8} d \right ) \textit {\_R} +2 a^{2} c \,d^{7}+4 a b \,c^{3} d^{5}-30 b^{2} c^{5} d^{3}\right )\right )}{16}\)	\(653\)

Input:

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

Output:

1/(a*d^2+b*c^2)^3*(((1/8*a^2*d^5+3/4*d^3*a*c^2*b+5/8*b^2*c^4*d)*x^3+(-1/2* 
a*c^3*d^2*b-1/2*c^5*b^2)*x^2-1/8*a*d*(a^2*d^4-2*a*b*c^2*d^2-3*b^2*c^4)/b*x 
+1/4*a*c*(a^2*d^4-b^2*c^4)/b)/(b*x^2+a)^2+1/8*d/b*(4*b*c^3*d*ln(b*x^2+a)+( 
a^2*d^4+6*a*b*c^2*d^2-3*b^2*c^4)/(a*b)^(1/2)*arctan(b*x/(a*b)^(1/2))))-c^3 
*d^2*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. 520 vs. \(2 (200) = 400\).

Time = 1.30 (sec) , antiderivative size = 1061, normalized size of antiderivative = 4.93 \[ \int \frac {x^3}{(c+d x) \left (a+b x^2\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/16*(4*a^2*b^3*c^5 - 4*a^4*b*c*d^4 - 2*(5*a*b^4*c^4*d + 6*a^2*b^3*c^2*d 
^3 + a^3*b^2*d^5)*x^3 + 8*(a*b^4*c^5 + a^2*b^3*c^3*d^2)*x^2 - (3*a^2*b^2*c 
^4*d - 6*a^3*b*c^2*d^3 - a^4*d^5 + (3*b^4*c^4*d - 6*a*b^3*c^2*d^3 - a^2*b^ 
2*d^5)*x^4 + 2*(3*a*b^3*c^4*d - 6*a^2*b^2*c^2*d^3 - a^3*b*d^5)*x^2)*sqrt(- 
a*b)*log((b*x^2 - 2*sqrt(-a*b)*x - a)/(b*x^2 + a)) - 2*(3*a^2*b^3*c^4*d + 
2*a^3*b^2*c^2*d^3 - a^4*b*d^5)*x - 8*(a*b^4*c^3*d^2*x^4 + 2*a^2*b^3*c^3*d^ 
2*x^2 + a^3*b^2*c^3*d^2)*log(b*x^2 + a) + 16*(a*b^4*c^3*d^2*x^4 + 2*a^2*b^ 
3*c^3*d^2*x^2 + a^3*b^2*c^3*d^2)*log(d*x + c))/(a^3*b^5*c^6 + 3*a^4*b^4*c^ 
4*d^2 + 3*a^5*b^3*c^2*d^4 + a^6*b^2*d^6 + (a*b^7*c^6 + 3*a^2*b^6*c^4*d^2 + 
 3*a^3*b^5*c^2*d^4 + a^4*b^4*d^6)*x^4 + 2*(a^2*b^6*c^6 + 3*a^3*b^5*c^4*d^2 
 + 3*a^4*b^4*c^2*d^4 + a^5*b^3*d^6)*x^2), -1/8*(2*a^2*b^3*c^5 - 2*a^4*b*c* 
d^4 - (5*a*b^4*c^4*d + 6*a^2*b^3*c^2*d^3 + a^3*b^2*d^5)*x^3 + 4*(a*b^4*c^5 
 + a^2*b^3*c^3*d^2)*x^2 + (3*a^2*b^2*c^4*d - 6*a^3*b*c^2*d^3 - a^4*d^5 + ( 
3*b^4*c^4*d - 6*a*b^3*c^2*d^3 - a^2*b^2*d^5)*x^4 + 2*(3*a*b^3*c^4*d - 6*a^ 
2*b^2*c^2*d^3 - a^3*b*d^5)*x^2)*sqrt(a*b)*arctan(sqrt(a*b)*x/a) - (3*a^2*b 
^3*c^4*d + 2*a^3*b^2*c^2*d^3 - a^4*b*d^5)*x - 4*(a*b^4*c^3*d^2*x^4 + 2*a^2 
*b^3*c^3*d^2*x^2 + a^3*b^2*c^3*d^2)*log(b*x^2 + a) + 8*(a*b^4*c^3*d^2*x^4 
+ 2*a^2*b^3*c^3*d^2*x^2 + a^3*b^2*c^3*d^2)*log(d*x + c))/(a^3*b^5*c^6 + 3* 
a^4*b^4*c^4*d^2 + 3*a^5*b^3*c^2*d^4 + a^6*b^2*d^6 + (a*b^7*c^6 + 3*a^2*b^6 
*c^4*d^2 + 3*a^3*b^5*c^2*d^4 + a^4*b^4*d^6)*x^4 + 2*(a^2*b^6*c^6 + 3*a^...
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 379, normalized size of antiderivative = 1.76 \[ \int \frac {x^3}{(c+d x) \left (a+b x^2\right )^3} \, dx=\frac {c^{3} d^{2} \log \left (b x^{2} + a\right )}{2 \, {\left (b^{3} c^{6} + 3 \, a b^{2} c^{4} d^{2} + 3 \, a^{2} b c^{2} d^{4} + a^{3} d^{6}\right )}} - \frac {c^{3} d^{2} \log \left (d x + c\right )}{b^{3} c^{6} + 3 \, a b^{2} c^{4} d^{2} + 3 \, a^{2} b c^{2} d^{4} + a^{3} d^{6}} - \frac {{\left (3 \, b^{2} c^{4} d - 6 \, a b c^{2} d^{3} - a^{2} d^{5}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, {\left (b^{4} c^{6} + 3 \, a b^{3} c^{4} d^{2} + 3 \, a^{2} b^{2} c^{2} d^{4} + a^{3} b d^{6}\right )} \sqrt {a b}} - \frac {4 \, b^{2} c^{3} x^{2} + 2 \, a b c^{3} - 2 \, a^{2} c d^{2} - {\left (5 \, b^{2} c^{2} d + a b d^{3}\right )} x^{3} - {\left (3 \, a b c^{2} d - a^{2} d^{3}\right )} x}{8 \, {\left (a^{2} b^{3} c^{4} + 2 \, a^{3} b^{2} c^{2} d^{2} + a^{4} b d^{4} + {\left (b^{5} c^{4} + 2 \, a b^{4} c^{2} d^{2} + a^{2} b^{3} d^{4}\right )} x^{4} + 2 \, {\left (a b^{4} c^{4} + 2 \, a^{2} b^{3} c^{2} d^{2} + a^{3} b^{2} d^{4}\right )} x^{2}\right )}} \] Input:

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

Output:

1/2*c^3*d^2*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^3*d^2*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*(3*b^2*c^4*d - 6*a*b*c^2*d^3 - a^2*d^5)*arctan(b*x/sqr 
t(a*b))/((b^4*c^6 + 3*a*b^3*c^4*d^2 + 3*a^2*b^2*c^2*d^4 + a^3*b*d^6)*sqrt( 
a*b)) - 1/8*(4*b^2*c^3*x^2 + 2*a*b*c^3 - 2*a^2*c*d^2 - (5*b^2*c^2*d + a*b* 
d^3)*x^3 - (3*a*b*c^2*d - a^2*d^3)*x)/(a^2*b^3*c^4 + 2*a^3*b^2*c^2*d^2 + a 
^4*b*d^4 + (b^5*c^4 + 2*a*b^4*c^2*d^2 + a^2*b^3*d^4)*x^4 + 2*(a*b^4*c^4 + 
2*a^2*b^3*c^2*d^2 + a^3*b^2*d^4)*x^2)
 

Giac [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 342, normalized size of antiderivative = 1.59 \[ \int \frac {x^3}{(c+d x) \left (a+b x^2\right )^3} \, dx=-\frac {c^{3} d^{3} \log \left ({\left | d x + c \right |}\right )}{b^{3} c^{6} d + 3 \, a b^{2} c^{4} d^{3} + 3 \, a^{2} b c^{2} d^{5} + a^{3} d^{7}} + \frac {c^{3} d^{2} \log \left (b x^{2} + a\right )}{2 \, {\left (b^{3} c^{6} + 3 \, a b^{2} c^{4} d^{2} + 3 \, a^{2} b c^{2} d^{4} + a^{3} d^{6}\right )}} - \frac {{\left (3 \, b^{2} c^{4} d - 6 \, a b c^{2} d^{3} - a^{2} d^{5}\right )} \arctan \left (\frac {b x}{\sqrt {a b}}\right )}{8 \, {\left (b^{4} c^{6} + 3 \, a b^{3} c^{4} d^{2} + 3 \, a^{2} b^{2} c^{2} d^{4} + a^{3} b d^{6}\right )} \sqrt {a b}} - \frac {2 \, a b^{2} c^{5} - 2 \, a^{3} c d^{4} - {\left (5 \, b^{3} c^{4} d + 6 \, a b^{2} c^{2} d^{3} + a^{2} b d^{5}\right )} x^{3} + 4 \, {\left (b^{3} c^{5} + a b^{2} c^{3} d^{2}\right )} x^{2} - {\left (3 \, a b^{2} c^{4} d + 2 \, a^{2} b c^{2} d^{3} - a^{3} d^{5}\right )} x}{8 \, {\left (b c^{2} + a d^{2}\right )}^{3} {\left (b x^{2} + a\right )}^{2} b} \] Input:

