\(\int (a+b x)^{3/2} (c+d x)^{5/2} \, dx\) [332]

Optimal result

Integrand size = 19, antiderivative size = 224 \[ \int (a+b x)^{3/2} (c+d x)^{5/2} \, dx=-\frac {3 (b c-a d)^4 \sqrt {a+b x} \sqrt {c+d x}}{128 b^3 d^2}+\frac {(b c-a d)^3 (a+b x)^{3/2} \sqrt {c+d x}}{64 b^3 d}+\frac {(b c-a d)^2 (a+b x)^{5/2} \sqrt {c+d x}}{16 b^3}+\frac {(b c-a d) (a+b x)^{5/2} (c+d x)^{3/2}}{8 b^2}+\frac {(a+b x)^{5/2} (c+d x)^{5/2}}{5 b}+\frac {3 (b c-a d)^5 \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{128 b^{7/2} d^{5/2}} \] Output:

-3/128*(-a*d+b*c)^4*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^3/d^2+1/64*(-a*d+b*c)^3* 
(b*x+a)^(3/2)*(d*x+c)^(1/2)/b^3/d+1/16*(-a*d+b*c)^2*(b*x+a)^(5/2)*(d*x+c)^ 
(1/2)/b^3+1/8*(-a*d+b*c)*(b*x+a)^(5/2)*(d*x+c)^(3/2)/b^2+1/5*(b*x+a)^(5/2) 
*(d*x+c)^(5/2)/b+3/128*(-a*d+b*c)^5*arctanh(d^(1/2)*(b*x+a)^(1/2)/b^(1/2)/ 
(d*x+c)^(1/2))/b^(7/2)/d^(5/2)
 

Mathematica [A] (verified)

Time = 0.35 (sec) , antiderivative size = 220, normalized size of antiderivative = 0.98 \[ \int (a+b x)^{3/2} (c+d x)^{5/2} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 a^4 d^4-10 a^3 b d^3 (7 c+d x)+2 a^2 b^2 d^2 \left (64 c^2+23 c d x+4 d^2 x^2\right )+2 a b^3 d \left (35 c^3+233 c^2 d x+256 c d^2 x^2+88 d^3 x^3\right )+b^4 \left (-15 c^4+10 c^3 d x+248 c^2 d^2 x^2+336 c d^3 x^3+128 d^4 x^4\right )\right )}{640 b^3 d^2}+\frac {3 (b c-a d)^5 \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{128 b^{7/2} d^{5/2}} \] Input:

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

Output:

(Sqrt[a + b*x]*Sqrt[c + d*x]*(15*a^4*d^4 - 10*a^3*b*d^3*(7*c + d*x) + 2*a^ 
2*b^2*d^2*(64*c^2 + 23*c*d*x + 4*d^2*x^2) + 2*a*b^3*d*(35*c^3 + 233*c^2*d* 
x + 256*c*d^2*x^2 + 88*d^3*x^3) + b^4*(-15*c^4 + 10*c^3*d*x + 248*c^2*d^2* 
x^2 + 336*c*d^3*x^3 + 128*d^4*x^4)))/(640*b^3*d^2) + (3*(b*c - a*d)^5*ArcT 
anh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt[a + b*x])])/(128*b^(7/2)*d^(5/2) 
)
 

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 237, normalized size of antiderivative = 1.06, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.368, Rules used = {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.

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 60

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

\(\Big \downarrow \) 66

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

\(\Big \downarrow \) 221

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

Input:

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

Output:

((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)
 

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 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 [A] (verified)

Time = 0.14 (sec) , antiderivative size = 239, normalized size of antiderivative = 1.07

method	result	size
default	\(\frac {\left (b x +a \right )^{\frac {3}{2}} \left (x d +c \right )^{\frac {7}{2}}}{5 d}-\frac {3 \left (-a d +b c \right ) \left (\frac {\sqrt {b x +a}\, \left (x d +c \right )^{\frac {7}{2}}}{4 d}-\frac {\left (-a d +b c \right ) \left (\frac {\sqrt {b x +a}\, \left (x d +c \right )^{\frac {5}{2}}}{3 b}-\frac {5 \left (a d -b c \right ) \left (\frac {\sqrt {b x +a}\, \left (x d +c \right )^{\frac {3}{2}}}{2 b}-\frac {3 \left (a d -b c \right ) \left (\frac {\sqrt {b x +a}\, \sqrt {x d +c}}{b}-\frac {\left (a d -b c \right ) \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \ln \left (\frac {\frac {1}{2} a d +\frac {1}{2} b c +b d x}{\sqrt {d b}}+\sqrt {b d \,x^{2}+\left (a d +b c \right ) x +a c}\right )}{2 b \sqrt {x d +c}\, \sqrt {b x +a}\, \sqrt {d b}}\right )}{4 b}\right )}{6 b}\right )}{8 d}\right )}{10 d}\)	\(239\)

Input:

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

Output:

