\(\int \frac {x^4}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx\) [413]

Optimal result

Integrand size = 22, antiderivative size = 252 \[ \int \frac {x^4}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=-\frac {2 a^4}{3 b^4 (b c-a d) (a+b x)^{3/2} (c+d x)^{3/2}}+\frac {2 \left (b^4 c^4+a^4 d^4\right ) \sqrt {a+b x}}{3 b^4 d^2 (b c-a d)^3 (c+d x)^{3/2}}+\frac {4 a^3 (2 b c-a d)}{b^3 (b c-a d)^3 \sqrt {a+b x} \sqrt {c+d x}}-\frac {8 \left (b^4 c^4-3 a b^3 c^3 d-3 a^3 b c d^3+a^4 d^4\right ) \sqrt {a+b x}}{3 b^3 d^2 (b c-a d)^4 \sqrt {c+d x}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{5/2} d^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.27 (sec) , antiderivative size = 200, normalized size of antiderivative = 0.79 \[ \int \frac {x^4}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=-\frac {2 (a+b x)^{3/2} \left (b^2 c^4 d+\frac {3 b^3 c^4 (c+d x)}{a+b x}-\frac {12 a b^2 c^3 d (c+d x)}{a+b x}-\frac {12 a^3 b c d^2 (c+d x)^2}{(a+b x)^2}+\frac {3 a^4 d^3 (c+d x)^2}{(a+b x)^2}+\frac {a^4 b d^2 (c+d x)^3}{(a+b x)^3}\right )}{3 b^2 d^2 (b c-a d)^4 (c+d x)^{3/2}}+\frac {2 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{b^{5/2} d^{5/2}} \] Input:

Integrate[x^4/((a + b*x)^(5/2)*(c + d*x)^(5/2)),x]
 

Output:

(-2*(a + b*x)^(3/2)*(b^2*c^4*d + (3*b^3*c^4*(c + d*x))/(a + b*x) - (12*a*b 
^2*c^3*d*(c + d*x))/(a + b*x) - (12*a^3*b*c*d^2*(c + d*x)^2)/(a + b*x)^2 + 
 (3*a^4*d^3*(c + d*x)^2)/(a + b*x)^2 + (a^4*b*d^2*(c + d*x)^3)/(a + b*x)^3 
))/(3*b^2*d^2*(b*c - a*d)^4*(c + d*x)^(3/2)) + (2*ArcTanh[(Sqrt[b]*Sqrt[c 
+ d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(b^(5/2)*d^(5/2))
 

Rubi [A] (verified)

Time = 0.38 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.11, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.318, Rules used = {109, 27, 167, 27, 162, 66, 221}

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.

\(\displaystyle \int \frac {x^4}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx\)

\(\Big \downarrow \) 109

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 167

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 162

\(\displaystyle \frac {2 a x^3}{3 b (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}-\frac {\frac {\frac {2 c \sqrt {a+b x} \left (c (a d+b c) \left (3 a^2 d^2-14 a b c d+3 b^2 c^2\right )+2 d x \left (3 a^3 d^3-12 a^2 b c d^2-a b^2 c^2 d+2 b^3 c^3\right )\right )}{3 d^2 (c+d x)^{3/2} (b c-a d)^2}-\frac {(b c-a d)^2 \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{d^2}}{b (b c-a d)}-\frac {2 a x^2 (3 b c-a d)}{b \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}}{b (b c-a d)}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {2 a x^3}{3 b (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}-\frac {\frac {\frac {2 c \sqrt {a+b x} \left (c (a d+b c) \left (3 a^2 d^2-14 a b c d+3 b^2 c^2\right )+2 d x \left (3 a^3 d^3-12 a^2 b c d^2-a b^2 c^2 d+2 b^3 c^3\right )\right )}{3 d^2 (c+d x)^{3/2} (b c-a d)^2}-\frac {2 (b c-a d)^2 \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{d^2}}{b (b c-a d)}-\frac {2 a x^2 (3 b c-a d)}{b \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}}{b (b c-a d)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 a x^3}{3 b (a+b x)^{3/2} (c+d x)^{3/2} (b c-a d)}-\frac {\frac {\frac {2 c \sqrt {a+b x} \left (c (a d+b c) \left (3 a^2 d^2-14 a b c d+3 b^2 c^2\right )+2 d x \left (3 a^3 d^3-12 a^2 b c d^2-a b^2 c^2 d+2 b^3 c^3\right )\right )}{3 d^2 (c+d x)^{3/2} (b c-a d)^2}-\frac {2 (b c-a d)^2 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{5/2}}}{b (b c-a d)}-\frac {2 a x^2 (3 b c-a d)}{b \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}}{b (b c-a d)}\)

