\(\int \frac {(A+B x) (c+d x)^3}{(a+b x^2)^{5/2}} \, dx\) [188]

Optimal result

Integrand size = 24, antiderivative size = 163 \[ \int \frac {(A+B x) (c+d x)^3}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {(c+d x)^2 (a (B c+A d)-(A b c-a B d) x)}{3 a b \left (a+b x^2\right )^{3/2}}-\frac {2 a d \left (A b c^2+3 a B c d+a A d^2\right )-\left (3 a B d \left (b c^2-a d^2\right )+2 A b c \left (b c^2+a d^2\right )\right ) x}{3 a^2 b^2 \sqrt {a+b x^2}}+\frac {B d^3 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{5/2}} \] Output:

-1/3*(d*x+c)^2*(a*(A*d+B*c)-(A*b*c-B*a*d)*x)/a/b/(b*x^2+a)^(3/2)-1/3*(2*a* 
d*(A*a*d^2+A*b*c^2+3*B*a*c*d)-(3*a*B*d*(-a*d^2+b*c^2)+2*A*b*c*(a*d^2+b*c^2 
))*x)/a^2/b^2/(b*x^2+a)^(1/2)+B*d^3*arctanh(b^(1/2)*x/(b*x^2+a)^(1/2))/b^( 
5/2)
 

Mathematica [A] (verified)

Time = 2.25 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.00 \[ \int \frac {(A+B x) (c+d x)^3}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {2 A b^3 c^3 x^3-a^3 d^2 (6 B c+2 A d+3 B d x)+3 a b^2 c x \left (B c d x^2+A \left (c^2+d^2 x^2\right )\right )-a^2 b \left (3 A d \left (c^2+d^2 x^2\right )+B \left (c^3+9 c d^2 x^2+4 d^3 x^3\right )\right )}{3 a^2 b^2 \left (a+b x^2\right )^{3/2}}-\frac {B d^3 \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{b^{5/2}} \] Input:

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

Output:

(2*A*b^3*c^3*x^3 - a^3*d^2*(6*B*c + 2*A*d + 3*B*d*x) + 3*a*b^2*c*x*(B*c*d* 
x^2 + A*(c^2 + d^2*x^2)) - a^2*b*(3*A*d*(c^2 + d^2*x^2) + B*(c^3 + 9*c*d^2 
*x^2 + 4*d^3*x^3)))/(3*a^2*b^2*(a + b*x^2)^(3/2)) - (B*d^3*Log[-(Sqrt[b]*x 
) + Sqrt[a + b*x^2]])/b^(5/2)
 

Rubi [A] (verified)

Time = 0.32 (sec) , antiderivative size = 176, normalized size of antiderivative = 1.08, number of steps used = 5, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {684, 675, 224, 219}

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.

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

\(\Big \downarrow \) 684

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

\(\Big \downarrow \) 675

\(\displaystyle \frac {\frac {3 a B d^3 \int \frac {1}{\sqrt {b x^2+a}}dx}{b}+\frac {x \left (\frac {2 A b c^3}{a}-\frac {3 a B d^3}{b}+c d (2 A d+3 B c)\right )}{\sqrt {a+b x^2}}-\frac {2 d \left (a A d^2+3 a B c d+A b c^2\right )}{b \sqrt {a+b x^2}}}{3 a b}-\frac {(c+d x)^2 (a (A d+B c)-x (A b c-a B d))}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {\frac {3 a B d^3 \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{b}+\frac {x \left (\frac {2 A b c^3}{a}-\frac {3 a B d^3}{b}+c d (2 A d+3 B c)\right )}{\sqrt {a+b x^2}}-\frac {2 d \left (a A d^2+3 a B c d+A b c^2\right )}{b \sqrt {a+b x^2}}}{3 a b}-\frac {(c+d x)^2 (a (A d+B c)-x (A b c-a B d))}{3 a b \left (a+b x^2\right )^{3/2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\frac {x \left (\frac {2 A b c^3}{a}-\frac {3 a B d^3}{b}+c d (2 A d+3 B c)\right )}{\sqrt {a+b x^2}}-\frac {2 d \left (a A d^2+3 a B c d+A b c^2\right )}{b \sqrt {a+b x^2}}+\frac {3 a B d^3 \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{3/2}}}{3 a b}-\frac {(c+d x)^2 (a (A d+B c)-x (A b c-a B d))}{3 a b \left (a+b x^2\right )^{3/2}}\)

