\(\int x (a+b x)^{3/2} (c+d x)^{5/2} \, dx\) [233]

Optimal result

Integrand size = 20, antiderivative size = 315 \[ \int x (a+b x)^{3/2} (c+d x)^{5/2} \, dx=\frac {(b c-a d)^4 (5 b c+7 a d) \sqrt {a+b x} \sqrt {c+d x}}{512 b^4 d^3}-\frac {(b c-a d)^3 (5 b c+7 a d) (a+b x)^{3/2} \sqrt {c+d x}}{768 b^4 d^2}-\frac {(b c-a d)^2 (5 b c+7 a d) (a+b x)^{5/2} \sqrt {c+d x}}{192 b^4 d}-\frac {(b c-a d) (5 b c+7 a d) (a+b x)^{5/2} (c+d x)^{3/2}}{96 b^3 d}-\frac {(5 b c+7 a d) (a+b x)^{5/2} (c+d x)^{5/2}}{60 b^2 d}+\frac {(a+b x)^{5/2} (c+d x)^{7/2}}{6 b d}-\frac {(b c-a d)^5 (5 b c+7 a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{512 b^{9/2} d^{7/2}} \] Output:

1/512*(-a*d+b*c)^4*(7*a*d+5*b*c)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^4/d^3-1/768 
*(-a*d+b*c)^3*(7*a*d+5*b*c)*(b*x+a)^(3/2)*(d*x+c)^(1/2)/b^4/d^2-1/192*(-a* 
d+b*c)^2*(7*a*d+5*b*c)*(b*x+a)^(5/2)*(d*x+c)^(1/2)/b^4/d-1/96*(-a*d+b*c)*( 
7*a*d+5*b*c)*(b*x+a)^(5/2)*(d*x+c)^(3/2)/b^3/d-1/60*(7*a*d+5*b*c)*(b*x+a)^ 
(5/2)*(d*x+c)^(5/2)/b^2/d+1/6*(b*x+a)^(5/2)*(d*x+c)^(7/2)/b/d-1/512*(-a*d+ 
b*c)^5*(7*a*d+5*b*c)*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/(d*x+c)^(1/2))/ 
b^(9/2)/d^(7/2)
 

Mathematica [A] (verified)

Time = 0.63 (sec) , antiderivative size = 293, normalized size of antiderivative = 0.93 \[ \int x (a+b x)^{3/2} (c+d x)^{5/2} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (-105 a^5 d^5+5 a^4 b d^4 (83 c+14 d x)-2 a^3 b^2 d^3 \left (273 c^2+136 c d x+28 d^2 x^2\right )+6 a^2 b^3 d^2 \left (25 c^3+58 c^2 d x+36 c d^2 x^2+8 d^3 x^3\right )+a b^4 d \left (-245 c^4+160 c^3 d x+3384 c^2 d^2 x^2+4448 c d^3 x^3+1664 d^4 x^4\right )+5 b^5 \left (15 c^5-10 c^4 d x+8 c^3 d^2 x^2+432 c^2 d^3 x^3+640 c d^4 x^4+256 d^5 x^5\right )\right )}{7680 b^4 d^3}-\frac {(b c-a d)^5 (5 b c+7 a d) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{512 b^{9/2} d^{7/2}} \] Input:

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

Output:

(Sqrt[a + b*x]*Sqrt[c + d*x]*(-105*a^5*d^5 + 5*a^4*b*d^4*(83*c + 14*d*x) - 
 2*a^3*b^2*d^3*(273*c^2 + 136*c*d*x + 28*d^2*x^2) + 6*a^2*b^3*d^2*(25*c^3 
+ 58*c^2*d*x + 36*c*d^2*x^2 + 8*d^3*x^3) + a*b^4*d*(-245*c^4 + 160*c^3*d*x 
 + 3384*c^2*d^2*x^2 + 4448*c*d^3*x^3 + 1664*d^4*x^4) + 5*b^5*(15*c^5 - 10* 
c^4*d*x + 8*c^3*d^2*x^2 + 432*c^2*d^3*x^3 + 640*c*d^4*x^4 + 256*d^5*x^5))) 
/(7680*b^4*d^3) - ((b*c - a*d)^5*(5*b*c + 7*a*d)*ArcTanh[(Sqrt[b]*Sqrt[c + 
 d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(512*b^(9/2)*d^(7/2))
 

