Integrand size = 26, antiderivative size = 238 \[ \int \frac {(A+B x) (d+e x)^3}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {2 (d+e x)^2 \left (A b c d+\left (2 A c^2 d+b^2 B e-b c (B d+A e)\right ) x\right )}{b^2 c \sqrt {b x+c x^2}}+\frac {e \left (32 A c^3 d^2-15 b^3 B e^2+12 b^2 c e (3 B d+A e)-8 b c^2 d (2 B d+3 A e)+2 c e \left (8 A c^2 d+5 b^2 B e-4 b c (B d+A e)\right ) x\right ) \sqrt {b x+c x^2}}{4 b^2 c^3}+\frac {3 e \left (4 A c e (2 c d-b e)+B \left (8 c^2 d^2-12 b c d e+5 b^2 e^2\right )\right ) \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right )}{4 c^{7/2}} \]
3/4*e*(4*A*c*e*(-b*e+2*c*d)+B*(5*b^2*e^2-12*b*c*d*e+8*c^2*d^2))*arctanh(x* c^(1/2)/(c*x^2+b*x)^(1/2))/c^(7/2)-2*(e*x+d)^2*(A*b*c*d+(2*A*c^2*d+b^2*B*e -b*c*(A*e+B*d))*x)/b^2/c/(c*x^2+b*x)^(1/2)+1/4*e*(32*A*c^3*d^2-15*b^3*B*e^ 2+12*b^2*c*e*(A*e+3*B*d)-8*b*c^2*d*(3*A*e+2*B*d)+2*c*e*(8*A*c^2*d+5*b^2*B* e-4*b*c*(A*e+B*d))*x)*(c*x^2+b*x)^(1/2)/b^2/c^3
Time = 0.95 (sec) , antiderivative size = 287, normalized size of antiderivative = 1.21 \[ \int \frac {(A+B x) (d+e x)^3}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {\sqrt {c} \left (4 A c \left (-4 c^3 d^3 x+3 b^3 e^3 x-2 b c^2 d^2 (d-3 e x)+b^2 c e^2 x (-6 d+e x)\right )+b B x \left (8 c^3 d^3-15 b^3 e^3+b^2 c e^2 (36 d-5 e x)+2 b c^2 e \left (-12 d^2+6 d e x+e^2 x^2\right )\right )\right )+24 b^3 c e^2 (3 B d+A e) \sqrt {x} \sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{\sqrt {b}-\sqrt {b+c x}}\right )+6 b^2 e \left (8 B c^2 d^2+8 A c^2 d e+5 b^2 B e^2\right ) \sqrt {x} \sqrt {b+c x} \text {arctanh}\left (\frac {\sqrt {c} \sqrt {x}}{-\sqrt {b}+\sqrt {b+c x}}\right )}{4 b^2 c^{7/2} \sqrt {x (b+c x)}} \]
(Sqrt[c]*(4*A*c*(-4*c^3*d^3*x + 3*b^3*e^3*x - 2*b*c^2*d^2*(d - 3*e*x) + b^ 2*c*e^2*x*(-6*d + e*x)) + b*B*x*(8*c^3*d^3 - 15*b^3*e^3 + b^2*c*e^2*(36*d - 5*e*x) + 2*b*c^2*e*(-12*d^2 + 6*d*e*x + e^2*x^2))) + 24*b^3*c*e^2*(3*B*d + A*e)*Sqrt[x]*Sqrt[b + c*x]*ArcTanh[(Sqrt[c]*Sqrt[x])/(Sqrt[b] - Sqrt[b + c*x])] + 6*b^2*e*(8*B*c^2*d^2 + 8*A*c^2*d*e + 5*b^2*B*e^2)*Sqrt[x]*Sqrt[ b + c*x]*ArcTanh[(Sqrt[c]*Sqrt[x])/(-Sqrt[b] + Sqrt[b + c*x])])/(4*b^2*c^( 7/2)*Sqrt[x*(b + c*x)])
Time = 0.45 (sec) , antiderivative size = 245, normalized size of antiderivative = 1.03, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {1233, 27, 1225, 1091, 219}
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.