1/5/d*(b*x+a)^(3/2)*(d*x+c)^(7/2)-3/10*(-a*d+b*c)/d*(1/4/d*(b*x+a)^(1/2)*( 
d*x+c)^(7/2)-1/8*(-a*d+b*c)/d*(1/3*(b*x+a)^(1/2)*(d*x+c)^(5/2)/b-5/6*(a*d- 
b*c)/b*(1/2*(b*x+a)^(1/2)*(d*x+c)^(3/2)/b-3/4*(a*d-b*c)/b*((b*x+a)^(1/2)*( 
d*x+c)^(1/2)/b-1/2*(a*d-b*c)/b*((b*x+a)*(d*x+c))^(1/2)/(d*x+c)^(1/2)/(b*x+ 
a)^(1/2)*ln((1/2*a*d+1/2*b*c+b*d*x)/(d*b)^(1/2)+(b*d*x^2+(a*d+b*c)*x+a*c)^ 
(1/2))/(d*b)^(1/2)))))
 

Fricas [A] (verification not implemented)

Time = 0.13 (sec) , antiderivative size = 702, normalized size of antiderivative = 3.13 \[ \int (a+b x)^{3/2} (c+d x)^{5/2} \, dx=\left [-\frac {15 \, {\left (b^{5} c^{5} - 5 \, a b^{4} c^{4} d + 10 \, a^{2} b^{3} c^{3} d^{2} - 10 \, a^{3} b^{2} c^{2} d^{3} + 5 \, a^{4} b c d^{4} - a^{5} d^{5}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) - 4 \, {\left (128 \, b^{5} d^{5} x^{4} - 15 \, b^{5} c^{4} d + 70 \, a b^{4} c^{3} d^{2} + 128 \, a^{2} b^{3} c^{2} d^{3} - 70 \, a^{3} b^{2} c d^{4} + 15 \, a^{4} b d^{5} + 16 \, {\left (21 \, b^{5} c d^{4} + 11 \, a b^{4} d^{5}\right )} x^{3} + 8 \, {\left (31 \, b^{5} c^{2} d^{3} + 64 \, a b^{4} c d^{4} + a^{2} b^{3} d^{5}\right )} x^{2} + 2 \, {\left (5 \, b^{5} c^{3} d^{2} + 233 \, a b^{4} c^{2} d^{3} + 23 \, a^{2} b^{3} c d^{4} - 5 \, a^{3} b^{2} d^{5}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{2560 \, b^{4} d^{3}}, -\frac {15 \, {\left (b^{5} c^{5} - 5 \, a b^{4} c^{4} d + 10 \, a^{2} b^{3} c^{3} d^{2} - 10 \, a^{3} b^{2} c^{2} d^{3} + 5 \, a^{4} b c d^{4} - a^{5} d^{5}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) - 2 \, {\left (128 \, b^{5} d^{5} x^{4} - 15 \, b^{5} c^{4} d + 70 \, a b^{4} c^{3} d^{2} + 128 \, a^{2} b^{3} c^{2} d^{3} - 70 \, a^{3} b^{2} c d^{4} + 15 \, a^{4} b d^{5} + 16 \, {\left (21 \, b^{5} c d^{4} + 11 \, a b^{4} d^{5}\right )} x^{3} + 8 \, {\left (31 \, b^{5} c^{2} d^{3} + 64 \, a b^{4} c d^{4} + a^{2} b^{3} d^{5}\right )} x^{2} + 2 \, {\left (5 \, b^{5} c^{3} d^{2} + 233 \, a b^{4} c^{2} d^{3} + 23 \, a^{2} b^{3} c d^{4} - 5 \, a^{3} b^{2} d^{5}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{1280 \, b^{4} d^{3}}\right ] \] Input:

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

Output:

[-1/2560*(15*(b^5*c^5 - 5*a*b^4*c^4*d + 10*a^2*b^3*c^3*d^2 - 10*a^3*b^2*c^ 
2*d^3 + 5*a^4*b*c*d^4 - a^5*d^5)*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*(128*b^5*d^5*x^4 - 15*b^5*c^4*d + 
70*a*b^4*c^3*d^2 + 128*a^2*b^3*c^2*d^3 - 70*a^3*b^2*c*d^4 + 15*a^4*b*d^5 + 
 16*(21*b^5*c*d^4 + 11*a*b^4*d^5)*x^3 + 8*(31*b^5*c^2*d^3 + 64*a*b^4*c*d^4 
 + a^2*b^3*d^5)*x^2 + 2*(5*b^5*c^3*d^2 + 233*a*b^4*c^2*d^3 + 23*a^2*b^3*c* 
d^4 - 5*a^3*b^2*d^5)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^4*d^3), -1/1280*(1 
5*(b^5*c^5 - 5*a*b^4*c^4*d + 10*a^2*b^3*c^3*d^2 - 10*a^3*b^2*c^2*d^3 + 5*a 
^4*b*c*d^4 - a^5*d^5)*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*(128*b^5*d^5*x^4 - 15*b^5*c^4*d + 70*a*b^4*c^3*d^2 + 128*a^2*b^3 
*c^2*d^3 - 70*a^3*b^2*c*d^4 + 15*a^4*b*d^5 + 16*(21*b^5*c*d^4 + 11*a*b^4*d 
^5)*x^3 + 8*(31*b^5*c^2*d^3 + 64*a*b^4*c*d^4 + a^2*b^3*d^5)*x^2 + 2*(5*b^5 
*c^3*d^2 + 233*a*b^4*c^2*d^3 + 23*a^2*b^3*c*d^4 - 5*a^3*b^2*d^5)*x)*sqrt(b 
*x + a)*sqrt(d*x + c))/(b^4*d^3)]
 

Sympy [F]

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

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

Output:

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

Maxima [F(-2)]

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

integrate((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. 1877 vs. \(2 (180) = 360\).

