\(\int \frac {x}{(c+d x)^2 (a+b x^2)^{5/2}} \, dx\) [1291]

Optimal result

Integrand size = 20, antiderivative size = 219 \[ \int \frac {x}{(c+d x)^2 \left (a+b x^2\right )^{5/2}} \, dx=-\frac {b c^2-a d^2-2 b c d x}{3 \left (b c^2+a d^2\right )^2 \left (a+b x^2\right )^{3/2}}-\frac {d \left (3 a d \left (3 b c^2-a d^2\right )+2 b c \left (b c^2-5 a d^2\right ) x\right )}{3 a \left (b c^2+a d^2\right )^3 \sqrt {a+b x^2}}+\frac {c d^4 \sqrt {a+b x^2}}{\left (b c^2+a d^2\right )^3 (c+d x)}+\frac {d^3 \left (4 b c^2-a d^2\right ) \text {arctanh}\left (\frac {a d-b c x}{\sqrt {b c^2+a d^2} \sqrt {a+b x^2}}\right )}{\left (b c^2+a d^2\right )^{7/2}} \] Output:

-1/3*(-2*b*c*d*x-a*d^2+b*c^2)/(a*d^2+b*c^2)^2/(b*x^2+a)^(3/2)-1/3*d*(3*a*d 
*(-a*d^2+3*b*c^2)+2*b*c*(-5*a*d^2+b*c^2)*x)/a/(a*d^2+b*c^2)^3/(b*x^2+a)^(1 
/2)+c*d^4*(b*x^2+a)^(1/2)/(a*d^2+b*c^2)^3/(d*x+c)+d^3*(-a*d^2+4*b*c^2)*arc 
tanh((-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)^(7/2)
 

Mathematica [A] (verified)

Time = 10.35 (sec) , antiderivative size = 212, normalized size of antiderivative = 0.97 \[ \int \frac {x}{(c+d x)^2 \left (a+b x^2\right )^{5/2}} \, dx=\frac {\frac {a (-c+d x)}{\left (a+b x^2\right )^{3/2}}+\frac {d \left (-2 b c^2 x+a d (-5 c+3 d x)\right )}{\left (b c^2+a d^2\right ) \sqrt {a+b x^2}}+\frac {d^2 \left (c \sqrt {b c^2+a d^2} \left (-2 b c^2+13 a d^2\right ) \sqrt {a+b x^2}-3 a d \left (-4 b c^2+a d^2\right ) (c+d x) \text {arctanh}\left (\frac {a d-b c x}{\sqrt {b c^2+a d^2} \sqrt {a+b x^2}}\right )\right )}{\left (b c^2+a d^2\right )^{5/2}}}{3 a \left (b c^2+a d^2\right ) (c+d x)} \] Input:

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

Output:

((a*(-c + d*x))/(a + b*x^2)^(3/2) + (d*(-2*b*c^2*x + a*d*(-5*c + 3*d*x)))/ 
((b*c^2 + a*d^2)*Sqrt[a + b*x^2]) + (d^2*(c*Sqrt[b*c^2 + a*d^2]*(-2*b*c^2 
+ 13*a*d^2)*Sqrt[a + b*x^2] - 3*a*d*(-4*b*c^2 + a*d^2)*(c + d*x)*ArcTanh[( 
a*d - b*c*x)/(Sqrt[b*c^2 + a*d^2]*Sqrt[a + b*x^2])]))/(b*c^2 + a*d^2)^(5/2 
))/(3*a*(b*c^2 + a*d^2)*(c + d*x))
 

Rubi [A] (verified)

Time = 0.70 (sec) , antiderivative size = 251, normalized size of antiderivative = 1.15, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {593, 25, 686, 25, 27, 679, 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.

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

\(\Big \downarrow \) 593

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 686

\(\displaystyle -\frac {d \left (\frac {x \left (2 b c^2-3 a d^2\right )+5 a c d}{a \sqrt {a+b x^2} (c+d x) \left (a d^2+b c^2\right )}-\frac {\int -\frac {b d \left (10 a c d+\left (2 b c^2-3 a d^2\right ) x\right )}{(c+d x)^2 \sqrt {b x^2+a}}dx}{a b \left (a d^2+b c^2\right )}\right )}{3 \left (a d^2+b c^2\right )}-\frac {c-d x}{3 \left (a+b x^2\right )^{3/2} (c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {d \left (\frac {\int \frac {b d \left (10 a c d+\left (2 b c^2-3 a d^2\right ) x\right )}{(c+d x)^2 \sqrt {b x^2+a}}dx}{a b \left (a d^2+b c^2\right )}+\frac {x \left (2 b c^2-3 a d^2\right )+5 a c d}{a \sqrt {a+b x^2} (c+d x) \left (a d^2+b c^2\right )}\right )}{3 \left (a d^2+b c^2\right )}-\frac {c-d x}{3 \left (a+b x^2\right )^{3/2} (c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 27