Input:

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

Output:

-1/3*((c + d*x)^2*(a*(B*c + A*d) - (A*b*c - a*B*d)*x))/(a*b*(a + b*x^2)^(3 
/2)) + ((-2*d*(A*b*c^2 + 3*a*B*c*d + a*A*d^2))/(b*Sqrt[a + b*x^2]) + (((2* 
A*b*c^3)/a - (3*a*B*d^3)/b + c*d*(3*B*c + 2*A*d))*x)/Sqrt[a + b*x^2] + (3* 
a*B*d^3*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/b^(3/2))/(3*a*b)
 

Defintions of rubi rules used

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 675

Int[((d_) + (e_.)*(x_))*((f_) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_S 
ymbol] :> Simp[a*(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] + (- 
Simp[(c*d*f - a*e*g)*x*((a + c*x^2)^(p + 1)/(2*a*c*(p + 1))), x] - Simp[(a* 
e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1))   Int[(a + c*x^2)^(p + 1), x], x]) / 
; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1] &&  !(IntegerQ[p] && NiceSqrtQ 
[(-a)*c])
 

rule 684

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] - Simp[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] && 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])
 

Maple [A] (verified)

Time = 1.15 (sec) , antiderivative size = 235, normalized size of antiderivative = 1.44

method	result	size
default	\(A \,c^{3} \left (\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\right )+d^{2} \left (A d +3 B c \right ) \left (-\frac {x^{2}}{b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}-\frac {2 a}{3 b^{2} \left (b \,x^{2}+a \right )^{\frac {3}{2}}}\right )+3 c d \left (A d +B c \right ) \left (-\frac {x}{2 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {a \left (\frac {x}{3 a \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {2 x}{3 a^{2} \sqrt {b \,x^{2}+a}}\right )}{2 b}\right )-\frac {c^{2} \left (3 A d +B c \right )}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+B \,d^{3} \left (-\frac {x^{3}}{3 b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}}{b}\right )\)	\(235\)

Input:

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

Output:

A*c^3*(1/3*x/a/(b*x^2+a)^(3/2)+2/3/a^2/(b*x^2+a)^(1/2)*x)+d^2*(A*d+3*B*c)* 
(-x^2/b/(b*x^2+a)^(3/2)-2/3*a/b^2/(b*x^2+a)^(3/2))+3*c*d*(A*d+B*c)*(-1/2*x 
/b/(b*x^2+a)^(3/2)+1/2*a/b*(1/3*x/a/(b*x^2+a)^(3/2)+2/3/a^2/(b*x^2+a)^(1/2 
)*x))-1/3*c^2*(3*A*d+B*c)/b/(b*x^2+a)^(3/2)+B*d^3*(-1/3*x^3/b/(b*x^2+a)^(3 
/2)+1/b*(-x/b/(b*x^2+a)^(1/2)+1/b^(3/2)*ln(b^(1/2)*x+(b*x^2+a)^(1/2))))
 