\(\displaystyle \int \frac {(A+B x) (d+e x)^3}{\left (b x+c x^2\right )^{3/2}} \, dx\) |
\(\Big \downarrow \) 1233 |
\(\displaystyle \frac {2 \int \frac {e (d+e x) \left (b (b B+4 A c) d+\left (5 B e b^2-4 c (B d+A e) b+8 A c^2 d\right ) x\right )}{2 \sqrt {c x^2+b x}}dx}{b^2 c}-\frac {2 (d+e x)^2 \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{b^2 c \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {e \int \frac {(d+e x) \left (b (b B+4 A c) d+\left (5 B e b^2-4 c (B d+A e) b+8 A c^2 d\right ) x\right )}{\sqrt {c x^2+b x}}dx}{b^2 c}-\frac {2 (d+e x)^2 \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{b^2 c \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 1225 |
\(\displaystyle \frac {e \left (\frac {3 b^2 \left (4 A c e (2 c d-b e)+B \left (5 b^2 e^2-12 b c d e+8 c^2 d^2\right )\right ) \int \frac {1}{\sqrt {c x^2+b x}}dx}{8 c^2}+\frac {\sqrt {b x+c x^2} \left (2 c e x \left (-4 b c (A e+B d)+8 A c^2 d+5 b^2 B e\right )+12 b^2 c e (A e+3 B d)-8 b c^2 d (3 A e+2 B d)+32 A c^3 d^2-15 b^3 B e^2\right )}{4 c^2}\right )}{b^2 c}-\frac {2 (d+e x)^2 \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{b^2 c \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 1091 |
\(\displaystyle \frac {e \left (\frac {3 b^2 \left (4 A c e (2 c d-b e)+B \left (5 b^2 e^2-12 b c d e+8 c^2 d^2\right )\right ) \int \frac {1}{1-\frac {c x^2}{c x^2+b x}}d\frac {x}{\sqrt {c x^2+b x}}}{4 c^2}+\frac {\sqrt {b x+c x^2} \left (2 c e x \left (-4 b c (A e+B d)+8 A c^2 d+5 b^2 B e\right )+12 b^2 c e (A e+3 B d)-8 b c^2 d (3 A e+2 B d)+32 A c^3 d^2-15 b^3 B e^2\right )}{4 c^2}\right )}{b^2 c}-\frac {2 (d+e x)^2 \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{b^2 c \sqrt {b x+c x^2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {e \left (\frac {3 b^2 \text {arctanh}\left (\frac {\sqrt {c} x}{\sqrt {b x+c x^2}}\right ) \left (4 A c e (2 c d-b e)+B \left (5 b^2 e^2-12 b c d e+8 c^2 d^2\right )\right )}{4 c^{5/2}}+\frac {\sqrt {b x+c x^2} \left (2 c e x \left (-4 b c (A e+B d)+8 A c^2 d+5 b^2 B e\right )+12 b^2 c e (A e+3 B d)-8 b c^2 d (3 A e+2 B d)+32 A c^3 d^2-15 b^3 B e^2\right )}{4 c^2}\right )}{b^2 c}-\frac {2 (d+e x)^2 \left (x \left (-b c (A e+B d)+2 A c^2 d+b^2 B e\right )+A b c d\right )}{b^2 c \sqrt {b x+c x^2}}\) |
