\(\int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx\) [393]

Optimal result

Integrand size = 22, antiderivative size = 232 \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=-\frac {2 a^4}{b^4 (b c-a d) \sqrt {a+b x} (c+d x)^{3/2}}-\frac {2 \left (b^4 c^4+3 a^4 d^4\right ) \sqrt {a+b x}}{3 b^4 d^3 (b c-a d)^2 (c+d x)^{3/2}}+\frac {2 \left (7 b^4 c^4-12 a b^3 c^3 d-3 a^4 d^4\right ) \sqrt {a+b x}}{3 b^3 d^3 (b c-a d)^3 \sqrt {c+d x}}+\frac {\sqrt {a+b x} \sqrt {c+d x}}{b^2 d^3}-\frac {(5 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{5/2} d^{7/2}} \] Output:

-2*a^4/b^4/(-a*d+b*c)/(b*x+a)^(1/2)/(d*x+c)^(3/2)-2/3*(3*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)^(3/2)+2/3*(-3*a^4*d^4-12*a*b^3* 
c^3*d+7*b^4*c^4)*(b*x+a)^(1/2)/b^3/d^3/(-a*d+b*c)^3/(d*x+c)^(1/2)+(b*x+a)^ 
(1/2)*(d*x+c)^(1/2)/b^2/d^3-(3*a*d+5*b*c)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^ 
(1/2)/(d*x+c)^(1/2))/b^(5/2)/d^(7/2)
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 234, normalized size of antiderivative = 1.01 \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\frac {9 a^4 d^3 (c+d x)^2+3 a^3 b d^2 (-3 c+d x) (c+d x)^2-b^4 c^3 x \left (15 c^2+20 c d x+3 d^2 x^2\right )+a^2 b^2 c d \left (31 c^3+33 c^2 d x-9 c d^2 x^2-9 d^3 x^3\right )+a b^3 c^2 \left (-15 c^3+11 c^2 d x+39 c d^2 x^2+9 d^3 x^3\right )}{3 b^2 d^3 (-b c+a d)^3 \sqrt {a+b x} (c+d x)^{3/2}}-\frac {(5 b c+3 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{b^{5/2} d^{7/2}} \] Input:

Integrate[x^4/((a + b*x)^(3/2)*(c + d*x)^(5/2)),x]
 

Output:

(9*a^4*d^3*(c + d*x)^2 + 3*a^3*b*d^2*(-3*c + d*x)*(c + d*x)^2 - b^4*c^3*x* 
(15*c^2 + 20*c*d*x + 3*d^2*x^2) + a^2*b^2*c*d*(31*c^3 + 33*c^2*d*x - 9*c*d 
^2*x^2 - 9*d^3*x^3) + a*b^3*c^2*(-15*c^3 + 11*c^2*d*x + 39*c*d^2*x^2 + 9*d 
^3*x^3))/(3*b^2*d^3*(-(b*c) + a*d)^3*Sqrt[a + b*x]*(c + d*x)^(3/2)) - ((5* 
b*c + 3*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(b^ 
(5/2)*d^(7/2))
 

Rubi [A] (verified)

Time = 0.40 (sec) , antiderivative size = 289, normalized size of antiderivative = 1.25, 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, 160, 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.

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

\(\Big \downarrow \) 109

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 167

\(\displaystyle \frac {2 a x^3}{b \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}-\frac {\frac {2 c x^2 \sqrt {a+b x} (3 a d+b c)}{3 d (c+d x)^{3/2} (b c-a d)}-\frac {2 \int \frac {x \left (4 a c (b c+3 a d)+\left (5 b^2 c^2-6 a b d c+9 a^2 d^2\right ) x\right )}{2 \sqrt {a+b x} (c+d x)^{3/2}}dx}{3 d (b c-a d)}}{b (b c-a d)}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 a x^3}{b \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}-\frac {\frac {2 c x^2 \sqrt {a+b x} (3 a d+b c)}{3 d (c+d x)^{3/2} (b c-a d)}-\frac {\int \frac {x \left (4 a c (b c+3 a d)+\left (5 b^2 c^2-6 a b d c+9 a^2 d^2\right ) x\right )}{\sqrt {a+b x} (c+d x)^{3/2}}dx}{3 d (b c-a d)}}{b (b c-a d)}\)

