\(\int \frac {x^3}{\sqrt {a+b x} \sqrt {c+d x}} \, dx\) [341]

Optimal result

Integrand size = 22, antiderivative size = 191 \[ \int \frac {x^3}{\sqrt {a+b x} \sqrt {c+d x}} \, dx=\frac {\left (5 b^2 c^2+8 a b c d+11 a^2 d^2\right ) \sqrt {a+b x} \sqrt {c+d x}}{8 b^3 d^3}-\frac {(5 b c+13 a d) (a+b x)^{3/2} \sqrt {c+d x}}{12 b^3 d^2}+\frac {(a+b x)^{5/2} \sqrt {c+d x}}{3 b^3 d}-\frac {(b c+a d) \left (5 b^2 c^2-2 a b c d+5 a^2 d^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{8 b^{7/2} d^{7/2}} \] Output:

1/8*(11*a^2*d^2+8*a*b*c*d+5*b^2*c^2)*(b*x+a)^(1/2)*(d*x+c)^(1/2)/b^3/d^3-1 
/12*(13*a*d+5*b*c)*(b*x+a)^(3/2)*(d*x+c)^(1/2)/b^3/d^2+1/3*(b*x+a)^(5/2)*( 
d*x+c)^(1/2)/b^3/d-1/8*(a*d+b*c)*(5*a^2*d^2-2*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 = 0.37 (sec) , antiderivative size = 155, normalized size of antiderivative = 0.81 \[ \int \frac {x^3}{\sqrt {a+b x} \sqrt {c+d x}} \, dx=\frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 a^2 d^2+2 a b d (7 c-5 d x)+b^2 \left (15 c^2-10 c d x+8 d^2 x^2\right )\right )}{24 b^3 d^3}-\frac {\left (5 b^3 c^3+3 a b^2 c^2 d+3 a^2 b c d^2+5 a^3 d^3\right ) \text {arctanh}\left (\frac {\sqrt {b} \sqrt {c+d x}}{\sqrt {d} \sqrt {a+b x}}\right )}{8 b^{7/2} d^{7/2}} \] Input:

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

Output:

(Sqrt[a + b*x]*Sqrt[c + d*x]*(15*a^2*d^2 + 2*a*b*d*(7*c - 5*d*x) + b^2*(15 
*c^2 - 10*c*d*x + 8*d^2*x^2)))/(24*b^3*d^3) - ((5*b^3*c^3 + 3*a*b^2*c^2*d 
+ 3*a^2*b*c*d^2 + 5*a^3*d^3)*ArcTanh[(Sqrt[b]*Sqrt[c + d*x])/(Sqrt[d]*Sqrt 
[a + b*x])])/(8*b^(7/2)*d^(7/2))
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 180, normalized size of antiderivative = 0.94, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.227, Rules used = {111, 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.

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

\(\Big \downarrow \) 111

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 164

\(\displaystyle \frac {x^2 \sqrt {a+b x} \sqrt {c+d x}}{3 b d}-\frac {\frac {3 (a d+b c) \left (5 a^2 d^2-2 a b c d+5 b^2 c^2\right ) \int \frac {1}{\sqrt {a+b x} \sqrt {c+d x}}dx}{8 b^2 d^2}-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 a^2 d^2-10 b d x (a d+b c)+14 a b c d+15 b^2 c^2\right )}{4 b^2 d^2}}{6 b d}\)

\(\Big \downarrow \) 66

\(\displaystyle \frac {x^2 \sqrt {a+b x} \sqrt {c+d x}}{3 b d}-\frac {\frac {3 (a d+b c) \left (5 a^2 d^2-2 a b c d+5 b^2 c^2\right ) \int \frac {1}{b-\frac {d (a+b x)}{c+d x}}d\frac {\sqrt {a+b x}}{\sqrt {c+d x}}}{4 b^2 d^2}-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 a^2 d^2-10 b d x (a d+b c)+14 a b c d+15 b^2 c^2\right )}{4 b^2 d^2}}{6 b d}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {x^2 \sqrt {a+b x} \sqrt {c+d x}}{3 b d}-\frac {\frac {3 (a d+b c) \left (5 a^2 d^2-2 a b c d+5 b^2 c^2\right ) \text {arctanh}\left (\frac {\sqrt {d} \sqrt {a+b x}}{\sqrt {b} \sqrt {c+d x}}\right )}{4 b^{5/2} d^{5/2}}-\frac {\sqrt {a+b x} \sqrt {c+d x} \left (15 a^2 d^2-10 b d x (a d+b c)+14 a b c d+15 b^2 c^2\right )}{4 b^2 d^2}}{6 b d}\)

Input:

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

Output:

(x^2*Sqrt[a + b*x]*Sqrt[c + d*x])/(3*b*d) - (-1/4*(Sqrt[a + b*x]*Sqrt[c + 
d*x]*(15*b^2*c^2 + 14*a*b*c*d + 15*a^2*d^2 - 10*b*d*(b*c + a*d)*x))/(b^2*d 
^2) + (3*(b*c + a*d)*(5*b^2*c^2 - 2*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)))/(6*b*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 111

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

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 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. \(394\) vs. \(2(159)=318\).