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

Output:

-c^3*d^3*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^3*d^2*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*(3*b^2*c^4*d - 6*a*b*c^2*d^3 - a^2*d^5)*arctan 
(b*x/sqrt(a*b))/((b^4*c^6 + 3*a*b^3*c^4*d^2 + 3*a^2*b^2*c^2*d^4 + a^3*b*d^ 
6)*sqrt(a*b)) - 1/8*(2*a*b^2*c^5 - 2*a^3*c*d^4 - (5*b^3*c^4*d + 6*a*b^2*c^ 
2*d^3 + a^2*b*d^5)*x^3 + 4*(b^3*c^5 + a*b^2*c^3*d^2)*x^2 - (3*a*b^2*c^4*d 
+ 2*a^2*b*c^2*d^3 - a^3*d^5)*x)/((b*c^2 + a*d^2)^3*(b*x^2 + a)^2*b)
 

Mupad [B] (verification not implemented)

Time = 8.31 (sec) , antiderivative size = 913, normalized size of antiderivative = 4.25 \[ \int \frac {x^3}{(c+d x) \left (a+b x^2\right )^3} \, dx =\text {Too large to display} \] Input:

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

Output:

(log(a^7*b^4*d^12*x - 9*b^5*c^12*(-a*b^3)^(3/2) + 1119*a*c^8*d^4*(-a*b^3)^ 
(5/2) + 558*b*c^10*d^2*(-a*b^3)^(5/2) + a^7*b^2*d^12*(-a*b^3)^(1/2) - 14*a 
^5*c^2*d^10*(-a*b^3)^(3/2) + 9*a*b^10*c^12*x - 612*a^3*b^2*c^6*d^6*(-a*b^3 
)^(3/2) + 558*a^2*b^9*c^10*d^2*x + 1119*a^3*b^8*c^8*d^4*x + 612*a^4*b^7*c^ 
6*d^6*x + 55*a^5*b^6*c^4*d^8*x + 14*a^6*b^5*c^2*d^10*x - 55*a^4*b*c^4*d^8* 
(-a*b^3)^(3/2))*(a*((b^3*c^3*d^2)/2 + (3*b*c^2*d^3*(-a*b^3)^(1/2))/8) + (a 
^2*d^5*(-a*b^3)^(1/2))/16 - (3*b^2*c^4*d*(-a*b^3)^(1/2))/16))/(a*b^6*c^6 + 
 a^4*b^3*d^6 + 3*a^2*b^5*c^4*d^2 + 3*a^3*b^4*c^2*d^4) - ((a*(b*c^3 - a*c*d 
^2))/(4*b*(a*d^2 + b*c^2)^2) - (x^3*(a*d^3 + 5*b*c^2*d))/(8*(a^2*d^4 + b^2 
*c^4 + 2*a*b*c^2*d^2)) + (b*c^3*x^2)/(2*(a*d^2 + b*c^2)^2) + (a*d*x*(a*d^2 
 - 3*b*c^2))/(8*(b^3*c^4 + a^2*b*d^4 + 2*a*b^2*c^2*d^2)))/(a^2 + b^2*x^4 + 
 2*a*b*x^2) + (log(9*b^5*c^12*(-a*b^3)^(3/2) + a^7*b^4*d^12*x - 1119*a*c^8 
*d^4*(-a*b^3)^(5/2) - 558*b*c^10*d^2*(-a*b^3)^(5/2) - a^7*b^2*d^12*(-a*b^3 
)^(1/2) + 14*a^5*c^2*d^10*(-a*b^3)^(3/2) + 9*a*b^10*c^12*x + 612*a^3*b^2*c 
^6*d^6*(-a*b^3)^(3/2) + 558*a^2*b^9*c^10*d^2*x + 1119*a^3*b^8*c^8*d^4*x + 
612*a^4*b^7*c^6*d^6*x + 55*a^5*b^6*c^4*d^8*x + 14*a^6*b^5*c^2*d^10*x + 55* 
a^4*b*c^4*d^8*(-a*b^3)^(3/2))*(a*((b^3*c^3*d^2)/2 - (3*b*c^2*d^3*(-a*b^3)^ 
(1/2))/8) - (a^2*d^5*(-a*b^3)^(1/2))/16 + (3*b^2*c^4*d*(-a*b^3)^(1/2))/16) 
)/(a*b^6*c^6 + a^4*b^3*d^6 + 3*a^2*b^5*c^4*d^2 + 3*a^3*b^4*c^2*d^4) - (c^3 
*d^2*log(c + d*x))/(a^3*d^6 + b^3*c^6 + 3*a*b^2*c^4*d^2 + 3*a^2*b*c^2*d...
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 690, normalized size of antiderivative = 3.21 \[ \int \frac {x^3}{(c+d x) \left (a+b x^2\right )^3} \, dx=\frac {2 b^{5} c^{5} x^{4}+2 a^{4} b c \,d^{4}+12 \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}-6 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{3} c^{4} d \,x^{2}+6 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a \,b^{3} c^{2} d^{3} x^{4}+6 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} b \,c^{2} d^{3}+2 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{3} b \,d^{5} x^{2}-3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} c^{4} d +\sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{2} b^{2} d^{5} x^{4}-3 \sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) b^{4} c^{4} d \,x^{4}+\sqrt {b}\, \sqrt {a}\, \mathit {atan} \left (\frac {b x}{\sqrt {b}\, \sqrt {a}}\right ) a^{4} d^{5}+2 a^{3} b^{2} c^{2} d^{3} x +3 a^{2} b^{3} c^{4} d x +6 a^{2} b^{3} c^{2} d^{3} x^{3}+5 a \,b^{4} c^{4} d \,x^{3}+2 a \,b^{4} c^{3} d^{2} x^{4}+8 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{2} b^{3} c^{3} d^{2} x^{2}+4 \,\mathrm {log}\left (b \,x^{2}+a \right ) a \,b^{4} c^{3} d^{2} x^{4}-16 \,\mathrm {log}\left (d x +c \right ) a^{2} b^{3} c^{3} d^{2} x^{2}-8 \,\mathrm {log}\left (d x +c \right ) a \,b^{4} c^{3} d^{2} x^{4}-a^{4} b \,d^{5} x +2 a^{3} b^{2} c^{3} d^{2}+a^{3} b^{2} d^{5} x^{3}+4 \,\mathrm {log}\left (b \,x^{2}+a \right ) a^{3} b^{2} c^{3} d^{2}-8 \,\mathrm {log}\left (d x +c \right ) a^{3} b^{2} c^{3} d^{2}}{8 a \,b^{2} \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^3/(d*x+c)/(b*x^2+a)^3,x)
 

