\(\int \frac {x^2}{(c+d x) (a+b x^2)^{5/2}} \, dx\) [1280]

Optimal result

Integrand size = 22, antiderivative size = 152 \[ \int \frac {x^2}{(c+d x) \left (a+b x^2\right )^{5/2}} \, dx=-\frac {a d+b c x}{3 b \left (b c^2+a d^2\right ) \left (a+b x^2\right )^{3/2}}+\frac {c \left (3 a c d+\left (b c^2-2 a d^2\right ) x\right )}{3 a \left (b c^2+a d^2\right )^2 \sqrt {a+b x^2}}-\frac {c^2 d^2 \text {arctanh}\left (\frac {a d-b c x}{\sqrt {b c^2+a d^2} \sqrt {a+b x^2}}\right )}{\left (b c^2+a d^2\right )^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.81 (sec) , antiderivative size = 157, normalized size of antiderivative = 1.03 \[ \int \frac {x^2}{(c+d x) \left (a+b x^2\right )^{5/2}} \, dx=\frac {-a^3 d^3+b^3 c^3 x^3+a^2 b c d (2 c-3 d x)+a b^2 c d x^2 (3 c-2 d x)}{3 a b \left (b c^2+a d^2\right )^2 \left (a+b x^2\right )^{3/2}}-\frac {2 c^2 d^2 \arctan \left (\frac {\sqrt {b} (c+d x)-d \sqrt {a+b x^2}}{\sqrt {-b c^2-a d^2}}\right )}{\left (-b c^2-a d^2\right )^{5/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.54 (sec) , antiderivative size = 165, normalized size of antiderivative = 1.09, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {601, 25, 27, 686, 27, 488, 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.

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

\(\Big \downarrow \) 601

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 686

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 488

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

-1/3*(a*d + b*c*x)/(b*(b*c^2 + a*d^2)*(a + b*x^2)^(3/2)) + (c*((3*a*c*d + 
(b*c^2 - 2*a*d^2)*x)/(a*(b*c^2 + a*d^2)*Sqrt[a + b*x^2]) - (3*c*d^2*ArcTan 
h[(a*d - b*c*x)/(Sqrt[b*c^2 + a*d^2]*Sqrt[a + b*x^2])])/(b*c^2 + a*d^2)^(3 
/2)))/(3*(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 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 488

Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ 
Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ 
[{a, b, c, d}, 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 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])
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(658\) vs. \(2(138)=276\).

Time = 0.36 (sec) , antiderivative size = 659, normalized size of antiderivative = 4.34

method	result	size
default	\(-\frac {1}{3 d b \left (b \,x^{2}+a \right )^{\frac {3}{2}}}+\frac {c^{2} \left (\frac {d^{2}}{3 \left (a \,d^{2}+b \,c^{2}\right ) \left (b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}\right )^{\frac {3}{2}}}+\frac {b c d \left (\frac {\frac {4 b \left (x +\frac {c}{d}\right )}{3}-\frac {4 b c}{3 d}}{\left (\frac {4 b \left (a \,d^{2}+b \,c^{2}\right )}{d^{2}}-\frac {4 b^{2} c^{2}}{d^{2}}\right ) \left (b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}\right )^{\frac {3}{2}}}+\frac {16 b \left (2 b \left (x +\frac {c}{d}\right )-\frac {2 b c}{d}\right )}{3 {\left (\frac {4 b \left (a \,d^{2}+b \,c^{2}\right )}{d^{2}}-\frac {4 b^{2} c^{2}}{d^{2}}\right )}^{2} \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}\right )}{a \,d^{2}+b \,c^{2}}+\frac {d^{2} \left (\frac {d^{2}}{\left (a \,d^{2}+b \,c^{2}\right ) \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}+\frac {2 b c d \left (2 b \left (x +\frac {c}{d}\right )-\frac {2 b c}{d}\right )}{\left (a \,d^{2}+b \,c^{2}\right ) \left (\frac {4 b \left (a \,d^{2}+b \,c^{2}\right )}{d^{2}}-\frac {4 b^{2} c^{2}}{d^{2}}\right ) \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}-\frac {d^{2} \ln \left (\frac {\frac {2 a \,d^{2}+2 b \,c^{2}}{d^{2}}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+2 \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}\, \sqrt {b \left (x +\frac {c}{d}\right )^{2}-\frac {2 b c \left (x +\frac {c}{d}\right )}{d}+\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}{x +\frac {c}{d}}\right )}{\left (a \,d^{2}+b \,c^{2}\right ) \sqrt {\frac {a \,d^{2}+b \,c^{2}}{d^{2}}}}\right )}{a \,d^{2}+b \,c^{2}}\right )}{d^{3}}-\frac {c \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}}\)	\(659\)

Input:

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

Output:

-1/3/d/b/(b*x^2+a)^(3/2)+c^2/d^3*(1/3/(a*d^2+b*c^2)*d^2/(b*(x+c/d)^2-2*b*c 
/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(3/2)+b*c*d/(a*d^2+b*c^2)*(2/3*(2*b*(x+c/d)- 
2*b*c/d)/(4*b*(a*d^2+b*c^2)/d^2-4*b^2*c^2/d^2)/(b*(x+c/d)^2-2*b*c/d*(x+c/d 
)+(a*d^2+b*c^2)/d^2)^(3/2)+16/3*b/(4*b*(a*d^2+b*c^2)/d^2-4*b^2*c^2/d^2)^2* 
(2*b*(x+c/d)-2*b*c/d)/(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2 
))+1/(a*d^2+b*c^2)*d^2*(1/(a*d^2+b*c^2)*d^2/(b*(x+c/d)^2-2*b*c/d*(x+c/d)+( 
a*d^2+b*c^2)/d^2)^(1/2)+2*b*c*d/(a*d^2+b*c^2)*(2*b*(x+c/d)-2*b*c/d)/(4*b*( 
a*d^2+b*c^2)/d^2-4*b^2*c^2/d^2)/(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2) 
/d^2)^(1/2)-1/(a*d^2+b*c^2)*d^2/((a*d^2+b*c^2)/d^2)^(1/2)*ln((2*(a*d^2+b*c 
^2)/d^2-2*b*c/d*(x+c/d)+2*((a*d^2+b*c^2)/d^2)^(1/2)*(b*(x+c/d)^2-2*b*c/d*( 
x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2))/(x+c/d))))-c/d^2*(1/3*x/a/(b*x^2+a)^(3/2) 
+2/3/a^2/(b*x^2+a)^(1/2)*x)
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 419 vs. \(2 (139) = 278\).

Time = 0.17 (sec) , antiderivative size = 864, normalized size of antiderivative = 5.68 \[ \int \frac {x^2}{(c+d x) \left (a+b x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 360 vs. \(2 (139) = 278\).

Time = 0.09 (sec) , antiderivative size = 360, normalized size of antiderivative = 2.37 \[ \int \frac {x^2}{(c+d x) \left (a+b x^2\right )^{5/2}} \, dx=\frac {b c^{3} x}{\sqrt {b x^{2} + a} a b^{2} c^{4} + 2 \, \sqrt {b x^{2} + a} a^{2} b c^{2} d^{2} + \sqrt {b x^{2} + a} a^{3} d^{4}} + \frac {b c^{3} x}{3 \, {\left ({\left (b x^{2} + a\right )}^{\frac {3}{2}} a b c^{2} d^{2} + {\left (b x^{2} + a\right )}^{\frac {3}{2}} a^{2} d^{4}\right )}} + \frac {2 \, b c^{3} x}{3 \, {\left (\sqrt {b x^{2} + a} a^{2} b c^{2} d^{2} + \sqrt {b x^{2} + a} a^{3} d^{4}\right )}} + \frac {c^{2}}{\frac {\sqrt {b x^{2} + a} b^{2} c^{4}}{d} + 2 \, \sqrt {b x^{2} + a} a b c^{2} d + \sqrt {b x^{2} + a} a^{2} d^{3}} + \frac {c^{2}}{3 \, {\left ({\left (b x^{2} + a\right )}^{\frac {3}{2}} b c^{2} d + {\left (b x^{2} + a\right )}^{\frac {3}{2}} a d^{3}\right )}} - \frac {2 \, c x}{3 \, \sqrt {b x^{2} + a} a^{2} d^{2}} - \frac {c x}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} a d^{2}} + \frac {c^{2} \operatorname {arsinh}\left (\frac {b c x}{\sqrt {a b} {\left | d x + c \right |}} - \frac {a d}{\sqrt {a b} {\left | d x + c \right |}}\right )}{{\left (a + \frac {b c^{2}}{d^{2}}\right )}^{\frac {5}{2}} d^{3}} - \frac {1}{3 \, {\left (b x^{2} + a\right )}^{\frac {3}{2}} b d} \] Input:

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

Output:

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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 977 vs. \(2 (139) = 278\).

Time = 0.15 (sec) , antiderivative size = 977, normalized size of antiderivative = 6.43 \[ \int \frac {x^2}{(c+d x) \left (a+b x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