Time = 0.27 (sec) , antiderivative size = 395, normalized size of antiderivative = 2.07

method	result	size
default	\(-\frac {\left (-16 b^{2} d^{2} x^{2} \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{3} d^{3}+9 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a^{2} b c \,d^{2}+9 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) a \,b^{2} c^{2} d +15 \ln \left (\frac {2 b d x +2 \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, \sqrt {d b}+a d +b c}{2 \sqrt {d b}}\right ) b^{3} c^{3}+20 \sqrt {d b}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, a b \,d^{2} x +20 \sqrt {d b}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, b^{2} c d x -30 \sqrt {d b}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, a^{2} d^{2}-28 \sqrt {d b}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, a b c d -30 \sqrt {d b}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}\, b^{2} c^{2}\right ) \sqrt {b x +a}\, \sqrt {x d +c}}{48 b^{3} d^{3} \sqrt {d b}\, \sqrt {\left (b x +a \right ) \left (x d +c \right )}}\)	\(395\)

Input:

int(x^3/(b*x+a)^(1/2)/(d*x+c)^(1/2),x,method=_RETURNVERBOSE)
 

Output:

-1/48*(-16*b^2*d^2*x^2*((b*x+a)*(d*x+c))^(1/2)*(d*b)^(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^3*d^3+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^2*b*c*d^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*b^2*c^2*d+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^3*c^3+20*(d*b)^(1/2)*((b*x+a)*(d*x+c) 
)^(1/2)*a*b*d^2*x+20*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^2*c*d*x-30*(d*b 
)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*a^2*d^2-28*(d*b)^(1/2)*((b*x+a)*(d*x+c))^( 
1/2)*a*b*c*d-30*(d*b)^(1/2)*((b*x+a)*(d*x+c))^(1/2)*b^2*c^2)*(b*x+a)^(1/2) 
*(d*x+c)^(1/2)/b^3/d^3/(d*b)^(1/2)/((b*x+a)*(d*x+c))^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.16 (sec) , antiderivative size = 410, normalized size of antiderivative = 2.15 \[ \int \frac {x^3}{\sqrt {a+b x} \sqrt {c+d x}} \, dx=\left [\frac {3 \, {\left (5 \, b^{3} c^{3} + 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3}\right )} \sqrt {b d} \log \left (8 \, b^{2} d^{2} x^{2} + b^{2} c^{2} + 6 \, a b c d + a^{2} d^{2} - 4 \, {\left (2 \, b d x + b c + a d\right )} \sqrt {b d} \sqrt {b x + a} \sqrt {d x + c} + 8 \, {\left (b^{2} c d + a b d^{2}\right )} x\right ) + 4 \, {\left (8 \, b^{3} d^{3} x^{2} + 15 \, b^{3} c^{2} d + 14 \, a b^{2} c d^{2} + 15 \, a^{2} b d^{3} - 10 \, {\left (b^{3} c d^{2} + a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{96 \, b^{4} d^{4}}, \frac {3 \, {\left (5 \, b^{3} c^{3} + 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3}\right )} \sqrt {-b d} \arctan \left (\frac {{\left (2 \, b d x + b c + a d\right )} \sqrt {-b d} \sqrt {b x + a} \sqrt {d x + c}}{2 \, {\left (b^{2} d^{2} x^{2} + a b c d + {\left (b^{2} c d + a b d^{2}\right )} x\right )}}\right ) + 2 \, {\left (8 \, b^{3} d^{3} x^{2} + 15 \, b^{3} c^{2} d + 14 \, a b^{2} c d^{2} + 15 \, a^{2} b d^{3} - 10 \, {\left (b^{3} c d^{2} + a b^{2} d^{3}\right )} x\right )} \sqrt {b x + a} \sqrt {d x + c}}{48 \, b^{4} d^{4}}\right ] \] Input:

integrate(x^3/(b*x+a)^(1/2)/(d*x+c)^(1/2),x, algorithm="fricas")
 