Rubi [A] (verified)

Time = 0.30 (sec) , antiderivative size = 285, normalized size of antiderivative = 0.90, number of steps used = 9, number of rules used = 8, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {90, 60, 60, 60, 60, 60, 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.

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

\(\Big \downarrow \) 90

\(\displaystyle \frac {(a+b x)^{5/2} (c+d x)^{7/2}}{6 b d}-\frac {(7 a d+5 b c) \int (a+b x)^{3/2} (c+d x)^{5/2}dx}{12 b d}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(a+b x)^{5/2} (c+d x)^{7/2}}{6 b d}-\frac {(7 a d+5 b c) \left (\frac {(b c-a d) \int (a+b x)^{3/2} (c+d x)^{3/2}dx}{2 b}+\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b}\right )}{12 b d}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(a+b x)^{5/2} (c+d x)^{7/2}}{6 b d}-\frac {(7 a d+5 b c) \left (\frac {(b c-a d) \left (\frac {3 (b c-a d) \int (a+b x)^{3/2} \sqrt {c+d x}dx}{8 b}+\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{4 b}\right )}{2 b}+\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b}\right )}{12 b d}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(a+b x)^{5/2} (c+d x)^{7/2}}{6 b d}-\frac {(7 a d+5 b c) \left (\frac {(b c-a d) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \int \frac {(a+b x)^{3/2}}{\sqrt {c+d x}}dx}{6 b}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b}\right )}{8 b}+\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{4 b}\right )}{2 b}+\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b}\right )}{12 b d}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(a+b x)^{5/2} (c+d x)^{7/2}}{6 b d}-\frac {(7 a d+5 b c) \left (\frac {(b c-a d) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \int \frac {\sqrt {a+b x}}{\sqrt {c+d x}}dx}{4 d}\right )}{6 b}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b}\right )}{8 b}+\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{4 b}\right )}{2 b}+\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b}\right )}{12 b d}\)

\(\Big \downarrow \) 60

\(\displaystyle \frac {(a+b x)^{5/2} (c+d x)^{7/2}}{6 b d}-\frac {(7 a d+5 b c) \left (\frac {(b c-a d) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{2 d}\right )}{4 d}\right )}{6 b}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b}\right )}{8 b}+\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{4 b}\right )}{2 b}+\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b}\right )}{12 b d}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {(a+b x)^{5/2} (c+d x)^{7/2}}{6 b d}-\frac {(7 a d+5 b c) \left (\frac {(b c-a d) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{d}\right )}{4 d}\right )}{6 b}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b}\right )}{8 b}+\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{4 b}\right )}{2 b}+\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b}\right )}{12 b d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {(a+b x)^{5/2} (c+d x)^{7/2}}{6 b d}-\frac {(7 a d+5 b c) \left (\frac {(b c-a d) \left (\frac {3 (b c-a d) \left (\frac {(b c-a d) \left (\frac {(a+b x)^{3/2} \sqrt {c+d x}}{2 d}-\frac {3 (b c-a d) \left (\frac {\sqrt {a+b x} \sqrt {c+d x}}{d}-\frac {(b c-a d) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{\sqrt {b} d^{3/2}}\right )}{4 d}\right )}{6 b}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b}\right )}{8 b}+\frac {(a+b x)^{5/2} (c+d x)^{3/2}}{4 b}\right )}{2 b}+\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b}\right )}{12 b d}\)

Input:

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

Output:

((a + b*x)^(5/2)*(c + d*x)^(7/2))/(6*b*d) - ((5*b*c + 7*a*d)*(((a + b*x)^( 
5/2)*(c + d*x)^(5/2))/(5*b) + ((b*c - a*d)*(((a + b*x)^(5/2)*(c + d*x)^(3/ 
2))/(4*b) + (3*(b*c - a*d)*(((a + b*x)^(5/2)*Sqrt[c + d*x])/(3*b) + ((b*c 
- a*d)*(((a + b*x)^(3/2)*Sqrt[c + d*x])/(2*d) - (3*(b*c - a*d)*((Sqrt[a + 
b*x]*Sqrt[c + d*x])/d - ((b*c - a*d)*ArcTanh[(Sqrt[d]*Sqrt[a + b*x])/(Sqrt 
[b]*Sqrt[c + d*x])])/(Sqrt[b]*d^(3/2))))/(4*d)))/(6*b)))/(8*b)))/(2*b)))/( 
12*b*d)
 

Defintions of rubi rules used

rule 60

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[ 
(a + b*x)^(m + 1)*((c + d*x)^n/(b*(m + n + 1))), x] + Simp[n*((b*c - a*d)/( 
b*(m + n + 1)))   Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a, b, 
 c, d}, x] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !Integer 
Q[n] || (GtQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinear 
Q[a, b, c, d, m, n, 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 90

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

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. \(1036\) vs. \(2(265)=530\).

Time = 0.22 (sec) , antiderivative size = 1037, normalized size of antiderivative = 3.29

method	result	size
default	\(\text {Expression too large to display}\)	\(1037\)

Input:

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

Output:

1/15360*(b*x+a)^(1/2)*(d*x+c)^(1/2)*(8896*a*b^4*c*d^4*x^3*((b*x+a)*(d*x+c) 
)^(1/2)*(d*b)^(1/2)+432*a^2*b^3*c*d^4*x^2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1 
/2)+6768*a*b^4*c^2*d^3*x^2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)-544*(d*b)^( 
1/2)*((b*x+a)*(d*x+c))^(1/2)*a^3*b^2*c*d^4*x+696*(d*b)^(1/2)*((b*x+a)*(d*x 
+c))^(1/2)*a^2*b^3*c^2*d^3*x+320*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a*b^4 
*c^3*d^2*x-450*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*b*c*d^5+675*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^4-300*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^3-2 
25*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+270*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^5*c^5*d+3328*a*b^4*d^5*x^4*((b*x+a)*(d*x+c) 
)^(1/2)*(d*b)^(1/2)+6400*b^5*c*d^4*x^4*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2) 
+96*a^2*b^3*d^5*x^3*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+4320*b^5*c^2*d^3*x 
^3*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(1/2)+2560*b^5*d^5*x^5*((b*x+a)*(d*x+c))^ 
(1/2)*(d*b)^(1/2)+105*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^6-75*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^6-210*(d*b)^(1/2)*((b*x+a)*(d*x 
+c))^(1/2)*a^5*d^5+150*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^5*c^5-1092*(d 
*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^3*b^2*c^2*d^3+300*(d*b)^(1/2)*((b*x...
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 894, normalized size of antiderivative = 2.84 \[ \int x (a+b x)^{3/2} (c+d x)^{5/2} \, dx =\text {Too large to display} \] Input:

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

Output:

[-1/30720*(15*(5*b^6*c^6 - 18*a*b^5*c^5*d + 15*a^2*b^4*c^4*d^2 + 20*a^3*b^ 
3*c^3*d^3 - 45*a^4*b^2*c^2*d^4 + 30*a^5*b*c*d^5 - 7*a^6*d^6)*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)*s 
qrt(b*d)*sqrt(b*x + a)*sqrt(d*x + c) + 8*(b^2*c*d + a*b*d^2)*x) - 4*(1280* 
b^6*d^6*x^5 + 75*b^6*c^5*d - 245*a*b^5*c^4*d^2 + 150*a^2*b^4*c^3*d^3 - 546 
*a^3*b^3*c^2*d^4 + 415*a^4*b^2*c*d^5 - 105*a^5*b*d^6 + 128*(25*b^6*c*d^5 + 
 13*a*b^5*d^6)*x^4 + 16*(135*b^6*c^2*d^4 + 278*a*b^5*c*d^5 + 3*a^2*b^4*d^6 
)*x^3 + 8*(5*b^6*c^3*d^3 + 423*a*b^5*c^2*d^4 + 27*a^2*b^4*c*d^5 - 7*a^3*b^ 
3*d^6)*x^2 - 2*(25*b^6*c^4*d^2 - 80*a*b^5*c^3*d^3 - 174*a^2*b^4*c^2*d^4 + 
136*a^3*b^3*c*d^5 - 35*a^4*b^2*d^6)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^5*d 
^4), 1/15360*(15*(5*b^6*c^6 - 18*a*b^5*c^5*d + 15*a^2*b^4*c^4*d^2 + 20*a^3 
*b^3*c^3*d^3 - 45*a^4*b^2*c^2*d^4 + 30*a^5*b*c*d^5 - 7*a^6*d^6)*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*(1280*b^6*d^6*x^5 + 75 
*b^6*c^5*d - 245*a*b^5*c^4*d^2 + 150*a^2*b^4*c^3*d^3 - 546*a^3*b^3*c^2*d^4 
 + 415*a^4*b^2*c*d^5 - 105*a^5*b*d^6 + 128*(25*b^6*c*d^5 + 13*a*b^5*d^6)*x 
^4 + 16*(135*b^6*c^2*d^4 + 278*a*b^5*c*d^5 + 3*a^2*b^4*d^6)*x^3 + 8*(5*b^6 
*c^3*d^3 + 423*a*b^5*c^2*d^4 + 27*a^2*b^4*c*d^5 - 7*a^3*b^3*d^6)*x^2 - 2*( 
25*b^6*c^4*d^2 - 80*a*b^5*c^3*d^3 - 174*a^2*b^4*c^2*d^4 + 136*a^3*b^3*c*d^ 
5 - 35*a^4*b^2*d^6)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^5*d^4)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate(x*(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. 2572 vs. \(2 (265) = 530\).

Time = 0.43 (sec) , antiderivative size = 2572, normalized size of antiderivative = 8.17 \[ \int x (a+b x)^{3/2} (c+d x)^{5/2} \, dx=\text {Too large to display} \] Input:

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

Output:

1/7680*(40*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(b*x + a)*(4*(b*x + a)* 
(6*(b*x + a)/b^3 + (b^12*c*d^5 - 25*a*b^11*d^6)/(b^14*d^6)) - (5*b^13*c^2* 
d^4 + 14*a*b^12*c*d^5 - 163*a^2*b^11*d^6)/(b^14*d^6)) + 3*(5*b^14*c^3*d^3 
+ 9*a*b^13*c^2*d^4 + 15*a^2*b^12*c*d^5 - 93*a^3*b^11*d^6)/(b^14*d^6))*sqrt 
(b*x + a) + 3*(5*b^4*c^4 + 4*a*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 + 20*a^3*b*c* 
d^3 - 35*a^4*d^4)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a 
)*b*d - a*b*d)))/(sqrt(b*d)*b^2*d^3))*c^2*abs(b) + 160*(sqrt(b^2*c + (b*x 
+ a)*b*d - a*b*d)*(2*(b*x + a)*(4*(b*x + a)*(6*(b*x + a)/b^3 + (b^12*c*d^5 
 - 25*a*b^11*d^6)/(b^14*d^6)) - (5*b^13*c^2*d^4 + 14*a*b^12*c*d^5 - 163*a^ 
2*b^11*d^6)/(b^14*d^6)) + 3*(5*b^14*c^3*d^3 + 9*a*b^13*c^2*d^4 + 15*a^2*b^ 
12*c*d^5 - 93*a^3*b^11*d^6)/(b^14*d^6))*sqrt(b*x + a) + 3*(5*b^4*c^4 + 4*a 
*b^3*c^3*d + 6*a^2*b^2*c^2*d^2 + 20*a^3*b*c*d^3 - 35*a^4*d^4)*log(abs(-sqr 
t(b*d)*sqrt(b*x + a) + sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*b^ 
2*d^3))*a*c*d*abs(b)/b + (sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*(4*(2*(b* 
x + a)*(8*(b*x + a)*(10*(b*x + a)/b^5 + (b^30*c*d^9 - 61*a*b^29*d^10)/(b^3 
4*d^10)) - 3*(3*b^31*c^2*d^8 + 14*a*b^30*c*d^9 - 417*a^2*b^29*d^10)/(b^34* 
d^10)) + (21*b^32*c^3*d^7 + 77*a*b^31*c^2*d^8 + 183*a^2*b^30*c*d^9 - 3481* 
a^3*b^29*d^10)/(b^34*d^10))*(b*x + a) - 5*(21*b^33*c^4*d^6 + 56*a*b^32*c^3 
*d^7 + 106*a^2*b^31*c^2*d^8 + 176*a^3*b^30*c*d^9 - 2279*a^4*b^29*d^10)/(b^ 
34*d^10))*(b*x + a) + 15*(21*b^34*c^5*d^5 + 35*a*b^33*c^4*d^6 + 50*a^2*...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.30 (sec) , antiderivative size = 860, normalized size of antiderivative = 2.73 \[ \int x (a+b x)^{3/2} (c+d x)^{5/2} \, dx =\text {Too large to display} \] Input:

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

Output:

( - 105*sqrt(c + d*x)*sqrt(a + b*x)*a**5*b*d**6 + 415*sqrt(c + d*x)*sqrt(a 
 + b*x)*a**4*b**2*c*d**5 + 70*sqrt(c + d*x)*sqrt(a + b*x)*a**4*b**2*d**6*x 
 - 546*sqrt(c + d*x)*sqrt(a + b*x)*a**3*b**3*c**2*d**4 - 272*sqrt(c + d*x) 
*sqrt(a + b*x)*a**3*b**3*c*d**5*x - 56*sqrt(c + d*x)*sqrt(a + b*x)*a**3*b* 
*3*d**6*x**2 + 150*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b**4*c**3*d**3 + 348*s 
qrt(c + d*x)*sqrt(a + b*x)*a**2*b**4*c**2*d**4*x + 216*sqrt(c + d*x)*sqrt( 
a + b*x)*a**2*b**4*c*d**5*x**2 + 48*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b**4* 
d**6*x**3 - 245*sqrt(c + d*x)*sqrt(a + b*x)*a*b**5*c**4*d**2 + 160*sqrt(c 
+ d*x)*sqrt(a + b*x)*a*b**5*c**3*d**3*x + 3384*sqrt(c + d*x)*sqrt(a + b*x) 
*a*b**5*c**2*d**4*x**2 + 4448*sqrt(c + d*x)*sqrt(a + b*x)*a*b**5*c*d**5*x* 
*3 + 1664*sqrt(c + d*x)*sqrt(a + b*x)*a*b**5*d**6*x**4 + 75*sqrt(c + d*x)* 
sqrt(a + b*x)*b**6*c**5*d - 50*sqrt(c + d*x)*sqrt(a + b*x)*b**6*c**4*d**2* 
x + 40*sqrt(c + d*x)*sqrt(a + b*x)*b**6*c**3*d**3*x**2 + 2160*sqrt(c + d*x 
)*sqrt(a + b*x)*b**6*c**2*d**4*x**3 + 3200*sqrt(c + d*x)*sqrt(a + b*x)*b** 
6*c*d**5*x**4 + 1280*sqrt(c + d*x)*sqrt(a + b*x)*b**6*d**6*x**5 + 105*sqrt 
(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - 
 b*c))*a**6*d**6 - 450*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b 
)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**5*b*c*d**5 + 675*sqrt(d)*sqrt(b)*log( 
(sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**4*b**2 
*c**2*d**4 - 300*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*s...