\(\Big \downarrow \) 160

\(\displaystyle \frac {2 a x^3}{b \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}-\frac {\frac {2 c x^2 \sqrt {a+b x} (3 a d+b c)}{3 d (c+d x)^{3/2} (b c-a d)}-\frac {\frac {\sqrt {a+b x} \left (d x (b c-a d) \left (9 a^2 d^2-6 a b c d+5 b^2 c^2\right )+c \left (-9 a^3 d^3+9 a^2 b c d^2-31 a b^2 c^2 d+15 b^3 c^3\right )\right )}{b d^2 \sqrt {c+d x} (b c-a d)}-\frac {3 (b c-a d)^2 (3 a d+5 b c) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 b d^2}}{3 d (b c-a d)}}{b (b c-a d)}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {2 a x^3}{b \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}-\frac {\frac {2 c x^2 \sqrt {a+b x} (3 a d+b c)}{3 d (c+d x)^{3/2} (b c-a d)}-\frac {\frac {\sqrt {a+b x} \left (d x (b c-a d) \left (9 a^2 d^2-6 a b c d+5 b^2 c^2\right )+c \left (-9 a^3 d^3+9 a^2 b c d^2-31 a b^2 c^2 d+15 b^3 c^3\right )\right )}{b d^2 \sqrt {c+d x} (b c-a d)}-\frac {3 (b c-a d)^2 (3 a d+5 b c) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{b d^2}}{3 d (b c-a d)}}{b (b c-a d)}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {2 a x^3}{b \sqrt {a+b x} (c+d x)^{3/2} (b c-a d)}-\frac {\frac {2 c x^2 \sqrt {a+b x} (3 a d+b c)}{3 d (c+d x)^{3/2} (b c-a d)}-\frac {\frac {\sqrt {a+b x} \left (d x (b c-a d) \left (9 a^2 d^2-6 a b c d+5 b^2 c^2\right )+c \left (-9 a^3 d^3+9 a^2 b c d^2-31 a b^2 c^2 d+15 b^3 c^3\right )\right )}{b d^2 \sqrt {c+d x} (b c-a d)}-\frac {3 (b c-a d)^2 (3 a d+5 b c) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{b^{3/2} d^{5/2}}}{3 d (b c-a d)}}{b (b c-a d)}\)

Input:

Int[x^4/((a + b*x)^(3/2)*(c + d*x)^(5/2)),x]
 

Output:

(2*a*x^3)/(b*(b*c - a*d)*Sqrt[a + b*x]*(c + d*x)^(3/2)) - ((2*c*(b*c + 3*a 
*d)*x^2*Sqrt[a + b*x])/(3*d*(b*c - a*d)*(c + d*x)^(3/2)) - ((Sqrt[a + b*x] 
*(c*(15*b^3*c^3 - 31*a*b^2*c^2*d + 9*a^2*b*c*d^2 - 9*a^3*d^3) + d*(b*c - a 
*d)*(5*b^2*c^2 - 6*a*b*c*d + 9*a^2*d^2)*x))/(b*d^2*(b*c - a*d)*Sqrt[c + d* 
x]) - (3*(b*c - a*d)^2*(5*b*c + 3*a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sq 
rt[b]*Sqrt[c + d*x])])/(b^(3/2)*d^(5/2)))/(3*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 160

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

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. \(1713\) vs. \(2(200)=400\).

Time = 0.29 (sec) , antiderivative size = 1714, normalized size of antiderivative = 7.39

method	result	size
default	\(\text {Expression too large to display}\)	\(1714\)

Input:

int(x^4/(b*x+a)^(3/2)/(d*x+c)^(5/2),x,method=_RETURNVERBOSE)
 

Output:

-1/6*(-22*a*b^3*c^4*d*x*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+36*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 
^5*d-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^3*c^2*d^4*x^3+36*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^3*x^3+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^4*b*c*d^5*x^2-4 
2*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^2*d^4*x^2+57*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^4*d^2*x^2+9*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^6*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))*b^5*c^ 
6*x+9*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^2*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^6-18*a^4*c^2*d^3*((b*x+a)*(d*x+c))^(1/2)* 
(d*b)^(1/2)+30*a*b^3*c^5*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+9*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^6 
*x^3-15*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^4*d^2*x^3-30*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*d*x^2+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^5*c*d^5*x-12*ln(1/2...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 807 vs. \(2 (200) = 400\).