(-2*(d + e*x)^2*(A*b*c*d + (2*A*c^2*d + b^2*B*e - b*c*(B*d + A*e))*x))/(b^ 2*c*Sqrt[b*x + c*x^2]) + (e*(((32*A*c^3*d^2 - 15*b^3*B*e^2 + 12*b^2*c*e*(3 *B*d + A*e) - 8*b*c^2*d*(2*B*d + 3*A*e) + 2*c*e*(8*A*c^2*d + 5*b^2*B*e - 4 *b*c*(B*d + A*e))*x)*Sqrt[b*x + c*x^2])/(4*c^2) + (3*b^2*(4*A*c*e*(2*c*d - b*e) + B*(8*c^2*d^2 - 12*b*c*d*e + 5*b^2*e^2))*ArcTanh[(Sqrt[c]*x)/Sqrt[b *x + c*x^2]])/(4*c^(5/2))))/(b^2*c)
3.12.99.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(b_.)*(x_) + (c_.)*(x_)^2], x_Symbol] :> Simp[2 Subst[Int[1/(1 - c*x^2), x], x, x/Sqrt[b*x + c*x^2]], x] /; FreeQ[{b, c}, x]
Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c_.)*( x_)^2)^(p_), x_Symbol] :> Simp[(-(b*e*g*(p + 2) - c*(e*f + d*g)*(2*p + 3) - 2*c*e*g*(p + 1)*x))*((a + b*x + c*x^2)^(p + 1)/(2*c^2*(p + 1)*(2*p + 3))), x] + Simp[(b^2*e*g*(p + 2) - 2*a*c*e*g + c*(2*c*d*f - b*(e*f + d*g))*(2*p + 3))/(2*c^2*(2*p + 3)) Int[(a + b*x + c*x^2)^p, x], x] /; FreeQ[{a, b, c , d, e, f, g, p}, x] && !LeQ[p, -1]
Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_.) + (b_.)*(x_) + (c _.)*(x_)^2)^(p_.), x_Symbol] :> Simp[(-(d + e*x)^(m - 1))*(a + b*x + c*x^2) ^(p + 1)*((2*a*c*(e*f + d*g) - b*(c*d*f + a*e*g) - (2*c^2*d*f + b^2*e*g - c *(b*e*f + b*d*g + 2*a*e*g))*x)/(c*(p + 1)*(b^2 - 4*a*c))), x] - Simp[1/(c*( p + 1)*(b^2 - 4*a*c)) Int[(d + e*x)^(m - 2)*(a + b*x + c*x^2)^(p + 1)*Sim p[2*c^2*d^2*f*(2*p + 3) + b*e*g*(a*e*(m - 1) + b*d*(p + 2)) - c*(2*a*e*(e*f *(m - 1) + d*g*m) + b*d*(d*g*(2*p + 3) - e*f*(m - 2*p - 4))) + e*(b^2*e*g*( m + p + 1) + 2*c^2*d*f*(m + 2*p + 2) - c*(2*a*e*g*m + b*(e*f + d*g)*(m + 2* p + 2)))*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && LtQ[p, -1] && GtQ[m, 1] && ((EqQ[m, 2] && EqQ[p, -3] && RationalQ[a, b, c, d, e, f, g]) | | !ILtQ[m + 2*p + 3, 0])
Time = 0.51 (sec) , antiderivative size = 218, normalized size of antiderivative = 0.92
method | result | size |
pseudoelliptic | \(-\frac {4 \left (\frac {3 \sqrt {x \left (c x +b \right )}\, \left (\left (-2 A d e -2 B \,d^{2}\right ) c^{2}+b e \left (A e +3 B d \right ) c -\frac {5 B \,b^{2} e^{2}}{4}\right ) e \,b^{2} \operatorname {arctanh}\left (\frac {\sqrt {x \left (c x +b \right )}}{x \sqrt {c}}\right )}{4}+\left (-\frac {B \,b^{2} c^{2} e^{3} x^{3}}{8}-\frac {c \left (\left (A e +3 B d \right ) c -\frac {5 B b e}{4}\right ) e^{2} b^{2} x^{2}}{4}+\left (A \,c^{4} d^{3}-\frac {3 \left (A e +\frac {B d}{3}\right ) d^{2} b \,c^{3}}{2}+\frac {3 b^{2} c^{2} d e \left (A e +B d \right )}{2}-\frac {3 b^{3} c \,e^{2} \left (A e +3 B d \right )}{4}+\frac {15 b^{4} B \,e^{3}}{16}\right ) x +\frac {A b \,c^{3} d^{3}}{2}\right ) \sqrt {c}\right )}{\sqrt {x \left (c x +b \right )}\, c^{\frac {7}{2}} b^{2}}\) | \(218\) |