Fricas [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 513, normalized size of antiderivative = 3.15 \[ \int \frac {(A+B x) (c+d x)^3}{\left (a+b x^2\right )^{5/2}} \, dx=\left [\frac {3 \, {\left (B a^{2} b^{2} d^{3} x^{4} + 2 \, B a^{3} b d^{3} x^{2} + B a^{4} d^{3}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (B a^{2} b^{2} c^{3} + 3 \, A a^{2} b^{2} c^{2} d + 6 \, B a^{3} b c d^{2} + 2 \, A a^{3} b d^{3} - {\left (2 \, A b^{4} c^{3} + 3 \, B a b^{3} c^{2} d + 3 \, A a b^{3} c d^{2} - 4 \, B a^{2} b^{2} d^{3}\right )} x^{3} + 3 \, {\left (3 \, B a^{2} b^{2} c d^{2} + A a^{2} b^{2} d^{3}\right )} x^{2} - 3 \, {\left (A a b^{3} c^{3} - B a^{3} b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{6 \, {\left (a^{2} b^{5} x^{4} + 2 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}}, -\frac {3 \, {\left (B a^{2} b^{2} d^{3} x^{4} + 2 \, B a^{3} b d^{3} x^{2} + B a^{4} d^{3}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) + {\left (B a^{2} b^{2} c^{3} + 3 \, A a^{2} b^{2} c^{2} d + 6 \, B a^{3} b c d^{2} + 2 \, A a^{3} b d^{3} - {\left (2 \, A b^{4} c^{3} + 3 \, B a b^{3} c^{2} d + 3 \, A a b^{3} c d^{2} - 4 \, B a^{2} b^{2} d^{3}\right )} x^{3} + 3 \, {\left (3 \, B a^{2} b^{2} c d^{2} + A a^{2} b^{2} d^{3}\right )} x^{2} - 3 \, {\left (A a b^{3} c^{3} - B a^{3} b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{3 \, {\left (a^{2} b^{5} x^{4} + 2 \, a^{3} b^{4} x^{2} + a^{4} b^{3}\right )}}\right ] \] Input:

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

Output:

[1/6*(3*(B*a^2*b^2*d^3*x^4 + 2*B*a^3*b*d^3*x^2 + B*a^4*d^3)*sqrt(b)*log(-2 
*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) - 2*(B*a^2*b^2*c^3 + 3*A*a^2*b^2 
*c^2*d + 6*B*a^3*b*c*d^2 + 2*A*a^3*b*d^3 - (2*A*b^4*c^3 + 3*B*a*b^3*c^2*d 
+ 3*A*a*b^3*c*d^2 - 4*B*a^2*b^2*d^3)*x^3 + 3*(3*B*a^2*b^2*c*d^2 + A*a^2*b^ 
2*d^3)*x^2 - 3*(A*a*b^3*c^3 - B*a^3*b*d^3)*x)*sqrt(b*x^2 + a))/(a^2*b^5*x^ 
4 + 2*a^3*b^4*x^2 + a^4*b^3), -1/3*(3*(B*a^2*b^2*d^3*x^4 + 2*B*a^3*b*d^3*x 
^2 + B*a^4*d^3)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) + (B*a^2*b^2*c 
^3 + 3*A*a^2*b^2*c^2*d + 6*B*a^3*b*c*d^2 + 2*A*a^3*b*d^3 - (2*A*b^4*c^3 + 
3*B*a*b^3*c^2*d + 3*A*a*b^3*c*d^2 - 4*B*a^2*b^2*d^3)*x^3 + 3*(3*B*a^2*b^2* 
c*d^2 + A*a^2*b^2*d^3)*x^2 - 3*(A*a*b^3*c^3 - B*a^3*b*d^3)*x)*sqrt(b*x^2 + 
 a))/(a^2*b^5*x^4 + 2*a^3*b^4*x^2 + a^4*b^3)]
 

Sympy [F]

\[ \int \frac {(A+B x) (c+d x)^3}{\left (a+b x^2\right )^{5/2}} \, dx=\int \frac {\left (A + B x\right ) \left (c + d x\right )^{3}}{\left (a + b x^{2}\right )^{\frac {5}{2}}}\, dx \] Input:

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

Output:

Integral((A + B*x)*(c + d*x)**3/(a + b*x**2)**(5/2), x)
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 268, normalized size of antiderivative = 1.64 \[ \int \frac {(A+B x) (c+d x)^3}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {1}{3} \, B d^{3} x {\left (\frac {3 \, x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {2 \, a}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}}\right )} + \frac {2 \, A c^{3} x}{3 \, \sqrt {b x^{2} + a} a^{2}} + \frac {A c^{3} x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a} - \frac {B d^{3} x}{3 \, \sqrt {b x^{2} + a} b^{2}} + \frac {B d^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {5}{2}}} - \frac {B c^{3}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b} - \frac {A c^{2} d}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} - \frac {{\left (3 \, B c d^{2} + A d^{3}\right )} x^{2}}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} - \frac {{\left (B c^{2} d + A c d^{2}\right )} x}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} b} + \frac {{\left (B c^{2} d + A c d^{2}\right )} x}{\sqrt {b x^{2} + a} a b} - \frac {2 \, {\left (3 \, B c d^{2} + A d^{3}\right )} a}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b^{2}} \] Input:

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

Output:

-1/3*B*d^3*x*(3*x^2/((b*x^2 + a)^(3/2)*b) + 2*a/((b*x^2 + a)^(3/2)*b^2)) + 
 2/3*A*c^3*x/(sqrt(b*x^2 + a)*a^2) + 1/3*A*c^3*x/((b*x^2 + a)^(3/2)*a) - 1 
/3*B*d^3*x/(sqrt(b*x^2 + a)*b^2) + B*d^3*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 
1/3*B*c^3/((b*x^2 + a)^(3/2)*b) - A*c^2*d/((b*x^2 + a)^(3/2)*b) - (3*B*c*d 
^2 + A*d^3)*x^2/((b*x^2 + a)^(3/2)*b) - (B*c^2*d + A*c*d^2)*x/((b*x^2 + a) 
^(3/2)*b) + (B*c^2*d + A*c*d^2)*x/(sqrt(b*x^2 + a)*a*b) - 2/3*(3*B*c*d^2 + 
 A*d^3)*a/((b*x^2 + a)^(3/2)*b^2)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 219, normalized size of antiderivative = 1.34 \[ \int \frac {(A+B x) (c+d x)^3}{\left (a+b x^2\right )^{5/2}} \, dx=-\frac {B d^{3} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{b^{\frac {5}{2}}} + \frac {{\left (x {\left (\frac {{\left (2 \, A b^{5} c^{3} + 3 \, B a b^{4} c^{2} d + 3 \, A a b^{4} c d^{2} - 4 \, B a^{2} b^{3} d^{3}\right )} x}{a^{2} b^{4}} - \frac {3 \, {\left (3 \, B a^{2} b^{3} c d^{2} + A a^{2} b^{3} d^{3}\right )}}{a^{2} b^{4}}\right )} + \frac {3 \, {\left (A a b^{4} c^{3} - B a^{3} b^{2} d^{3}\right )}}{a^{2} b^{4}}\right )} x - \frac {B a^{2} b^{3} c^{3} + 3 \, A a^{2} b^{3} c^{2} d + 6 \, B a^{3} b^{2} c d^{2} + 2 \, A a^{3} b^{2} d^{3}}{a^{2} b^{4}}}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}}} \] Input:

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

Output:

-B*d^3*log(abs(-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2) + 1/3*((x*((2*A*b^5* 
c^3 + 3*B*a*b^4*c^2*d + 3*A*a*b^4*c*d^2 - 4*B*a^2*b^3*d^3)*x/(a^2*b^4) - 3 
*(3*B*a^2*b^3*c*d^2 + A*a^2*b^3*d^3)/(a^2*b^4)) + 3*(A*a*b^4*c^3 - B*a^3*b 
^2*d^3)/(a^2*b^4))*x - (B*a^2*b^3*c^3 + 3*A*a^2*b^3*c^2*d + 6*B*a^3*b^2*c* 
d^2 + 2*A*a^3*b^2*d^3)/(a^2*b^4))/(b*x^2 + a)^(3/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 470, normalized size of antiderivative = 2.88 \[ \int \frac {(A+B x) (c+d x)^3}{\left (a+b x^2\right )^{5/2}} \, dx=\frac {-2 \sqrt {b \,x^{2}+a}\, a^{3} d^{3}-3 \sqrt {b \,x^{2}+a}\, a^{2} b \,c^{2} d -6 \sqrt {b \,x^{2}+a}\, a^{2} b c \,d^{2}-3 \sqrt {b \,x^{2}+a}\, a^{2} b \,d^{3} x^{2}-3 \sqrt {b \,x^{2}+a}\, a^{2} b \,d^{3} x +3 \sqrt {b \,x^{2}+a}\, a \,b^{2} c^{3} x -\sqrt {b \,x^{2}+a}\, a \,b^{2} c^{3}+3 \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,d^{2} x^{3}-9 \sqrt {b \,x^{2}+a}\, a \,b^{2} c \,d^{2} x^{2}-4 \sqrt {b \,x^{2}+a}\, a \,b^{2} d^{3} x^{3}+2 \sqrt {b \,x^{2}+a}\, b^{3} c^{3} x^{3}+3 \sqrt {b \,x^{2}+a}\, b^{3} c^{2} d \,x^{3}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{3} d^{3}+6 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} b \,d^{3} x^{2}+3 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a \,b^{2} d^{3} x^{4}+3 \sqrt {b}\, a^{3} c \,d^{2}-2 \sqrt {b}\, a^{2} b \,c^{3}+3 \sqrt {b}\, a^{2} b \,c^{2} d +6 \sqrt {b}\, a^{2} b c \,d^{2} x^{2}-4 \sqrt {b}\, a \,b^{2} c^{3} x^{2}+6 \sqrt {b}\, a \,b^{2} c^{2} d \,x^{2}+3 \sqrt {b}\, a \,b^{2} c \,d^{2} x^{4}-2 \sqrt {b}\, b^{3} c^{3} x^{4}+3 \sqrt {b}\, b^{3} c^{2} d \,x^{4}}{3 a \,b^{2} \left (b^{2} x^{4}+2 a b \,x^{2}+a^{2}\right )} \] Input:

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

Output:

( - 2*sqrt(a + b*x**2)*a**3*d**3 - 3*sqrt(a + b*x**2)*a**2*b*c**2*d - 6*sq 
rt(a + b*x**2)*a**2*b*c*d**2 - 3*sqrt(a + b*x**2)*a**2*b*d**3*x**2 - 3*sqr 
t(a + b*x**2)*a**2*b*d**3*x + 3*sqrt(a + b*x**2)*a*b**2*c**3*x - sqrt(a + 
b*x**2)*a*b**2*c**3 + 3*sqrt(a + b*x**2)*a*b**2*c*d**2*x**3 - 9*sqrt(a + b 
*x**2)*a*b**2*c*d**2*x**2 - 4*sqrt(a + b*x**2)*a*b**2*d**3*x**3 + 2*sqrt(a 
 + b*x**2)*b**3*c**3*x**3 + 3*sqrt(a + b*x**2)*b**3*c**2*d*x**3 + 3*sqrt(b 
)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**3*d**3 + 6*sqrt(b)*log((s 
qrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*b*d**3*x**2 + 3*sqrt(b)*log((sq 
rt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b**2*d**3*x**4 + 3*sqrt(b)*a**3*c*d 
**2 - 2*sqrt(b)*a**2*b*c**3 + 3*sqrt(b)*a**2*b*c**2*d + 6*sqrt(b)*a**2*b*c 
*d**2*x**2 - 4*sqrt(b)*a*b**2*c**3*x**2 + 6*sqrt(b)*a*b**2*c**2*d*x**2 + 3 
*sqrt(b)*a*b**2*c*d**2*x**4 - 2*sqrt(b)*b**3*c**3*x**4 + 3*sqrt(b)*b**3*c* 
*2*d*x**4)/(3*a*b**2*(a**2 + 2*a*b*x**2 + b**2*x**4))