-2*c^2*d^2*arctan(((sqrt(b)*x - sqrt(b*x^2 + a))*d + sqrt(b)*c)/sqrt(-b*c^ 
2 - a*d^2))/((b^2*c^4 + 2*a*b*c^2*d^2 + a^2*d^4)*sqrt(-b*c^2 - a*d^2)) + 1 
/3*((((b^9*c^15 + 4*a*b^8*c^13*d^2 + 3*a^2*b^7*c^11*d^4 - 10*a^3*b^6*c^9*d 
^6 - 25*a^4*b^5*c^7*d^8 - 24*a^5*b^4*c^5*d^10 - 11*a^6*b^3*c^3*d^12 - 2*a^ 
7*b^2*c*d^14)*x/(a*b^9*c^16 + 8*a^2*b^8*c^14*d^2 + 28*a^3*b^7*c^12*d^4 + 5 
6*a^4*b^6*c^10*d^6 + 70*a^5*b^5*c^8*d^8 + 56*a^6*b^4*c^6*d^10 + 28*a^7*b^3 
*c^4*d^12 + 8*a^8*b^2*c^2*d^14 + a^9*b*d^16) + 3*(a*b^8*c^14*d + 6*a^2*b^7 
*c^12*d^3 + 15*a^3*b^6*c^10*d^5 + 20*a^4*b^5*c^8*d^7 + 15*a^5*b^4*c^6*d^9 
+ 6*a^6*b^3*c^4*d^11 + a^7*b^2*c^2*d^13)/(a*b^9*c^16 + 8*a^2*b^8*c^14*d^2 
+ 28*a^3*b^7*c^12*d^4 + 56*a^4*b^6*c^10*d^6 + 70*a^5*b^5*c^8*d^8 + 56*a^6* 
b^4*c^6*d^10 + 28*a^7*b^3*c^4*d^12 + 8*a^8*b^2*c^2*d^14 + a^9*b*d^16))*x - 
 3*(a^2*b^7*c^13*d^2 + 6*a^3*b^6*c^11*d^4 + 15*a^4*b^5*c^9*d^6 + 20*a^5*b^ 
4*c^7*d^8 + 15*a^6*b^3*c^5*d^10 + 6*a^7*b^2*c^3*d^12 + a^8*b*c*d^14)/(a*b^ 
9*c^16 + 8*a^2*b^8*c^14*d^2 + 28*a^3*b^7*c^12*d^4 + 56*a^4*b^6*c^10*d^6 + 
70*a^5*b^5*c^8*d^8 + 56*a^6*b^4*c^6*d^10 + 28*a^7*b^3*c^4*d^12 + 8*a^8*b^2 
*c^2*d^14 + a^9*b*d^16))*x + (2*a^2*b^7*c^14*d + 11*a^3*b^6*c^12*d^3 + 24* 
a^4*b^5*c^10*d^5 + 25*a^5*b^4*c^8*d^7 + 10*a^6*b^3*c^6*d^9 - 3*a^7*b^2*c^4 
*d^11 - 4*a^8*b*c^2*d^13 - a^9*d^15)/(a*b^9*c^16 + 8*a^2*b^8*c^14*d^2 + 28 
*a^3*b^7*c^12*d^4 + 56*a^4*b^6*c^10*d^6 + 70*a^5*b^5*c^8*d^8 + 56*a^6*b^4* 
c^6*d^10 + 28*a^7*b^3*c^4*d^12 + 8*a^8*b^2*c^2*d^14 + a^9*b*d^16))/(b*x...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.51 (sec) , antiderivative size = 3786, normalized size of antiderivative = 24.91 \[ \int \frac {x^2}{(c+d x) \left (a+b x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 6*sqrt(b)*sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2)* 
sqrt(a*d**2 + b*c**2)*atan((sqrt(a + b*x**2)*d + sqrt(b)*d*x)/sqrt(2*sqrt( 
b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2))*a**2*b*c**3 - 12*sqrt(b)* 
sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2)*sqrt(a*d**2 + 
b*c**2)*atan((sqrt(a + b*x**2)*d + sqrt(b)*d*x)/sqrt(2*sqrt(b)*sqrt(a*d**2 
 + b*c**2)*c - a*d**2 - 2*b*c**2))*a*b**2*c**3*x**2 - 6*sqrt(b)*sqrt(2*sqr 
t(b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2)*sqrt(a*d**2 + b*c**2)*at 
an((sqrt(a + b*x**2)*d + sqrt(b)*d*x)/sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2) 
*c - a*d**2 - 2*b*c**2))*b**3*c**3*x**4 - 6*sqrt(2*sqrt(b)*sqrt(a*d**2 + b 
*c**2)*c - a*d**2 - 2*b*c**2)*atan((sqrt(a + b*x**2)*d + sqrt(b)*d*x)/sqrt 
(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2))*a**3*b*c**2*d**2 
- 6*sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2)*atan((sqrt 
(a + b*x**2)*d + sqrt(b)*d*x)/sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c - a*d 
**2 - 2*b*c**2))*a**2*b**2*c**4 - 12*sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)* 
c - a*d**2 - 2*b*c**2)*atan((sqrt(a + b*x**2)*d + sqrt(b)*d*x)/sqrt(2*sqrt 
(b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2))*a**2*b**2*c**2*d**2*x**2 
 - 12*sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c - a*d**2 - 2*b*c**2)*atan((sq 
rt(a + b*x**2)*d + sqrt(b)*d*x)/sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c**2)*c - a 
*d**2 - 2*b*c**2))*a*b**3*c**4*x**2 - 6*sqrt(2*sqrt(b)*sqrt(a*d**2 + b*c** 
2)*c - a*d**2 - 2*b*c**2)*atan((sqrt(a + b*x**2)*d + sqrt(b)*d*x)/sqrt(...