risch | \(\frac {\left (c x +b \right ) \left (2 B \,b^{2} c \,e^{3} x^{2}+4 A \,b^{2} c \,e^{3} x -7 B \,b^{3} e^{3} x +12 B \,b^{2} c d \,e^{2} x -8 A \,c^{3} d^{3}\right )}{4 b^{2} \sqrt {x \left (c x +b \right )}\, c^{3}}-\frac {-\frac {15 B \,b^{3} e^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{\sqrt {c}}+12 A \,b^{2} \sqrt {c}\, e^{3} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )-24 A b \,c^{\frac {3}{2}} d \,e^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )+36 B \,b^{2} \sqrt {c}\, d \,e^{2} \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )-24 B b \,c^{\frac {3}{2}} d^{2} e \ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )+\frac {2 \left (-8 A \,b^{3} c \,e^{3}+24 A \,b^{2} c^{2} d \,e^{2}-24 A b \,c^{3} d^{2} e +8 A \,c^{4} d^{3}+8 b^{4} B \,e^{3}-24 b^{3} B c d \,e^{2}+24 b^{2} B \,c^{2} d^{2} e -8 B b \,c^{3} d^{3}\right ) \sqrt {c \left (x +\frac {b}{c}\right )^{2}-b \left (x +\frac {b}{c}\right )}}{c b \left (x +\frac {b}{c}\right )}}{8 c^{3} b}\) | \(396\) |
default | \(-\frac {2 A \,d^{3} \left (2 c x +b \right )}{b^{2} \sqrt {c \,x^{2}+b x}}+B \,e^{3} \left (\frac {x^{3}}{2 c \sqrt {c \,x^{2}+b x}}-\frac {5 b \left (\frac {x^{2}}{c \sqrt {c \,x^{2}+b x}}-\frac {3 b \left (-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}\right )}{2 c}\right )}{4 c}\right )+\left (A \,e^{3}+3 B d \,e^{2}\right ) \left (\frac {x^{2}}{c \sqrt {c \,x^{2}+b x}}-\frac {3 b \left (-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}\right )}{2 c}\right )+\left (3 A \,d^{2} e +B \,d^{3}\right ) \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )+\left (3 A d \,e^{2}+3 B \,d^{2} e \right ) \left (-\frac {x}{c \sqrt {c \,x^{2}+b x}}-\frac {b \left (-\frac {1}{c \sqrt {c \,x^{2}+b x}}+\frac {2 c x +b}{b c \sqrt {c \,x^{2}+b x}}\right )}{2 c}+\frac {\ln \left (\frac {\frac {b}{2}+c x}{\sqrt {c}}+\sqrt {c \,x^{2}+b x}\right )}{c^{\frac {3}{2}}}\right )\) | \(473\) |