Output:

[1/96*(3*(5*b^3*c^3 + 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 + 5*a^3*d^3)*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*(8 
*b^3*d^3*x^2 + 15*b^3*c^2*d + 14*a*b^2*c*d^2 + 15*a^2*b*d^3 - 10*(b^3*c*d^ 
2 + a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt(d*x + c))/(b^4*d^4), 1/48*(3*(5*b^3*c 
^3 + 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 + 5*a^3*d^3)*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*(8*b^3*d^3*x^2 + 15*b^3*c^2*d + 14*a*b 
^2*c*d^2 + 15*a^2*b*d^3 - 10*(b^3*c*d^2 + a*b^2*d^3)*x)*sqrt(b*x + a)*sqrt 
(d*x + c))/(b^4*d^4)]
                                                                                    
                                                                                    
 

Sympy [F]

\[ \int \frac {x^3}{\sqrt {a+b x} \sqrt {c+d x}} \, dx=\int \frac {x^{3}}{\sqrt {a + b x} \sqrt {c + d x}}\, dx \] Input:

integrate(x**3/(b*x+a)**(1/2)/(d*x+c)**(1/2),x)
 

Output:

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

Maxima [F(-2)]

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

integrate(x^3/(b*x+a)^(1/2)/(d*x+c)^(1/2),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 [A] (verification not implemented)

Time = 0.14 (sec) , antiderivative size = 215, normalized size of antiderivative = 1.13 \[ \int \frac {x^3}{\sqrt {a+b x} \sqrt {c+d x}} \, dx=\frac {{\left (\sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \sqrt {b x + a} {\left (2 \, {\left (b x + a\right )} {\left (\frac {4 \, {\left (b x + a\right )}}{b^{4} d} - \frac {5 \, b^{12} c d^{3} + 13 \, a b^{11} d^{4}}{b^{15} d^{5}}\right )} + \frac {3 \, {\left (5 \, b^{13} c^{2} d^{2} + 8 \, a b^{12} c d^{3} + 11 \, a^{2} b^{11} d^{4}\right )}}{b^{15} d^{5}}\right )} + \frac {3 \, {\left (5 \, b^{3} c^{3} + 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} + 5 \, a^{3} d^{3}\right )} \log \left ({\left | -\sqrt {b d} \sqrt {b x + a} + \sqrt {b^{2} c + {\left (b x + a\right )} b d - a b d} \right |}\right )}{\sqrt {b d} b^{3} d^{3}}\right )} b}{24 \, {\left | b \right |}} \] Input:

integrate(x^3/(b*x+a)^(1/2)/(d*x+c)^(1/2),x, algorithm="giac")
 

Output:

1/24*(sqrt(b^2*c + (b*x + a)*b*d - a*b*d)*sqrt(b*x + a)*(2*(b*x + a)*(4*(b 
*x + a)/(b^4*d) - (5*b^12*c*d^3 + 13*a*b^11*d^4)/(b^15*d^5)) + 3*(5*b^13*c 
^2*d^2 + 8*a*b^12*c*d^3 + 11*a^2*b^11*d^4)/(b^15*d^5)) + 3*(5*b^3*c^3 + 3* 
a*b^2*c^2*d + 3*a^2*b*c*d^2 + 5*a^3*d^3)*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^3*d^3))*b/abs(b)
 

Mupad [B] (verification not implemented)

Time = 26.26 (sec) , antiderivative size = 900, normalized size of antiderivative = 4.71 \[ \int \frac {x^3}{\sqrt {a+b x} \sqrt {c+d x}} \, dx =\text {Too large to display} \] Input:

int(x^3/((a + b*x)^(1/2)*(c + d*x)^(1/2)),x)
 

Output:

- ((((a + b*x)^(1/2) - a^(1/2))^3*((85*b^4*c^3)/12 + (85*a^3*b*d^3)/12 + ( 
17*a^2*b^2*c*d^2)/4 + (17*a*b^3*c^2*d)/4))/(d^8*((c + d*x)^(1/2) - c^(1/2) 
)^3) - (((a + b*x)^(1/2) - a^(1/2))*((5*b^5*c^3)/4 + (5*a^3*b^2*d^3)/4 + ( 
3*a^2*b^3*c*d^2)/4 + (3*a*b^4*c^2*d)/4))/(d^9*((c + d*x)^(1/2) - c^(1/2))) 
 - (((a + b*x)^(1/2) - a^(1/2))^5*((33*a^3*d^3)/2 + (33*b^3*c^3)/2 + (327* 
a*b^2*c^2*d)/2 + (327*a^2*b*c*d^2)/2))/(d^7*((c + d*x)^(1/2) - c^(1/2))^5) 
 + (64*a^(3/2)*c^(3/2)*((a + b*x)^(1/2) - a^(1/2))^8)/(d^4*((c + d*x)^(1/2 
) - c^(1/2))^8) - (((a + b*x)^(1/2) - a^(1/2))^11*((5*a^3*d^3)/4 + (5*b^3* 
c^3)/4 + (3*a*b^2*c^2*d)/4 + (3*a^2*b*c*d^2)/4))/(b^3*d^4*((c + d*x)^(1/2) 
 - c^(1/2))^11) + (((a + b*x)^(1/2) - a^(1/2))^9*((85*a^3*d^3)/12 + (85*b^ 
3*c^3)/12 + (17*a*b^2*c^2*d)/4 + (17*a^2*b*c*d^2)/4))/(b^2*d^5*((c + d*x)^ 
(1/2) - c^(1/2))^9) - (((a + b*x)^(1/2) - a^(1/2))^7*((33*a^3*d^3)/2 + (33 
*b^3*c^3)/2 + (327*a*b^2*c^2*d)/2 + (327*a^2*b*c*d^2)/2))/(b*d^6*((c + d*x 
)^(1/2) - c^(1/2))^7) + (a^(1/2)*c^(1/2)*((a + b*x)^(1/2) - a^(1/2))^6*(12 
8*a^2*d^2 + 128*b^2*c^2 + (896*a*b*c*d)/3))/(d^6*((c + d*x)^(1/2) - c^(1/2 
))^6) + (64*a^(3/2)*b^2*c^(3/2)*((a + b*x)^(1/2) - a^(1/2))^4)/(d^6*((c + 
d*x)^(1/2) - c^(1/2))^4))/(((a + b*x)^(1/2) - a^(1/2))^12/((c + d*x)^(1/2) 
 - c^(1/2))^12 + b^6/d^6 - (6*b^5*((a + b*x)^(1/2) - a^(1/2))^2)/(d^5*((c 
+ d*x)^(1/2) - c^(1/2))^2) + (15*b^4*((a + b*x)^(1/2) - a^(1/2))^4)/(d^4*( 
(c + d*x)^(1/2) - c^(1/2))^4) - (20*b^3*((a + b*x)^(1/2) - a^(1/2))^6)/...
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 320, normalized size of antiderivative = 1.68 \[ \int \frac {x^3}{\sqrt {a+b x} \sqrt {c+d x}} \, dx=\frac {15 \sqrt {d x +c}\, \sqrt {b x +a}\, a^{2} b \,d^{3}+14 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{2} c \,d^{2}-10 \sqrt {d x +c}\, \sqrt {b x +a}\, a \,b^{2} d^{3} x +15 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{3} c^{2} d -10 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{3} c \,d^{2} x +8 \sqrt {d x +c}\, \sqrt {b x +a}\, b^{3} d^{3} x^{2}-15 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{3} d^{3}-9 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a^{2} b c \,d^{2}-9 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) a \,b^{2} c^{2} d -15 \sqrt {d}\, \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {d}\, \sqrt {b x +a}+\sqrt {b}\, \sqrt {d x +c}}{\sqrt {a d -b c}}\right ) b^{3} c^{3}}{24 b^{4} d^{4}} \] Input:

int(x^3/(b*x+a)^(1/2)/(d*x+c)^(1/2),x)
 

Output:

(15*sqrt(c + d*x)*sqrt(a + b*x)*a**2*b*d**3 + 14*sqrt(c + d*x)*sqrt(a + b* 
x)*a*b**2*c*d**2 - 10*sqrt(c + d*x)*sqrt(a + b*x)*a*b**2*d**3*x + 15*sqrt( 
c + d*x)*sqrt(a + b*x)*b**3*c**2*d - 10*sqrt(c + d*x)*sqrt(a + b*x)*b**3*c 
*d**2*x + 8*sqrt(c + d*x)*sqrt(a + b*x)*b**3*d**3*x**2 - 15*sqrt(d)*sqrt(b 
)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a** 
3*d**3 - 9*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt(a + b*x) + sqrt(b)*sqrt(c + d 
*x))/sqrt(a*d - b*c))*a**2*b*c*d**2 - 9*sqrt(d)*sqrt(b)*log((sqrt(d)*sqrt( 
a + b*x) + sqrt(b)*sqrt(c + d*x))/sqrt(a*d - b*c))*a*b**2*c**2*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**3*c**3)/(24*b**4*d**4)