Time = 0.46 (sec) , antiderivative size = 1628, normalized size of antiderivative = 7.02 \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\text {Too large to display} \] Input:

integrate(x^4/(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="fricas")
 

Output:

[1/12*(3*(5*a*b^4*c^6 - 12*a^2*b^3*c^5*d + 6*a^3*b^2*c^4*d^2 + 4*a^4*b*c^3 
*d^3 - 3*a^5*c^2*d^4 + (5*b^5*c^4*d^2 - 12*a*b^4*c^3*d^3 + 6*a^2*b^3*c^2*d 
^4 + 4*a^3*b^2*c*d^5 - 3*a^4*b*d^6)*x^3 + (10*b^5*c^5*d - 19*a*b^4*c^4*d^2 
 + 14*a^3*b^2*c^2*d^4 - 2*a^4*b*c*d^5 - 3*a^5*d^6)*x^2 + (5*b^5*c^6 - 2*a* 
b^4*c^5*d - 18*a^2*b^3*c^4*d^2 + 16*a^3*b^2*c^3*d^3 + 5*a^4*b*c^2*d^4 - 6* 
a^5*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*(15*a*b^4*c^5*d - 31*a^2*b^3*c^4*d^2 + 9*a^3*b^2*c^3* 
d^3 - 9*a^4*b*c^2*d^4 + 3*(b^5*c^3*d^3 - 3*a*b^4*c^2*d^4 + 3*a^2*b^3*c*d^5 
 - a^3*b^2*d^6)*x^3 + (20*b^5*c^4*d^2 - 39*a*b^4*c^3*d^3 + 9*a^2*b^3*c^2*d 
^4 + 3*a^3*b^2*c*d^5 - 9*a^4*b*d^6)*x^2 + (15*b^5*c^5*d - 11*a*b^4*c^4*d^2 
 - 33*a^2*b^3*c^3*d^3 + 15*a^3*b^2*c^2*d^4 - 18*a^4*b*c*d^5)*x)*sqrt(b*x + 
 a)*sqrt(d*x + c))/(a*b^6*c^5*d^4 - 3*a^2*b^5*c^4*d^5 + 3*a^3*b^4*c^3*d^6 
- a^4*b^3*c^2*d^7 + (b^7*c^3*d^6 - 3*a*b^6*c^2*d^7 + 3*a^2*b^5*c*d^8 - a^3 
*b^4*d^9)*x^3 + (2*b^7*c^4*d^5 - 5*a*b^6*c^3*d^6 + 3*a^2*b^5*c^2*d^7 + a^3 
*b^4*c*d^8 - a^4*b^3*d^9)*x^2 + (b^7*c^5*d^4 - a*b^6*c^4*d^5 - 3*a^2*b^5*c 
^3*d^6 + 5*a^3*b^4*c^2*d^7 - 2*a^4*b^3*c*d^8)*x), 1/6*(3*(5*a*b^4*c^6 - 12 
*a^2*b^3*c^5*d + 6*a^3*b^2*c^4*d^2 + 4*a^4*b*c^3*d^3 - 3*a^5*c^2*d^4 + (5* 
b^5*c^4*d^2 - 12*a*b^4*c^3*d^3 + 6*a^2*b^3*c^2*d^4 + 4*a^3*b^2*c*d^5 - 3*a 
^4*b*d^6)*x^3 + (10*b^5*c^5*d - 19*a*b^4*c^4*d^2 + 14*a^3*b^2*c^2*d^4 -...
 

Sympy [F]

\[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\int \frac {x^{4}}{\left (a + b x\right )^{\frac {3}{2}} \left (c + d x\right )^{\frac {5}{2}}}\, dx \] Input:

integrate(x**4/(b*x+a)**(3/2)/(d*x+c)**(5/2),x)
 

Output:

Integral(x**4/((a + b*x)**(3/2)*(c + d*x)**(5/2)), x)
 

Maxima [F(-2)]

Exception generated. \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\text {Exception raised: ValueError} \] Input:

integrate(x^4/(b*x+a)^(3/2)/(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. 974 vs. \(2 (200) = 400\).