-4/(x*(c*x+b))^(1/2)*(3/4*(x*(c*x+b))^(1/2)*((-2*A*d*e-2*B*d^2)*c^2+b*e*(A *e+3*B*d)*c-5/4*B*b^2*e^2)*e*b^2*arctanh((x*(c*x+b))^(1/2)/x/c^(1/2))+(-1/ 8*B*b^2*c^2*e^3*x^3-1/4*c*((A*e+3*B*d)*c-5/4*B*b*e)*e^2*b^2*x^2+(A*c^4*d^3 -3/2*(A*e+1/3*B*d)*d^2*b*c^3+3/2*b^2*c^2*d*e*(A*e+B*d)-3/4*b^3*c*e^2*(A*e+ 3*B*d)+15/16*b^4*B*e^3)*x+1/2*A*b*c^3*d^3)*c^(1/2))/c^(7/2)/b^2
Time = 0.41 (sec) , antiderivative size = 694, normalized size of antiderivative = 2.92 \[ \int \frac {(A+B x) (d+e x)^3}{\left (b x+c x^2\right )^{3/2}} \, dx=\left [\frac {3 \, {\left ({\left (8 \, B b^{2} c^{3} d^{2} e - 4 \, {\left (3 \, B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} d e^{2} + {\left (5 \, B b^{4} c - 4 \, A b^{3} c^{2}\right )} e^{3}\right )} x^{2} + {\left (8 \, B b^{3} c^{2} d^{2} e - 4 \, {\left (3 \, B b^{4} c - 2 \, A b^{3} c^{2}\right )} d e^{2} + {\left (5 \, B b^{5} - 4 \, A b^{4} c\right )} e^{3}\right )} x\right )} \sqrt {c} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right ) + 2 \, {\left (2 \, B b^{2} c^{3} e^{3} x^{3} - 8 \, A b c^{4} d^{3} + {\left (12 \, B b^{2} c^{3} d e^{2} - {\left (5 \, B b^{3} c^{2} - 4 \, A b^{2} c^{3}\right )} e^{3}\right )} x^{2} + {\left (8 \, {\left (B b c^{4} - 2 \, A c^{5}\right )} d^{3} - 24 \, {\left (B b^{2} c^{3} - A b c^{4}\right )} d^{2} e + 12 \, {\left (3 \, B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} d e^{2} - 3 \, {\left (5 \, B b^{4} c - 4 \, A b^{3} c^{2}\right )} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{8 \, {\left (b^{2} c^{5} x^{2} + b^{3} c^{4} x\right )}}, -\frac {3 \, {\left ({\left (8 \, B b^{2} c^{3} d^{2} e - 4 \, {\left (3 \, B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} d e^{2} + {\left (5 \, B b^{4} c - 4 \, A b^{3} c^{2}\right )} e^{3}\right )} x^{2} + {\left (8 \, B b^{3} c^{2} d^{2} e - 4 \, {\left (3 \, B b^{4} c - 2 \, A b^{3} c^{2}\right )} d e^{2} + {\left (5 \, B b^{5} - 4 \, A b^{4} c\right )} e^{3}\right )} x\right )} \sqrt {-c} \arctan \left (\frac {\sqrt {c x^{2} + b x} \sqrt {-c}}{c x}\right ) - {\left (2 \, B b^{2} c^{3} e^{3} x^{3} - 8 \, A b c^{4} d^{3} + {\left (12 \, B b^{2} c^{3} d e^{2} - {\left (5 \, B b^{3} c^{2} - 4 \, A b^{2} c^{3}\right )} e^{3}\right )} x^{2} + {\left (8 \, {\left (B b c^{4} - 2 \, A c^{5}\right )} d^{3} - 24 \, {\left (B b^{2} c^{3} - A b c^{4}\right )} d^{2} e + 12 \, {\left (3 \, B b^{3} c^{2} - 2 \, A b^{2} c^{3}\right )} d e^{2} - 3 \, {\left (5 \, B b^{4} c - 4 \, A b^{3} c^{2}\right )} e^{3}\right )} x\right )} \sqrt {c x^{2} + b x}}{4 \, {\left (b^{2} c^{5} x^{2} + b^{3} c^{4} x\right )}}\right ] \]