Input:

Int[x^4/((a + b*x)^(5/2)*(c + d*x)^(5/2)),x]
 

Output:

(2*a*x^3)/(3*b*(b*c - a*d)*(a + b*x)^(3/2)*(c + d*x)^(3/2)) - ((-2*a*(3*b* 
c - a*d)*x^2)/(b*(b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2)) + ((2*c*Sqrt[a 
 + b*x]*(c*(b*c + a*d)*(3*b^2*c^2 - 14*a*b*c*d + 3*a^2*d^2) + 2*d*(2*b^3*c 
^3 - a*b^2*c^2*d - 12*a^2*b*c*d^2 + 3*a^3*d^3)*x))/(3*d^2*(b*c - a*d)^2*(c 
 + d*x)^(3/2)) - (2*(b*c - a*d)^2*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b] 
*Sqrt[c + d*x])])/(Sqrt[b]*d^(5/2)))/(b*(b*c - a*d)))/(b*(b*c - a*d))
 

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 66

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

rule 109

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_), x_] :> Simp[(b*c - a*d)*(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*((e + f 
*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] + Simp[1/(b*(b*e - a*f)*(m + 1)) 
 Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 2)*(e + f*x)^p*Simp[a*d*(d*e*(n - 1) 
+ c*f*(p + 1)) + b*c*(d*e*(m - n + 2) - c*f*(m + p + 2)) + d*(a*d*f*(n + p) 
 + b*(d*e*(m + 1) - c*f*(m + n + p + 1)))*x, x], x], x] /; FreeQ[{a, b, c, 
d, e, f, p}, x] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ[2*m, 2*n, 2*p] || 
IntegersQ[m, n + p] || IntegersQ[p, m + n])
 

rule 162

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_) 
)*((g_.) + (h_.)*(x_)), x_] :> Simp[((b^3*c*e*g*(m + 2) - a^3*d*f*h*(n + 2) 
 - a^2*b*(c*f*h*m - d*(f*g + e*h)*(m + n + 3)) - a*b^2*(c*(f*g + e*h) + d*e 
*g*(2*m + n + 4)) + b*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d*(f*g + 
e*h)*(n + 1)) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2)))*x)/(b^2*(b 
*c - a*d)^2*(m + 1)*(m + 2)))*(a + b*x)^(m + 1)*(c + d*x)^(n + 1), x] + Sim 
p[(f*(h/b^2) - (d*(m + n + 3)*(a^2*d*f*h*(m - n) - a*b*(2*c*f*h*(m + 1) - d 
*(f*g + e*h)*(n + 1)) + b^2*(c*(f*g + e*h)*(m + 1) - d*e*g*(m + n + 2))))/( 
b^2*(b*c - a*d)^2*(m + 1)*(m + 2)))   Int[(a + b*x)^(m + 2)*(c + d*x)^n, x] 
, x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && (LtQ[m, -2] || (EqQ[m + 
 n + 3, 0] &&  !LtQ[n, -2]))
 