\(\displaystyle -\frac {d \left (\frac {d \int \frac {10 a c d+\left (2 b c^2-3 a d^2\right ) x}{(c+d x)^2 \sqrt {b x^2+a}}dx}{a \left (a d^2+b c^2\right )}+\frac {x \left (2 b c^2-3 a d^2\right )+5 a c d}{a \sqrt {a+b x^2} (c+d x) \left (a d^2+b c^2\right )}\right )}{3 \left (a d^2+b c^2\right )}-\frac {c-d x}{3 \left (a+b x^2\right )^{3/2} (c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 679

\(\displaystyle -\frac {d \left (\frac {d \left (\frac {3 a d \left (4 b c^2-a d^2\right ) \int \frac {1}{(c+d x) \sqrt {b x^2+a}}dx}{a d^2+b c^2}+\frac {c \sqrt {a+b x^2} \left (2 b c^2-13 a d^2\right )}{(c+d x) \left (a d^2+b c^2\right )}\right )}{a \left (a d^2+b c^2\right )}+\frac {x \left (2 b c^2-3 a d^2\right )+5 a c d}{a \sqrt {a+b x^2} (c+d x) \left (a d^2+b c^2\right )}\right )}{3 \left (a d^2+b c^2\right )}-\frac {c-d x}{3 \left (a+b x^2\right )^{3/2} (c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 488

\(\displaystyle -\frac {d \left (\frac {d \left (\frac {c \sqrt {a+b x^2} \left (2 b c^2-13 a d^2\right )}{(c+d x) \left (a d^2+b c^2\right )}-\frac {3 a d \left (4 b c^2-a d^2\right ) \int \frac {1}{b c^2+a d^2-\frac {(a d-b c x)^2}{b x^2+a}}d\frac {a d-b c x}{\sqrt {b x^2+a}}}{a d^2+b c^2}\right )}{a \left (a d^2+b c^2\right )}+\frac {x \left (2 b c^2-3 a d^2\right )+5 a c d}{a \sqrt {a+b x^2} (c+d x) \left (a d^2+b c^2\right )}\right )}{3 \left (a d^2+b c^2\right )}-\frac {c-d x}{3 \left (a+b x^2\right )^{3/2} (c+d x) \left (a d^2+b c^2\right )}\)

\(\Big \downarrow \) 219

\(\displaystyle -\frac {d \left (\frac {d \left (\frac {c \sqrt {a+b x^2} \left (2 b c^2-13 a d^2\right )}{(c+d x) \left (a d^2+b c^2\right )}-\frac {3 a d \left (4 b c^2-a d^2\right ) \text {arctanh}\left (\frac {a d-b c x}{\sqrt {a+b x^2} \sqrt {a d^2+b c^2}}\right )}{\left (a d^2+b c^2\right )^{3/2}}\right )}{a \left (a d^2+b c^2\right )}+\frac {x \left (2 b c^2-3 a d^2\right )+5 a c d}{a \sqrt {a+b x^2} (c+d x) \left (a d^2+b c^2\right )}\right )}{3 \left (a d^2+b c^2\right )}-\frac {c-d x}{3 \left (a+b x^2\right )^{3/2} (c+d x) \left (a d^2+b c^2\right )}\)

Input:

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

Output:

-1/3*(c - d*x)/((b*c^2 + a*d^2)*(c + d*x)*(a + b*x^2)^(3/2)) - (d*((5*a*c* 
d + (2*b*c^2 - 3*a*d^2)*x)/(a*(b*c^2 + a*d^2)*(c + d*x)*Sqrt[a + b*x^2]) + 
 (d*((c*(2*b*c^2 - 13*a*d^2)*Sqrt[a + b*x^2])/((b*c^2 + a*d^2)*(c + d*x)) 
- (3*a*d*(4*b*c^2 - a*d^2)*ArcTanh[(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)))/(a*(b*c^2 + a*d^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 593