[1/8*(3*((8*B*b^2*c^3*d^2*e - 4*(3*B*b^3*c^2 - 2*A*b^2*c^3)*d*e^2 + (5*B*b ^4*c - 4*A*b^3*c^2)*e^3)*x^2 + (8*B*b^3*c^2*d^2*e - 4*(3*B*b^4*c - 2*A*b^3 *c^2)*d*e^2 + (5*B*b^5 - 4*A*b^4*c)*e^3)*x)*sqrt(c)*log(2*c*x + b + 2*sqrt (c*x^2 + b*x)*sqrt(c)) + 2*(2*B*b^2*c^3*e^3*x^3 - 8*A*b*c^4*d^3 + (12*B*b^ 2*c^3*d*e^2 - (5*B*b^3*c^2 - 4*A*b^2*c^3)*e^3)*x^2 + (8*(B*b*c^4 - 2*A*c^5 )*d^3 - 24*(B*b^2*c^3 - A*b*c^4)*d^2*e + 12*(3*B*b^3*c^2 - 2*A*b^2*c^3)*d* e^2 - 3*(5*B*b^4*c - 4*A*b^3*c^2)*e^3)*x)*sqrt(c*x^2 + b*x))/(b^2*c^5*x^2 + b^3*c^4*x), -1/4*(3*((8*B*b^2*c^3*d^2*e - 4*(3*B*b^3*c^2 - 2*A*b^2*c^3)* d*e^2 + (5*B*b^4*c - 4*A*b^3*c^2)*e^3)*x^2 + (8*B*b^3*c^2*d^2*e - 4*(3*B*b ^4*c - 2*A*b^3*c^2)*d*e^2 + (5*B*b^5 - 4*A*b^4*c)*e^3)*x)*sqrt(-c)*arctan( sqrt(c*x^2 + b*x)*sqrt(-c)/(c*x)) - (2*B*b^2*c^3*e^3*x^3 - 8*A*b*c^4*d^3 + (12*B*b^2*c^3*d*e^2 - (5*B*b^3*c^2 - 4*A*b^2*c^3)*e^3)*x^2 + (8*(B*b*c^4 - 2*A*c^5)*d^3 - 24*(B*b^2*c^3 - A*b*c^4)*d^2*e + 12*(3*B*b^3*c^2 - 2*A*b^ 2*c^3)*d*e^2 - 3*(5*B*b^4*c - 4*A*b^3*c^2)*e^3)*x)*sqrt(c*x^2 + b*x))/(b^2 *c^5*x^2 + b^3*c^4*x)]
\[ \int \frac {(A+B x) (d+e x)^3}{\left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {\left (A + B x\right ) \left (d + e x\right )^{3}}{\left (x \left (b + c x\right )\right )^{\frac {3}{2}}}\, dx \]
Time = 0.20 (sec) , antiderivative size = 367, normalized size of antiderivative = 1.54 \[ \int \frac {(A+B x) (d+e x)^3}{\left (b x+c x^2\right )^{3/2}} \, dx=\frac {B e^{3} x^{3}}{2 \, \sqrt {c x^{2} + b x} c} - \frac {5 \, B b e^{3} x^{2}}{4 \, \sqrt {c x^{2} + b x} c^{2}} + \frac {2 \, B d^{3} x}{\sqrt {c x^{2} + b x} b} - \frac {4 \, A c d^{3} x}{\sqrt {c x^{2} + b x} b^{2}} + \frac {6 \, A d^{2} e x}{\sqrt {c x^{2} + b x} b} - \frac {15 \, B b^{2} e^{3} x}{4 \, \sqrt {c x^{2} + b x} c^{3}} + \frac {15 \, B b^{2} e^{3} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{8 \, c^{\frac {7}{2}}} - \frac {2 \, A d^{3}}{\sqrt {c x^{2} + b x} b} + \frac {{\left (3 \, B d e^{2} + A e^{3}\right )} x^{2}}{\sqrt {c x^{2} + b x} c} + \frac {3 \, {\left (3 \, B d e^{2} + A e^{3}\right )} b x}{\sqrt {c x^{2} + b x} c^{2}} - \frac {6 \, {\left (B d^{2} e + A d e^{2}\right )} x}{\sqrt {c x^{2} + b x} c} - \frac {3 \, {\left (3 \, B d e^{2} + A e^{3}\right )} b \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{2 \, c^{\frac {5}{2}}} + \frac {3 \, {\left (B d^{2} e + A d e^{2}\right )} \log \left (2 \, c x + b + 2 \, \sqrt {c x^{2} + b x} \sqrt {c}\right )}{c^{\frac {3}{2}}} \]