Time = 0.35 (sec) , antiderivative size = 974, normalized size of antiderivative = 4.20 \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx =\text {Too large to display} \] Input:

integrate(x^4/(b*x+a)^(3/2)/(d*x+c)^(5/2),x, algorithm="giac")
 

Output:

-1/6*(24*sqrt(b*d)*a^4*b^3/((b^2*c^2*abs(b) - 2*a*b*c*d*abs(b) + a^2*d^2*a 
bs(b))*(b^2*c - a*b*d - (sqrt(b*d)*sqrt(b*x + a) - sqrt(b^2*c + (b*x + a)* 
b*d - a*b*d))^2)) - 2*((b*x + a)*(3*(b^12*c^5*d^4*abs(b) - 5*a*b^11*c^4*d^ 
5*abs(b) + 10*a^2*b^10*c^3*d^6*abs(b) - 10*a^3*b^9*c^2*d^7*abs(b) + 5*a^4* 
b^8*c*d^8*abs(b) - a^5*b^7*d^9*abs(b))*(b*x + a)/(b^10*c^5*d^5 - 5*a*b^9*c 
^4*d^6 + 10*a^2*b^8*c^3*d^7 - 10*a^3*b^7*c^2*d^8 + 5*a^4*b^6*c*d^9 - a^5*b 
^5*d^10) + 2*(16*b^14*c^8*d^19*abs(b) - 80*a*b^13*c^7*d^20*abs(b) + 160*a^ 
2*b^12*c^6*d^21*abs(b) - 160*a^3*b^11*c^5*d^22*abs(b) + 80*a^4*b^10*c^4*d^ 
23*abs(b) - 16*a^5*b^9*c^3*d^24*abs(b) + 10*b^13*c^6*d^3*abs(b) - 60*a*b^1 
2*c^5*d^4*abs(b) + 156*a^2*b^11*c^4*d^5*abs(b) - 232*a^3*b^10*c^3*d^6*abs( 
b) + 205*a^4*b^9*c^2*d^7*abs(b) - 98*a^5*b^8*c*d^8*abs(b) + 19*a^6*b^7*d^9 
*abs(b))/(b^10*c^5*d^5 - 5*a*b^9*c^4*d^6 + 10*a^2*b^8*c^3*d^7 - 10*a^3*b^7 
*c^2*d^8 + 5*a^4*b^6*c*d^9 - a^5*b^5*d^10)) + 3*(8*b^15*c^9*d^18*abs(b) - 
48*a*b^14*c^8*d^19*abs(b) + 120*a^2*b^13*c^7*d^20*abs(b) - 160*a^3*b^12*c^ 
6*d^21*abs(b) + 120*a^4*b^11*c^5*d^22*abs(b) - 48*a^5*b^10*c^4*d^23*abs(b) 
 + 8*a^6*b^9*c^3*d^24*abs(b) + 5*b^14*c^7*d^2*abs(b) - 35*a*b^13*c^6*d^3*a 
bs(b) + 105*a^2*b^12*c^5*d^4*abs(b) - 183*a^3*b^11*c^4*d^5*abs(b) + 203*a^ 
4*b^10*c^3*d^6*abs(b) - 141*a^5*b^9*c^2*d^7*abs(b) + 55*a^6*b^8*c*d^8*abs( 
b) - 9*a^7*b^7*d^9*abs(b))/(b^10*c^5*d^5 - 5*a*b^9*c^4*d^6 + 10*a^2*b^8*c^ 
3*d^7 - 10*a^3*b^7*c^2*d^8 + 5*a^4*b^6*c*d^9 - a^5*b^5*d^10))*sqrt(b*x ...
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx=\int \frac {x^4}{{\left (a+b\,x\right )}^{3/2}\,{\left (c+d\,x\right )}^{5/2}} \,d x \] Input:

int(x^4/((a + b*x)^(3/2)*(c + d*x)^(5/2)),x)
 

Output:

int(x^4/((a + b*x)^(3/2)*(c + d*x)^(5/2)), x)
 

Reduce [B] (verification not implemented)

Time = 1.50 (sec) , antiderivative size = 1663, normalized size of antiderivative = 7.17 \[ \int \frac {x^4}{(a+b x)^{3/2} (c+d x)^{5/2}} \, dx =\text {Too large to display} \] Input:

int(x^4/(b*x+a)^(3/2)/(d*x+c)^(5/2),x)
 

Output:

( - 72*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*c**2*d**4 - 144*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*c*d**5*x - 72*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**6*x**2 + 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**3*d**3 + 192*sqrt(d)*sqrt(b)*sqrt(a + b*x)*lo 
g((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**3*b* 
c**2*d**4*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*d**5*x**2 + 144*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**4*d**2 + 288*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**3*x + 144*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**4*x**2 - 
 288*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*b**3*c**5*d - 576*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**4*d**2*x - 288*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**3*x**...