rule 167

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_)*((e_.) + (f_.)*(x_) 
)^(p_)*((g_.) + (h_.)*(x_)), x_] :> Simp[(b*g - a*h)*(a + b*x)^(m + 1)*(c + 
 d*x)^n*((e + f*x)^(p + 1)/(b*(b*e - a*f)*(m + 1))), x] - Simp[1/(b*(b*e - 
a*f)*(m + 1))   Int[(a + b*x)^(m + 1)*(c + d*x)^(n - 1)*(e + f*x)^p*Simp[b* 
c*(f*g - e*h)*(m + 1) + (b*g - a*h)*(d*e*n + c*f*(p + 1)) + d*(b*(f*g - e*h 
)*(m + 1) + f*(b*g - a*h)*(n + p + 1))*x, x], x], x] /; FreeQ[{a, b, c, d, 
e, f, g, h, p}, x] && LtQ[m, -1] && GtQ[n, 0] && IntegersQ[2*m, 2*n, 2*p]
 

rule 221

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2088\) vs. \(2(218)=436\).

Time = 0.30 (sec) , antiderivative size = 2089, normalized size of antiderivative = 8.29

method	result	size
default	\(\text {Expression too large to display}\)	\(2089\)

Input:

int(x^4/(b*x+a)^(5/2)/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

1/3*(12*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d* 
b)^(1/2))*a^3*b^3*c^2*d^4*x^3+12*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2) 
*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^2*b^4*c^3*d^3*x^3-8*(d*b)^(1/2)*((b*x 
+a)*(d*x+c))^(1/2)*a^4*b*d^5*x^3-8*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^5 
*c^4*d*x^3-27*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b* 
c)/(d*b)^(1/2))*a^4*b^2*c^2*d^4*x^2+48*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c)) 
^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^3*b^3*c^3*d^3*x^2-27*ln(1/2*(2* 
b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^2*b^4* 
c^4*d^2*x^2+12*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b 
*c)/(d*b)^(1/2))*a^4*b^2*c^3*d^3*x+12*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^ 
(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^3*b^3*c^4*d^2*x+6*(d*b)^(1/2)*(( 
b*x+a)*(d*x+c))^(1/2)*a^4*b*c*d^4*x^2+48*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/ 
2)*a^3*b^2*c^2*d^3*x^2+48*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*b^3*c^3* 
d^2*x^2+6*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b^4*c^4*d*x^2+3*ln(1/2*(2* 
b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^6*d^6* 
x^2+3*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b) 
^(1/2))*b^6*c^6*x^2+3*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2 
)+a*d+b*c)/(d*b)^(1/2))*a^6*c^2*d^4+3*ln(1/2*(2*b*d*x+2*((b*x+a)*(d*x+c))^ 
(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^2*b^4*c^6+18*ln(1/2*(2*b*d*x+2*( 
(b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+a*d+b*c)/(d*b)^(1/2))*a^2*b^4*c^2*d^...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1026 vs. \(2 (218) = 436\).

Time = 0.71 (sec) , antiderivative size = 2066, normalized size of antiderivative = 8.20 \[ \int \frac {x^4}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(x^4/(b*x+a)^(5/2)/(d*x+c)^(5/2),x, algorithm="fricas")
 

Output:

[1/6*(3*(a^2*b^4*c^6 - 4*a^3*b^3*c^5*d + 6*a^4*b^2*c^4*d^2 - 4*a^5*b*c^3*d 
^3 + a^6*c^2*d^4 + (b^6*c^4*d^2 - 4*a*b^5*c^3*d^3 + 6*a^2*b^4*c^2*d^4 - 4* 
a^3*b^3*c*d^5 + a^4*b^2*d^6)*x^4 + 2*(b^6*c^5*d - 3*a*b^5*c^4*d^2 + 2*a^2* 
b^4*c^3*d^3 + 2*a^3*b^3*c^2*d^4 - 3*a^4*b^2*c*d^5 + a^5*b*d^6)*x^3 + (b^6* 
c^6 - 9*a^2*b^4*c^4*d^2 + 16*a^3*b^3*c^3*d^3 - 9*a^4*b^2*c^2*d^4 + a^6*d^6 
)*x^2 + 2*(a*b^5*c^6 - 3*a^2*b^4*c^5*d + 2*a^3*b^3*c^4*d^2 + 2*a^4*b^2*c^3 
*d^3 - 3*a^5*b*c^2*d^4 + a^6*c*d^5)*x)*sqrt(b*d)*log(8*b^2*d^2*x^2 + b^2*c 
^2 + 6*a*b*c*d + a^2*d^2 + 4*(2*b*d*x + b*c + a*d)*sqrt(b*d)*sqrt(b*x + a) 
*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(3*a^2*b^4*c^5*d - 11*a^3*b^ 
3*c^4*d^2 - 11*a^4*b^2*c^3*d^3 + 3*a^5*b*c^2*d^4 + 4*(b^6*c^4*d^2 - 3*a*b^ 
5*c^3*d^3 - 3*a^3*b^3*c*d^5 + a^4*b^2*d^6)*x^3 + 3*(b^6*c^5*d - a*b^5*c^4* 
d^2 - 8*a^2*b^4*c^3*d^3 - 8*a^3*b^3*c^2*d^4 - a^4*b^2*c*d^5 + a^5*b*d^6)*x 
^2 + 6*(a*b^5*c^5*d - 3*a^2*b^4*c^4*d^2 - 4*a^3*b^3*c^3*d^3 - 3*a^4*b^2*c^ 
2*d^4 + a^5*b*c*d^5)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(a^2*b^7*c^6*d^3 - 4* 
a^3*b^6*c^5*d^4 + 6*a^4*b^5*c^4*d^5 - 4*a^5*b^4*c^3*d^6 + a^6*b^3*c^2*d^7 
+ (b^9*c^4*d^5 - 4*a*b^8*c^3*d^6 + 6*a^2*b^7*c^2*d^7 - 4*a^3*b^6*c*d^8 + a 
^4*b^5*d^9)*x^4 + 2*(b^9*c^5*d^4 - 3*a*b^8*c^4*d^5 + 2*a^2*b^7*c^3*d^6 + 2 
*a^3*b^6*c^2*d^7 - 3*a^4*b^5*c*d^8 + a^5*b^4*d^9)*x^3 + (b^9*c^6*d^3 - 9*a 
^2*b^7*c^4*d^5 + 16*a^3*b^6*c^3*d^6 - 9*a^4*b^5*c^2*d^7 + a^6*b^3*d^9)*x^2 
 + 2*(a*b^8*c^6*d^3 - 3*a^2*b^7*c^5*d^4 + 2*a^3*b^6*c^4*d^5 + 2*a^4*b^5...
 

Sympy [F]

\[ \int \frac {x^4}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\int \frac {x^{4}}{\left (a + b x\right )^{\frac {5}{2}} \left (c + d x\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(x**4/(b*x+a)**(5/2)/(d*x+c)**(5/2),x)
 

Output:

Integral(x**4/((a + b*x)**(5/2)*(c + d*x)**(5/2)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^4}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^4/(b*x+a)^(5/2)/(d*x+c)^(5/2),x, algorithm="maxima")
 

Output:

Exception raised: ValueError >> Computation failed since Maxima requested 
additional constraints; using the 'assume' command before evaluation *may* 
 help (example of legal syntax is 'assume(a*d-b*c>0)', see `assume?` for m 
ore detail
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 924 vs. \(2 (218) = 436\).

Time = 0.49 (sec) , antiderivative size = 924, normalized size of antiderivative = 3.67 \[ \int \frac {x^4}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(x^4/(b*x+a)^(5/2)/(d*x+c)^(5/2),x, algorithm="giac")
 

Output:

-1/3*(2*sqrt(b*x + a)*(4*(b^13*c^7*d^2 - 6*a*b^12*c^6*d^3 + 12*a^2*b^11*c^ 
5*d^4 - 10*a^3*b^10*c^4*d^5 + 3*a^4*b^9*c^3*d^6)*(b*x + a)/(b^9*c^7*d^3*ab 
s(b) - 7*a*b^8*c^6*d^4*abs(b) + 21*a^2*b^7*c^5*d^5*abs(b) - 35*a^3*b^6*c^4 
*d^6*abs(b) + 35*a^4*b^5*c^3*d^7*abs(b) - 21*a^5*b^4*c^2*d^8*abs(b) + 7*a^ 
6*b^3*c*d^9*abs(b) - a^7*b^2*d^10*abs(b)) + 3*(b^14*c^8*d - 8*a*b^13*c^7*d 
^2 + 22*a^2*b^12*c^6*d^3 - 28*a^3*b^11*c^5*d^4 + 17*a^4*b^10*c^4*d^5 - 4*a 
^5*b^9*c^3*d^6)/(b^9*c^7*d^3*abs(b) - 7*a*b^8*c^6*d^4*abs(b) + 21*a^2*b^7* 
c^5*d^5*abs(b) - 35*a^3*b^6*c^4*d^6*abs(b) + 35*a^4*b^5*c^3*d^7*abs(b) - 2 
1*a^5*b^4*c^2*d^8*abs(b) + 7*a^6*b^3*c*d^9*abs(b) - a^7*b^2*d^10*abs(b)))/ 
(b^2*c + (b*x + a)*b*d - a*b*d)^(3/2) + 3*sqrt(b*d)*b^2*log((sqrt(b*d)*sqr 
t(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(d^3*abs(b)) - 8*(6*s 
qrt(b*d)*a^3*b^8*c^3 - 14*sqrt(b*d)*a^4*b^7*c^2*d + 10*sqrt(b*d)*a^5*b^6*c 
*d^2 - 2*sqrt(b*d)*a^6*b^5*d^3 - 12*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - s 
qrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^3*b^6*c^2 + 15*sqrt(b*d)*(sqrt(b*d 
)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2*a^4*b^5*c*d - 3*s 
qrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2 
*a^5*b^4*d^2 + 6*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + 
a)*b*d - a*b*d))^4*a^3*b^4*c - 3*sqrt(b*d)*(sqrt(b*d)*sqrt(b*x + a) - sqrt 
(b^2*c + (b*x + a)*b*d - a*b*d))^4*a^4*b^3*d)/((b^3*c^3*abs(b) - 3*a*b^2*c 
^2*d*abs(b) + 3*a^2*b*c*d^2*abs(b) - a^3*d^3*abs(b))*(b^2*c - a*b*d - (...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx=\int \frac {x^4}{{\left (a+b\,x\right )}^{5/2}\,{\left (c+d\,x\right )}^{5/2}} \,d x \] Input:

int(x^4/((a + b*x)^(5/2)*(c + d*x)^(5/2)),x)
 

Output:

int(x^4/((a + b*x)^(5/2)*(c + d*x)^(5/2)), x)
 

Reduce [B] (verification not implemented)

Time = 16.72 (sec) , antiderivative size = 2353, normalized size of antiderivative = 9.34 \[ \int \frac {x^4}{(a+b x)^{5/2} (c+d x)^{5/2}} \, dx =\text {Too large to display} \] Input:

int(x^4/(b*x+a)^(5/2)/(d*x+c)^(5/2),x)
 

Output:

(2*(3*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*s 
qrt(c + d*x))/sqrt(a*d - b*c))*a**5*c**2*d**4 + 6*sqrt(d)*sqrt(b)*sqrt(a + 
 b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c)) 
*a**5*c*d**5*x + 3*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x 
) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**5*d**6*x**2 - 12*sqrt(d)*sq 
rt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sq 
rt(a*d - b*c))*a**4*b*c**3*d**3 - 21*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sq 
rt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**4*b*c**2* 
d**4*x - 6*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt 
(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**4*b*c*d**5*x**2 + 3*sqrt(d)*sqrt(b) 
*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a* 
d - b*c))*a**4*b*d**6*x**3 + 18*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d) 
*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**3*b**2*c**4*d* 
*2 + 24*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b) 
*sqrt(c + d*x))/sqrt(a*d - b*c))*a**3*b**2*c**3*d**3*x - 6*sqrt(d)*sqrt(b) 
*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a* 
d - b*c))*a**3*b**2*c**2*d**4*x**2 - 12*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log( 
(sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**3*b**2 
*c*d**5*x**3 - 12*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) 
 + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**2*b**3*c**5*d - 6*sqrt(d)...