1/2*B*e^3*x^3/(sqrt(c*x^2 + b*x)*c) - 5/4*B*b*e^3*x^2/(sqrt(c*x^2 + b*x)*c ^2) + 2*B*d^3*x/(sqrt(c*x^2 + b*x)*b) - 4*A*c*d^3*x/(sqrt(c*x^2 + b*x)*b^2 ) + 6*A*d^2*e*x/(sqrt(c*x^2 + b*x)*b) - 15/4*B*b^2*e^3*x/(sqrt(c*x^2 + b*x )*c^3) + 15/8*B*b^2*e^3*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(7/ 2) - 2*A*d^3/(sqrt(c*x^2 + b*x)*b) + (3*B*d*e^2 + A*e^3)*x^2/(sqrt(c*x^2 + b*x)*c) + 3*(3*B*d*e^2 + A*e^3)*b*x/(sqrt(c*x^2 + b*x)*c^2) - 6*(B*d^2*e + A*d*e^2)*x/(sqrt(c*x^2 + b*x)*c) - 3/2*(3*B*d*e^2 + A*e^3)*b*log(2*c*x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(5/2) + 3*(B*d^2*e + A*d*e^2)*log(2*c* x + b + 2*sqrt(c*x^2 + b*x)*sqrt(c))/c^(3/2)
Time = 0.29 (sec) , antiderivative size = 258, normalized size of antiderivative = 1.08 \[ \int \frac {(A+B x) (d+e x)^3}{\left (b x+c x^2\right )^{3/2}} \, dx=-\frac {\frac {8 \, A d^{3}}{b} - {\left ({\left (\frac {2 \, B e^{3} x}{c} + \frac {12 \, B b^{2} c^{2} d e^{2} - 5 \, B b^{3} c e^{3} + 4 \, A b^{2} c^{2} e^{3}}{b^{2} c^{3}}\right )} x + \frac {8 \, B b c^{3} d^{3} - 16 \, A c^{4} d^{3} - 24 \, B b^{2} c^{2} d^{2} e + 24 \, A b c^{3} d^{2} e + 36 \, B b^{3} c d e^{2} - 24 \, A b^{2} c^{2} d e^{2} - 15 \, B b^{4} e^{3} + 12 \, A b^{3} c e^{3}}{b^{2} c^{3}}\right )} x}{4 \, \sqrt {c x^{2} + b x}} - \frac {3 \, {\left (8 \, B c^{2} d^{2} e - 12 \, B b c d e^{2} + 8 \, A c^{2} d e^{2} + 5 \, B b^{2} e^{3} - 4 \, A b c e^{3}\right )} \log \left ({\left | 2 \, {\left (\sqrt {c} x - \sqrt {c x^{2} + b x}\right )} \sqrt {c} + b \right |}\right )}{8 \, c^{\frac {7}{2}}} \]
-1/4*(8*A*d^3/b - ((2*B*e^3*x/c + (12*B*b^2*c^2*d*e^2 - 5*B*b^3*c*e^3 + 4* A*b^2*c^2*e^3)/(b^2*c^3))*x + (8*B*b*c^3*d^3 - 16*A*c^4*d^3 - 24*B*b^2*c^2 *d^2*e + 24*A*b*c^3*d^2*e + 36*B*b^3*c*d*e^2 - 24*A*b^2*c^2*d*e^2 - 15*B*b ^4*e^3 + 12*A*b^3*c*e^3)/(b^2*c^3))*x)/sqrt(c*x^2 + b*x) - 3/8*(8*B*c^2*d^ 2*e - 12*B*b*c*d*e^2 + 8*A*c^2*d*e^2 + 5*B*b^2*e^3 - 4*A*b*c*e^3)*log(abs( 2*(sqrt(c)*x - sqrt(c*x^2 + b*x))*sqrt(c) + b))/c^(7/2)
Timed out. \[ \int \frac {(A+B x) (d+e x)^3}{\left (b x+c x^2\right )^{3/2}} \, dx=\int \frac {\left (A+B\,x\right )\,{\left (d+e\,x\right )}^3}{{\left (c\,x^2+b\,x\right )}^{3/2}} \,d x \]