Output:

(sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**4*d**5 + 6*sqrt(b)*sqrt( 
a)*atan((b*x)/(sqrt(b)*sqrt(a)))*a**3*b*c**2*d**3 + 2*sqrt(b)*sqrt(a)*atan 
((b*x)/(sqrt(b)*sqrt(a)))*a**3*b*d**5*x**2 - 3*sqrt(b)*sqrt(a)*atan((b*x)/ 
(sqrt(b)*sqrt(a)))*a**2*b**2*c**4*d + 12*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt( 
b)*sqrt(a)))*a**2*b**2*c**2*d**3*x**2 + sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b 
)*sqrt(a)))*a**2*b**2*d**5*x**4 - 6*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sq 
rt(a)))*a*b**3*c**4*d*x**2 + 6*sqrt(b)*sqrt(a)*atan((b*x)/(sqrt(b)*sqrt(a) 
))*a*b**3*c**2*d**3*x**4 - 3*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**3*b**2*c**3*d**2 + 8*log(a + b*x* 
*2)*a**2*b**3*c**3*d**2*x**2 + 4*log(a + b*x**2)*a*b**4*c**3*d**2*x**4 - 8 
*log(c + d*x)*a**3*b**2*c**3*d**2 - 16*log(c + d*x)*a**2*b**3*c**3*d**2*x* 
*2 - 8*log(c + d*x)*a*b**4*c**3*d**2*x**4 + 2*a**4*b*c*d**4 - a**4*b*d**5* 
x + 2*a**3*b**2*c**3*d**2 + 2*a**3*b**2*c**2*d**3*x + a**3*b**2*d**5*x**3 
+ 3*a**2*b**3*c**4*d*x + 6*a**2*b**3*c**2*d**3*x**3 + 5*a*b**4*c**4*d*x**3 
 + 2*a*b**4*c**3*d**2*x**4 + 2*b**5*c**5*x**4)/(8*a*b**2*(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))