\(\int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{3/2}} \, dx\) [384]

Optimal result

Integrand size = 22, antiderivative size = 220 \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{3/2}} \, dx=-\frac {2 a^4}{b^4 (b c-a d) \sqrt {a+b x} \sqrt {c+d x}}-\frac {2 \left (b^4 c^4+a^4 d^4\right ) \sqrt {a+b x}}{b^4 d^3 (b c-a d)^2 \sqrt {c+d x}}-\frac {(9 b c+7 a d) \sqrt {a+b x} \sqrt {c+d x}}{4 b^3 d^3}+\frac {\sqrt {a+b x} (c+d x)^{3/2}}{2 b^2 d^3}+\frac {3 \left (5 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{7/2} d^{7/2}} \] Output:

-2*a^4/b^4/(-a*d+b*c)/(b*x+a)^(1/2)/(d*x+c)^(1/2)-2*(a^4*d^4+b^4*c^4)*(b*x 
+a)^(1/2)/b^4/d^3/(-a*d+b*c)^2/(d*x+c)^(1/2)-1/4*(7*a*d+9*b*c)*(b*x+a)^(1/ 
2)*(d*x+c)^(1/2)/b^3/d^3+1/2*(b*x+a)^(1/2)*(d*x+c)^(3/2)/b^2/d^3+3/4*(5*a^ 
2*d^2+6*a*b*c*d+5*b^2*c^2)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^( 
1/2))/b^(7/2)/d^(7/2)
 

Mathematica [A] (verified)

Time = 0.41 (sec) , antiderivative size = 248, normalized size of antiderivative = 1.13 \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{3/2}} \, dx=\frac {-15 a^4 d^3 (c+d x)+a^3 b d^2 \left (7 c^2+2 c d x-5 d^2 x^2\right )+b^4 c^2 x \left (-15 c^2-5 c d x+2 d^2 x^2\right )+a b^3 c \left (-15 c^3+2 c^2 d x+5 c d^2 x^2-4 d^3 x^3\right )+a^2 b^2 d \left (7 c^3+10 c^2 d x+5 c d^2 x^2+2 d^3 x^3\right )}{4 b^3 d^3 (b c-a d)^2 \sqrt {a+b x} \sqrt {c+d x}}+\frac {3 \left (5 b^2 c^2+6 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{4 b^{7/2} d^{7/2}} \] Input:

Integrate[x^4/((a + b*x)^(3/2)*(c + d*x)^(3/2)),x]
 

Output:

(-15*a^4*d^3*(c + d*x) + a^3*b*d^2*(7*c^2 + 2*c*d*x - 5*d^2*x^2) + b^4*c^2 
*x*(-15*c^2 - 5*c*d*x + 2*d^2*x^2) + a*b^3*c*(-15*c^3 + 2*c^2*d*x + 5*c*d^ 
2*x^2 - 4*d^3*x^3) + a^2*b^2*d*(7*c^3 + 10*c^2*d*x + 5*c*d^2*x^2 + 2*d^3*x 
^3))/(4*b^3*d^3*(b*c - a*d)^2*Sqrt[a + b*x]*Sqrt[c + d*x]) + (3*(5*b^2*c^2 
 + 6*a*b*c*d + 5*a^2*d^2)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a 
+ b*x])])/(4*b^(7/2)*d^(7/2))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 279, normalized size of antiderivative = 1.27, 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, 164, 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.

\(\displaystyle \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{3/2}} \, dx\)

\(\Big \downarrow \) 109

\(\displaystyle \frac {2 a x^3}{b \sqrt {a+b x} \sqrt {c+d x} (b c-a d)}-\frac {2 \int \frac {x^2 (6 a c-(b c-5 a d) x)}{2 \sqrt {a+b x} (c+d x)^{3/2}}dx}{b (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 a x^3}{b \sqrt {a+b x} \sqrt {c+d x} (b c-a d)}-\frac {\int \frac {x^2 (6 a c-(b c-5 a d) x)}{\sqrt {a+b x} (c+d x)^{3/2}}dx}{b (b c-a d)}\)