Int[(x_)*((c_) + (d_.)*(x_))^(n_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 Simp[(c + d*x)^(n + 1)*(c - d*x)*((a + b*x^2)^(p + 1)/(2*(p + 1)*(b*c^2 + 
a*d^2))), x] - Simp[d/(2*(p + 1)*(b*c^2 + a*d^2))   Int[(c + d*x)^n*(a + b* 
x^2)^(p + 1)*(c*n - d*(n + 2*p + 4)*x), x], x] /; FreeQ[{a, b, c, d, n}, x] 
 && LtQ[p, -1] && NeQ[b*c^2 + a*d^2, 0]
 

rule 679

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[(-(e*f - d*g))*(d + e*x)^(m + 1)*((a + c*x^2)^(p + 1 
)/(2*(p + 1)*(c*d^2 + a*e^2))), x] + Simp[(c*d*f + a*e*g)/(c*d^2 + a*e^2) 
 Int[(d + e*x)^(m + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, 
 p}, x] && EqQ[Simplify[m + 2*p + 3], 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. \(1501\) vs. \(2(203)=406\).

Time = 0.42 (sec) , antiderivative size = 1502, normalized size of antiderivative = 6.86

method	result	size
default	\(\text {Expression too large to display}\)	\(1502\)

Input:

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

Output:

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

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 869 vs. \(2 (204) = 408\).

Time = 0.40 (sec) , antiderivative size = 1764, normalized size of antiderivative = 8.05 \[ \int \frac {x}{(c+d x)^2 \left (a+b x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

[-1/6*(3*(4*a^3*b*c^3*d^3 - a^4*c*d^5 + (4*a*b^3*c^2*d^4 - a^2*b^2*d^6)*x^ 
5 + (4*a*b^3*c^3*d^3 - a^2*b^2*c*d^5)*x^4 + 2*(4*a^2*b^2*c^2*d^4 - a^3*b*d 
^6)*x^3 + 2*(4*a^2*b^2*c^3*d^3 - a^3*b*c*d^5)*x^2 + (4*a^3*b*c^2*d^4 - a^4 
*d^6)*x)*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*(a*b^3*c^7 + 10*a^2*b^2*c^5*d^2 + 2*a^3* 
b*c^3*d^4 - 7*a^4*c*d^6 + (2*b^4*c^5*d^2 - 11*a*b^3*c^3*d^4 - 13*a^2*b^2*c 
*d^6)*x^4 + (2*b^4*c^6*d + a*b^3*c^4*d^3 - 4*a^2*b^2*c^2*d^5 - 3*a^3*b*d^7 
)*x^3 + 3*(3*a*b^3*c^5*d^2 - 4*a^2*b^2*c^3*d^4 - 7*a^3*b*c*d^6)*x^2 + (a*b 
^3*c^6*d - 2*a^2*b^2*c^4*d^3 - 7*a^3*b*c^2*d^5 - 4*a^4*d^7)*x)*sqrt(b*x^2 
+ a))/(a^3*b^4*c^9 + 4*a^4*b^3*c^7*d^2 + 6*a^5*b^2*c^5*d^4 + 4*a^6*b*c^3*d 
^6 + a^7*c*d^8 + (a*b^6*c^8*d + 4*a^2*b^5*c^6*d^3 + 6*a^3*b^4*c^4*d^5 + 4* 
a^4*b^3*c^2*d^7 + a^5*b^2*d^9)*x^5 + (a*b^6*c^9 + 4*a^2*b^5*c^7*d^2 + 6*a^ 
3*b^4*c^5*d^4 + 4*a^4*b^3*c^3*d^6 + a^5*b^2*c*d^8)*x^4 + 2*(a^2*b^5*c^8*d 
+ 4*a^3*b^4*c^6*d^3 + 6*a^4*b^3*c^4*d^5 + 4*a^5*b^2*c^2*d^7 + a^6*b*d^9)*x 
^3 + 2*(a^2*b^5*c^9 + 4*a^3*b^4*c^7*d^2 + 6*a^4*b^3*c^5*d^4 + 4*a^5*b^2*c^ 
3*d^6 + a^6*b*c*d^8)*x^2 + (a^3*b^4*c^8*d + 4*a^4*b^3*c^6*d^3 + 6*a^5*b^2* 
c^4*d^5 + 4*a^6*b*c^2*d^7 + a^7*d^9)*x), 1/3*(3*(4*a^3*b*c^3*d^3 - a^4*c*d 
^5 + (4*a*b^3*c^2*d^4 - a^2*b^2*d^6)*x^5 + (4*a*b^3*c^3*d^3 - a^2*b^2*c*d^ 
5)*x^4 + 2*(4*a^2*b^2*c^2*d^4 - a^3*b*d^6)*x^3 + 2*(4*a^2*b^2*c^3*d^3 -...
 