Time = 0.38 (sec) , antiderivative size = 1877, normalized size of antiderivative = 8.38 \[ \int (a+b x)^{3/2} (c+d x)^{5/2} \, dx=\text {Too large to display} \] Input:

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

Output:

-1/1920*(1920*((b^2*c - a*b*d)*log(abs(-sqrt(b*d)*sqrt(b*x + a) + sqrt(b^2 
*c + (b*x + a)*b*d - a*b*d)))/sqrt(b*d) - sqrt(b^2*c + (b*x + a)*b*d - a*b 
*d)*sqrt(b*x + a))*a^2*c^2*abs(b)/b^2 - 20*(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 - 9 
3*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*d* 
abs(b) - 20*(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))*sqr 
t(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))*a*d^2*abs(b)/b - 960*(sqrt(b^2*c + 
(b*x + a)*b*d - a*b*d)*(2*b*x + 2*a + (b*c*d - 5*a*d^2)/d^2)*sqrt(b*x + a) 
 + (b^3*c^2 + 2*a*b^2*c*d - 3*a^2*b*d^2)*log(abs(-sqrt(b*d)*sqrt(b*x + a) 
+ sqrt(b^2*c + (b*x + a)*b*d - a*b*d)))/(sqrt(b*d)*d))*a*c^2*abs(b)/b^2 - 
960*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*(2*b*x + 2*a + (b*c*d - 5*a*d^...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 651, normalized size of antiderivative = 2.91 \[ \int (a+b x)^{3/2} (c+d x)^{5/2} \, dx=\frac {15 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{4} b \,d^{5}-70 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{3} b^{2} c \,d^{4}-10 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{3} b^{2} d^{5} x +128 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{2} b^{3} c^{2} d^{3}+46 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{2} b^{3} c \,d^{4} x +8 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{2} b^{3} d^{5} x^{2}+70 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{4} c^{3} d^{2}+466 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{4} c^{2} d^{3} x +512 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{4} c \,d^{4} x^{2}+176 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{4} d^{5} x^{3}-15 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{5} c^{4} d +10 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{5} c^{3} d^{2} x +248 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{5} c^{2} d^{3} x^{2}+336 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{5} c \,d^{4} x^{3}+128 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{5} d^{5} x^{4}-15 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{5} d^{5}+75 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{4} b c \,d^{4}-150 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{3} b^{2} c^{2} d^{3}+150 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{2} b^{3} c^{3} d^{2}-75 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a \,b^{4} c^{4} d +15 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) b^{5} c^{5}}{640 b^{4} d^{3}} \] Input:

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

Output:

(15*sqrt(c + d*x)*sqrt(a + b*x)*a**4*b*d**5 - 70*sqrt(c + d*x)*sqrt(a + b* 
x)*a**3*b**2*c*d**4 - 10*sqrt(c + d*x)*sqrt(a + b*x)*a**3*b**2*d**5*x + 12 
8*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b**3*c**2*d**3 + 46*sqrt(c + d*x)*sqrt( 
a + b*x)*a**2*b**3*c*d**4*x + 8*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b**3*d**5 
*x**2 + 70*sqrt(c + d*x)*sqrt(a + b*x)*a*b**4*c**3*d**2 + 466*sqrt(c + d*x 
)*sqrt(a + b*x)*a*b**4*c**2*d**3*x + 512*sqrt(c + d*x)*sqrt(a + b*x)*a*b** 
4*c*d**4*x**2 + 176*sqrt(c + d*x)*sqrt(a + b*x)*a*b**4*d**5*x**3 - 15*sqrt 
(c + d*x)*sqrt(a + b*x)*b**5*c**4*d + 10*sqrt(c + d*x)*sqrt(a + b*x)*b**5* 
c**3*d**2*x + 248*sqrt(c + d*x)*sqrt(a + b*x)*b**5*c**2*d**3*x**2 + 336*sq 
rt(c + d*x)*sqrt(a + b*x)*b**5*c*d**4*x**3 + 128*sqrt(c + d*x)*sqrt(a + b* 
x)*b**5*d**5*x**4 - 15*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b 
)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**5*d**5 + 75*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*c*d**4 
- 150*sqrt(d)*sqrt(b)*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**3 + 150*sqrt(d)*sqrt(b)*log((sqrt(d)*sq 
rt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a**2*b**3*c**3*d**2 
- 75*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/s 
qrt(a*d - b*c))*a*b**4*c**4*d + 15*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b 
*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*b**5*c**5)/(640*b**4*d**3)