\(\Big \downarrow \) 167

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 164

\(\displaystyle \frac {2 a x^3}{b \sqrt {a+b x} \sqrt {c+d x} (b c-a d)}-\frac {\frac {2 c x^2 \sqrt {a+b x} (a d+b c)}{d \sqrt {c+d x} (b c-a d)}-\frac {\frac {3 (b c-a d)^2 \left (5 a^2 d^2+6 a b c d+5 b^2 c^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{8 b^2 d^2}-\frac {\sqrt {a+b x} \sqrt {c+d x} \left ((a d+b c) \left (15 a^2 d^2-22 a b c d+15 b^2 c^2\right )-2 b d x \left (5 a^2 d^2-2 a b c d+5 b^2 c^2\right )\right )}{4 b^2 d^2}}{d (b c-a d)}}{b (b c-a d)}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {2 a x^3}{b \sqrt {a+b x} \sqrt {c+d x} (b c-a d)}-\frac {\frac {2 c x^2 \sqrt {a+b x} (a d+b c)}{d \sqrt {c+d x} (b c-a d)}-\frac {\frac {3 (b c-a d)^2 \left (5 a^2 d^2+6 a b c d+5 b^2 c^2\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{4 b^2 d^2}-\frac {\sqrt {a+b x} \sqrt {c+d x} \left ((a d+b c) \left (15 a^2 d^2-22 a b c d+15 b^2 c^2\right )-2 b d x \left (5 a^2 d^2-2 a b c d+5 b^2 c^2\right )\right )}{4 b^2 d^2}}{d (b c-a d)}}{b (b c-a d)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 a x^3}{b \sqrt {a+b x} \sqrt {c+d x} (b c-a d)}-\frac {\frac {2 c x^2 \sqrt {a+b x} (a d+b c)}{d \sqrt {c+d x} (b c-a d)}-\frac {\frac {3 (b c-a d)^2 \left (5 a^2 d^2+6 a b c d+5 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2} d^{5/2}}-\frac {\sqrt {a+b x} \sqrt {c+d x} \left ((a d+b c) \left (15 a^2 d^2-22 a b c d+15 b^2 c^2\right )-2 b d x \left (5 a^2 d^2-2 a b c d+5 b^2 c^2\right )\right )}{4 b^2 d^2}}{d (b c-a d)}}{b (b c-a d)}\)

Input:

Int[x^4/((a + b*x)^(3/2)*(c + d*x)^(3/2)),x]
 

Output:

(2*a*x^3)/(b*(b*c - a*d)*Sqrt[a + b*x]*Sqrt[c + d*x]) - ((2*c*(b*c + a*d)* 
x^2*Sqrt[a + b*x])/(d*(b*c - a*d)*Sqrt[c + d*x]) - (-1/4*(Sqrt[a + b*x]*Sq 
rt[c + d*x]*((b*c + a*d)*(15*b^2*c^2 - 22*a*b*c*d + 15*a^2*d^2) - 2*b*d*(5 
*b^2*c^2 - 2*a*b*c*d + 5*a^2*d^2)*x))/(b^2*d^2) + (3*(b*c - a*d)^2*(5*b^2* 
c^2 + 6*a*b*c*d + 5*a^2*d^2)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt[b]*Sqrt 
[c + d*x])])/(4*b^(5/2)*d^(5/2)))/(d*(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 164

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

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. \(1368\) vs. \(2(186)=372\).

Time = 0.29 (sec) , antiderivative size = 1369, normalized size of antiderivative = 6.22

method	result	size
default	\(\text {Expression too large to display}\)	\(1369\)

Input:

int(x^4/(b*x+a)^(3/2)/(d*x+c)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/8*(4*b^4*c^2*d^2*x^3*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-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*b^4*c^3* 
d^2*x^2-10*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^3*b*d^4*x^2-10*(d*b)^(1/2 
)*((b*x+a)*(d*x+c))^(1/2)*b^4*c^3*d*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^3*b^2*c*d^4*x^2-6*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^3 
*c^2*d^3*x^2+15*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^5*d^5*x+15*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^5*c^5*x+15*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^5*c*d^4+15*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*b^4*c^5+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^4*b*c*d^4*x-8*a*b^3*c*d^3*x^3*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)+4*( 
d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^3*b*c*d^3*x+20*(d*b)^(1/2)*((b*x+a)*( 
d*x+c))^(1/2)*a^2*b^2*c^2*d^2*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^2*b^3*c^4*d-30*a^4*d^4*x*(d*b)^(1/2 
)*((b*x+a)*(d*x+c))^(1/2)-30*b^4*c^4*x*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2) 
-30*a^4*c*d^3*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)-30*a*b^3*c^4*(d*b)^(1/2) 
*((b*x+a)*(d*x+c))^(1/2)+15*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*d^5*x^2+15*ln(1/2*(2*b*d*x+2*((b*x+...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 593 vs. \(2 (186) = 372\).