Sympy [F]

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

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

Output:

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 775 vs. \(2 (204) = 408\).

Time = 0.11 (sec) , antiderivative size = 775, normalized size of antiderivative = 3.54 \[ \int \frac {x}{(c+d x)^2 \left (a+b x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

-5*b^2*c^3*x/(sqrt(b*x^2 + a)*a*b^3*c^6/d + 3*sqrt(b*x^2 + a)*a^2*b^2*c^4* 
d + 3*sqrt(b*x^2 + a)*a^3*b*c^2*d^3 + sqrt(b*x^2 + a)*a^4*d^5) - 5/3*b^2*c 
^3*x/((b*x^2 + a)^(3/2)*a*b^2*c^4*d + 2*(b*x^2 + a)^(3/2)*a^2*b*c^2*d^3 + 
(b*x^2 + a)^(3/2)*a^3*d^5) - 10/3*b^2*c^3*x/(sqrt(b*x^2 + a)*a^2*b^2*c^4*d 
 + 2*sqrt(b*x^2 + a)*a^3*b*c^2*d^3 + sqrt(b*x^2 + a)*a^4*d^5) - 5/3*b*c^2/ 
((b*x^2 + a)^(3/2)*b^2*c^4 + 2*(b*x^2 + a)^(3/2)*a*b*c^2*d^2 + (b*x^2 + a) 
^(3/2)*a^2*d^4) - 5*b*c^2/(3*sqrt(b*x^2 + a)*a*b^2*c^4 + sqrt(b*x^2 + a)*b 
^3*c^6/d^2 + 3*sqrt(b*x^2 + a)*a^2*b*c^2*d^2 + sqrt(b*x^2 + a)*a^3*d^4) + 
b*c*x/(sqrt(b*x^2 + a)*a*b^2*c^4/d + 2*sqrt(b*x^2 + a)*a^2*b*c^2*d + sqrt( 
b*x^2 + a)*a^3*d^3) + 5/3*b*c*x/((b*x^2 + a)^(3/2)*a*b*c^2*d + (b*x^2 + a) 
^(3/2)*a^2*d^3) + 10/3*b*c*x/(sqrt(b*x^2 + a)*a^2*b*c^2*d + sqrt(b*x^2 + a 
)*a^3*d^3) + c/((b*x^2 + a)^(3/2)*b*c^2*d*x + (b*x^2 + a)^(3/2)*a*d^3*x + 
(b*x^2 + a)^(3/2)*b*c^3 + (b*x^2 + a)^(3/2)*a*c*d^2) + 1/3/((b*x^2 + a)^(3 
/2)*b*c^2 + (b*x^2 + a)^(3/2)*a*d^2) + 1/(2*sqrt(b*x^2 + a)*a*b*c^2 + sqrt 
(b*x^2 + a)*b^2*c^4/d^2 + sqrt(b*x^2 + a)*a^2*d^2) - 5*b*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) 
^(7/2)*d^4) + 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^2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1409 vs. \(2 (204) = 408\).