Time = 0.40 (sec) , antiderivative size = 1200, normalized size of antiderivative = 5.45 \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{3/2}} \, dx =\text {Too large to display} \] Input:

integrate(x^4/(b*x+a)^(3/2)/(d*x+c)^(3/2),x, algorithm="fricas")
 

Output:

[1/16*(3*(5*a*b^4*c^5 - 4*a^2*b^3*c^4*d - 2*a^3*b^2*c^3*d^2 - 4*a^4*b*c^2* 
d^3 + 5*a^5*c*d^4 + (5*b^5*c^4*d - 4*a*b^4*c^3*d^2 - 2*a^2*b^3*c^2*d^3 - 4 
*a^3*b^2*c*d^4 + 5*a^4*b*d^5)*x^2 + (5*b^5*c^5 + a*b^4*c^4*d - 6*a^2*b^3*c 
^3*d^2 - 6*a^3*b^2*c^2*d^3 + a^4*b*c*d^4 + 5*a^5*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*(15*a*b^4* 
c^4*d - 7*a^2*b^3*c^3*d^2 - 7*a^3*b^2*c^2*d^3 + 15*a^4*b*c*d^4 - 2*(b^5*c^ 
2*d^3 - 2*a*b^4*c*d^4 + a^2*b^3*d^5)*x^3 + 5*(b^5*c^3*d^2 - a*b^4*c^2*d^3 
- a^2*b^3*c*d^4 + a^3*b^2*d^5)*x^2 + (15*b^5*c^4*d - 2*a*b^4*c^3*d^2 - 10* 
a^2*b^3*c^2*d^3 - 2*a^3*b^2*c*d^4 + 15*a^4*b*d^5)*x)*sqrt(b*x + a)*sqrt(d* 
x + c))/(a*b^6*c^3*d^4 - 2*a^2*b^5*c^2*d^5 + a^3*b^4*c*d^6 + (b^7*c^2*d^5 
- 2*a*b^6*c*d^6 + a^2*b^5*d^7)*x^2 + (b^7*c^3*d^4 - a*b^6*c^2*d^5 - a^2*b^ 
5*c*d^6 + a^3*b^4*d^7)*x), -1/8*(3*(5*a*b^4*c^5 - 4*a^2*b^3*c^4*d - 2*a^3* 
b^2*c^3*d^2 - 4*a^4*b*c^2*d^3 + 5*a^5*c*d^4 + (5*b^5*c^4*d - 4*a*b^4*c^3*d 
^2 - 2*a^2*b^3*c^2*d^3 - 4*a^3*b^2*c*d^4 + 5*a^4*b*d^5)*x^2 + (5*b^5*c^5 + 
 a*b^4*c^4*d - 6*a^2*b^3*c^3*d^2 - 6*a^3*b^2*c^2*d^3 + a^4*b*c*d^4 + 5*a^5 
*d^5)*x)*sqrt(-b*d)*arctan(1/2*(2*b*d*x + b*c + a*d)*sqrt(-b*d)*sqrt(b*x + 
 a)*sqrt(d*x + c)/(b^2*d^2*x^2 + a*b*c*d + (b^2*c*d + a*b*d^2)*x)) + 2*(15 
*a*b^4*c^4*d - 7*a^2*b^3*c^3*d^2 - 7*a^3*b^2*c^2*d^3 + 15*a^4*b*c*d^4 - 2* 
(b^5*c^2*d^3 - 2*a*b^4*c*d^4 + a^2*b^3*d^5)*x^3 + 5*(b^5*c^3*d^2 - a*b^...
 

Sympy [F]

\[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{3/2}} \, dx=\int \frac {x^{4}}{\left (a + b x\right )^{\frac {3}{2}} \left (c + d x\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(x**4/(b*x+a)**(3/2)/(d*x+c)**(3/2),x)
 

Output:

Integral(x**4/((a + b*x)**(3/2)*(c + d*x)**(3/2)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{3/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^4/(b*x+a)^(3/2)/(d*x+c)^(3/2),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. 487 vs. \(2 (186) = 372\).

Time = 0.25 (sec) , antiderivative size = 487, normalized size of antiderivative = 2.21 \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{3/2}} \, dx=-\frac {\frac {32 \, \sqrt {b d} a^{4} b^{2}}{{\left (b^{2} c - a b d - {\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )} {\left (b c {\left | b \right |} - a d {\left | b \right |}\right )}} - \frac {2 \, {\left ({\left (b x + a\right )} {\left (\frac {2 \, {\left (b^{6} c^{2} d^{4} {\left | b \right |} - 2 \, a b^{5} c d^{5} {\left | b \right |} + a^{2} b^{4} d^{6} {\left | b \right |}\right )} {\left (b x + a\right )}}{b^{6} c^{2} d^{5} - 2 \, a b^{5} c d^{6} + a^{2} b^{4} d^{7}} - \frac {5 \, b^{7} c^{3} d^{3} {\left | b \right |} + a b^{6} c^{2} d^{4} {\left | b \right |} - 17 \, a^{2} b^{5} c d^{5} {\left | b \right |} + 11 \, a^{3} b^{4} d^{6} {\left | b \right |}}{b^{6} c^{2} d^{5} - 2 \, a b^{5} c d^{6} + a^{2} b^{4} d^{7}}\right )} - \frac {15 \, b^{8} c^{4} d^{2} {\left | b \right |} - 12 \, a b^{7} c^{3} d^{3} {\left | b \right |} - 6 \, a^{2} b^{6} c^{2} d^{4} {\left | b \right |} + 20 \, a^{3} b^{5} c d^{5} {\left | b \right |} - 9 \, a^{4} b^{4} d^{6} {\left | b \right |}}{b^{6} c^{2} d^{5} - 2 \, a b^{5} c d^{6} + a^{2} b^{4} d^{7}}\right )} \sqrt {b x + a}}{\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}} + \frac {3 \, {\left (5 \, \sqrt {b d} b^{3} c^{2} + 6 \, \sqrt {b d} a b^{2} c d + 5 \, \sqrt {b d} a^{2} b d^{2}\right )} \log \left ({\left (\sqrt {b d} \sqrt {b x + a} - \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d}\right )}^{2}\right )}{d^{4} {\left | b \right |}}}{8 \, b^{4}} \] Input:

integrate(x^4/(b*x+a)^(3/2)/(d*x+c)^(3/2),x, algorithm="giac")
 

Output:

-1/8*(32*sqrt(b*d)*a^4*b^2/((b^2*c - a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sq 
rt(b^2*c + (b*x + a)*b*d - a*b*d))^2)*(b*c*abs(b) - a*d*abs(b))) - 2*((b*x 
 + a)*(2*(b^6*c^2*d^4*abs(b) - 2*a*b^5*c*d^5*abs(b) + a^2*b^4*d^6*abs(b))* 
(b*x + a)/(b^6*c^2*d^5 - 2*a*b^5*c*d^6 + a^2*b^4*d^7) - (5*b^7*c^3*d^3*abs 
(b) + a*b^6*c^2*d^4*abs(b) - 17*a^2*b^5*c*d^5*abs(b) + 11*a^3*b^4*d^6*abs( 
b))/(b^6*c^2*d^5 - 2*a*b^5*c*d^6 + a^2*b^4*d^7)) - (15*b^8*c^4*d^2*abs(b) 
- 12*a*b^7*c^3*d^3*abs(b) - 6*a^2*b^6*c^2*d^4*abs(b) + 20*a^3*b^5*c*d^5*ab 
s(b) - 9*a^4*b^4*d^6*abs(b))/(b^6*c^2*d^5 - 2*a*b^5*c*d^6 + a^2*b^4*d^7))* 
sqrt(b*x + a)/sqrt(b^2*c + (b*x + a)*b*d - a*b*d) + 3*(5*sqrt(b*d)*b^3*c^2 
 + 6*sqrt(b*d)*a*b^2*c*d + 5*sqrt(b*d)*a^2*b*d^2)*log((sqrt(b*d)*sqrt(b*x 
+ a) - sqrt(b^2*c + (b*x + a)*b*d - a*b*d))^2)/(d^4*abs(b)))/b^4
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{3/2}} \, dx=\int \frac {x^4}{{\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{3/2}} \,d x \] Input:

int(x^4/((a + b*x)^(3/2)*(c + d*x)^(3/2)),x)
 

Output:

int(x^4/((a + b*x)^(3/2)*(c + d*x)^(3/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.31 (sec) , antiderivative size = 1115, normalized size of antiderivative = 5.07 \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{3/2}} \, dx =\text {Too large to display} \] Input:

int(x^4/(b*x+a)^(3/2)/(d*x+c)^(3/2),x)
 

Output:

(120*sqrt(d)*sqrt(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sq 
rt(c + d*x))/sqrt(a*d - b*c))*a**4*c*d**4 + 120*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*d**5*x - 96*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*c**2*d**3 - 96*sqrt(d)*sqr 
t(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqr 
t(a*d - b*c))*a**3*b*c*d**4*x - 48*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**2*c**3 
*d**2 - 48*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**2*c**2*d**3*x - 96*sqrt(d)*sqr 
t(b)*sqrt(a + b*x)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqr 
t(a*d - b*c))*a*b**3*c**4*d - 96*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*b**3*c**3*d**2 
*x + 120*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))*b**4*c**5 + 120*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)) 
*b**4*c**4*d*x - 65*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**4*c*d**4 - 65*sqrt(d) 
*sqrt(b)*sqrt(a + b*x)*a**4*d**5*x + 4*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**3* 
b*c**2*d**3 + 4*sqrt(d)*sqrt(b)*sqrt(a + b*x)*a**3*b*c*d**4*x - 6*sqrt(d)* 
sqrt(b)*sqrt(a + b*x)*a**2*b**2*c**3*d**2 - 6*sqrt(d)*sqrt(b)*sqrt(a + ...