Time = 0.28 (sec) , antiderivative size = 1409, normalized size of antiderivative = 6.43 \[ \int \frac {x}{(c+d x)^2 \left (a+b x^2\right )^{5/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/3*((12*a*b^(3/2)*c^2*d^7*log(abs(-b*c*d + sqrt(b*c^2 + a*d^2)*sqrt(b)*a 
bs(d))) - 3*a^2*sqrt(b)*d^9*log(abs(-b*c*d + sqrt(b*c^2 + a*d^2)*sqrt(b)*a 
bs(d))) - 2*sqrt(b*c^2 + a*d^2)*b^2*c^3*d^4*abs(d) + 13*sqrt(b*c^2 + a*d^2 
)*a*b*c*d^6*abs(d))*sgn(1/(d*x + c))*sgn(d)/(sqrt(b*c^2 + a*d^2)*a*b^(7/2) 
*c^6*abs(d) + 3*sqrt(b*c^2 + a*d^2)*a^2*b^(5/2)*c^4*d^2*abs(d) + 3*sqrt(b* 
c^2 + a*d^2)*a^3*b^(3/2)*c^2*d^4*abs(d) + sqrt(b*c^2 + a*d^2)*a^4*sqrt(b)* 
d^6*abs(d)) + ((2*a*b^4*c^3*d^15*sgn(1/(d*x + c))*sgn(d) - 13*a^2*b^3*c*d^ 
17*sgn(1/(d*x + c))*sgn(d))/(a^2*b^4*c^6*d^11*sgn(1/(d*x + c))^2*sgn(d)^2 
+ 3*a^3*b^3*c^4*d^13*sgn(1/(d*x + c))^2*sgn(d)^2 + 3*a^4*b^2*c^2*d^15*sgn( 
1/(d*x + c))^2*sgn(d)^2 + a^5*b*d^17*sgn(1/(d*x + c))^2*sgn(d)^2) - (3*(2* 
a*b^4*c^4*d^16*sgn(1/(d*x + c))*sgn(d) - 17*a^2*b^3*c^2*d^18*sgn(1/(d*x + 
c))*sgn(d) + a^3*b^2*d^20*sgn(1/(d*x + c))*sgn(d))/(a^2*b^4*c^6*d^11*sgn(1 
/(d*x + c))^2*sgn(d)^2 + 3*a^3*b^3*c^4*d^13*sgn(1/(d*x + c))^2*sgn(d)^2 + 
3*a^4*b^2*c^2*d^15*sgn(1/(d*x + c))^2*sgn(d)^2 + a^5*b*d^17*sgn(1/(d*x + c 
))^2*sgn(d)^2) - (6*(a*b^4*c^5*d^17*sgn(1/(d*x + c))*sgn(d) - 11*a^2*b^3*c 
^3*d^19*sgn(1/(d*x + c))*sgn(d) - 2*a^3*b^2*c*d^21*sgn(1/(d*x + c))*sgn(d) 
)/(a^2*b^4*c^6*d^11*sgn(1/(d*x + c))^2*sgn(d)^2 + 3*a^3*b^3*c^4*d^13*sgn(1 
/(d*x + c))^2*sgn(d)^2 + 3*a^4*b^2*c^2*d^15*sgn(1/(d*x + c))^2*sgn(d)^2 + 
a^5*b*d^17*sgn(1/(d*x + c))^2*sgn(d)^2) - (2*(a*b^4*c^6*d^18*sgn(1/(d*x + 
c))*sgn(d) - 16*a^2*b^3*c^4*d^20*sgn(1/(d*x + c))*sgn(d) - 15*a^3*b^2*c...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.26 (sec) , antiderivative size = 1863, normalized size of antiderivative = 8.51 \[ \int \frac {x}{(c+d x)^2 \left (a+b x^2\right )^{5/2}} \, dx =\text {Too large to display} \] Input:

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

Output:

(3*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d 
+ b*c*x)*a**4*c*d**5 + 3*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)*sqrt(a 
*d**2 + b*c**2) - a*d + b*c*x)*a**4*d**6*x - 12*sqrt(a*d**2 + b*c**2)*log( 
sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a**3*b*c**3*d**3 - 1 
2*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + 
 b*c*x)*a**3*b*c**2*d**4*x + 6*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)* 
sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a**3*b*c*d**5*x**2 + 6*sqrt(a*d**2 + 
b*c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a**3*b*d 
**6*x**3 - 24*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 + b*c 
**2) - a*d + b*c*x)*a**2*b**2*c**3*d**3*x**2 - 24*sqrt(a*d**2 + b*c**2)*lo 
g(sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a**2*b**2*c**2*d** 
4*x**3 + 3*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2 
) - a*d + b*c*x)*a**2*b**2*c*d**5*x**4 + 3*sqrt(a*d**2 + b*c**2)*log(sqrt( 
a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a**2*b**2*d**6*x**5 - 12* 
sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2)*sqrt(a*d**2 + b*c**2) - a*d + b 
*c*x)*a*b**3*c**3*d**3*x**4 - 12*sqrt(a*d**2 + b*c**2)*log(sqrt(a + b*x**2 
)*sqrt(a*d**2 + b*c**2) - a*d + b*c*x)*a*b**3*c**2*d**4*x**5 - 3*sqrt(a*d* 
*2 + b*c**2)*log(c + d*x)*a**4*c*d**5 - 3*sqrt(a*d**2 + b*c**2)*log(c + d* 
x)*a**4*d**6*x + 12*sqrt(a*d**2 + b*c**2)*log(c + d*x)*a**3*b*c**3*d**3 + 
12*sqrt(a*d**2 + b*c**2)*log(c + d*x)*a**3*b*c**2*d**4*x - 6*sqrt(a*d**...