\(\int \frac {(A+B x) (c+d x)^3}{(a+b x^2)^{3/2}} \, dx\) [179]

Optimal result

Integrand size = 24, antiderivative size = 156 \[ \int \frac {(A+B x) (c+d x)^3}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {(c+d x)^2 (a (B c+A d)-(A b c-a B d) x)}{a b \sqrt {a+b x^2}}-\frac {d \left (4 \left (A b c^2-3 a B c d-a A d^2\right )+d (2 A b c-3 a B d) x\right ) \sqrt {a+b x^2}}{2 a b^2}-\frac {3 d \left (a B d^2-2 b c (B c+A d)\right ) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{2 b^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.85 (sec) , antiderivative size = 161, normalized size of antiderivative = 1.03 \[ \int \frac {(A+B x) (c+d x)^3}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {2 A b^2 c^3 x+a^2 d^2 (4 A d+3 B (4 c+d x))+a b \left (2 A d \left (-3 c^2-3 c d x+d^2 x^2\right )+B \left (-2 c^3-6 c^2 d x+6 c d^2 x^2+d^3 x^3\right )\right )}{2 a b^2 \sqrt {a+b x^2}}+\frac {3 d \left (a B d^2-2 b c (B c+A d)\right ) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{2 b^{5/2}} \] Input:

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

Output:

(2*A*b^2*c^3*x + a^2*d^2*(4*A*d + 3*B*(4*c + d*x)) + a*b*(2*A*d*(-3*c^2 - 
3*c*d*x + d^2*x^2) + B*(-2*c^3 - 6*c^2*d*x + 6*c*d^2*x^2 + d^3*x^3)))/(2*a 
*b^2*Sqrt[a + b*x^2]) + (3*d*(a*B*d^2 - 2*b*c*(B*c + A*d))*Log[-(Sqrt[b]*x 
) + Sqrt[a + b*x^2]])/(2*b^(5/2))
 

Rubi [A] (verified)

Time = 0.33 (sec) , antiderivative size = 173, normalized size of antiderivative = 1.11, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {684, 27, 676, 224, 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.

\(\displaystyle \int \frac {(A+B x) (c+d x)^3}{\left (a+b x^2\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 684

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 676

\(\displaystyle \frac {d \left (-\frac {3 a \left (a B d^2-2 b c (A d+B c)\right ) \int \frac {1}{\sqrt {b x^2+a}}dx}{2 b}-\frac {2 \sqrt {a+b x^2} \left (-a A d^2-3 a B c d+A b c^2\right )}{b}-\frac {d x \sqrt {a+b x^2} (2 A b c-3 a B d)}{2 b}\right )}{a b}-\frac {(c+d x)^2 (a (A d+B c)-x (A b c-a B d))}{a b \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {d \left (-\frac {3 a \left (a B d^2-2 b c (A d+B c)\right ) \int \frac {1}{1-\frac {b x^2}{b x^2+a}}d\frac {x}{\sqrt {b x^2+a}}}{2 b}-\frac {2 \sqrt {a+b x^2} \left (-a A d^2-3 a B c d+A b c^2\right )}{b}-\frac {d x \sqrt {a+b x^2} (2 A b c-3 a B d)}{2 b}\right )}{a b}-\frac {(c+d x)^2 (a (A d+B c)-x (A b c-a B d))}{a b \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {d \left (-\frac {3 a \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right ) \left (a B d^2-2 b c (A d+B c)\right )}{2 b^{3/2}}-\frac {2 \sqrt {a+b x^2} \left (-a A d^2-3 a B c d+A b c^2\right )}{b}-\frac {d x \sqrt {a+b x^2} (2 A b c-3 a B d)}{2 b}\right )}{a b}-\frac {(c+d x)^2 (a (A d+B c)-x (A b c-a B d))}{a b \sqrt {a+b x^2}}\)

Input:

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

Output:

-(((c + d*x)^2*(a*(B*c + A*d) - (A*b*c - a*B*d)*x))/(a*b*Sqrt[a + b*x^2])) 
 + (d*((-2*(A*b*c^2 - 3*a*B*c*d - a*A*d^2)*Sqrt[a + b*x^2])/b - (d*(2*A*b* 
c - 3*a*B*d)*x*Sqrt[a + b*x^2])/(2*b) - (3*a*(a*B*d^2 - 2*b*c*(B*c + A*d)) 
*ArcTanh[(Sqrt[b]*x)/Sqrt[a + b*x^2]])/(2*b^(3/2))))/(a*b)
 

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 219

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])
 

rule 224

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], 
x, x/Sqrt[a + b*x^2]] /; FreeQ[{a, b}, x] &&  !GtQ[a, 0]
 

rule 676

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x 
_Symbol] :> Simp[(e*f + d*g)*((a + c*x^2)^(p + 1)/(2*c*(p + 1))), x] + (Sim 
p[e*g*x*((a + c*x^2)^(p + 1)/(c*(2*p + 3))), x] - Simp[(a*e*g - c*d*f*(2*p 
+ 3))/(c*(2*p + 3))   Int[(a + c*x^2)^p, x], x]) /; FreeQ[{a, c, d, e, f, g 
, p}, x] &&  !LeQ[p, -1]
 

rule 684

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p 
_.), x_Symbol] :> Simp[(d + e*x)^(m - 1)*(a + c*x^2)^(p + 1)*((a*(e*f + d*g 
) - (c*d*f - a*e*g)*x)/(2*a*c*(p + 1))), x] - Simp[1/(2*a*c*(p + 1))   Int[ 
(d + e*x)^(m - 2)*(a + c*x^2)^(p + 1)*Simp[a*e*(e*f*(m - 1) + d*g*m) - c*d^ 
2*f*(2*p + 3) + e*(a*e*g*m - c*d*f*(m + 2*p + 2))*x, x], x], x] /; FreeQ[{a 
, c, d, e, f, g}, x] && LtQ[p, -1] && GtQ[m, 1] && (EqQ[d, 0] || (EqQ[m, 2] 
 && EqQ[p, -3] && RationalQ[a, c, d, e, f, g]) ||  !ILtQ[m + 2*p + 3, 0])
 

Maple [A] (verified)

Time = 1.19 (sec) , antiderivative size = 179, normalized size of antiderivative = 1.15

method	result	size
risch	\(\frac {d^{2} \left (B d x +2 A d +6 B c \right ) \sqrt {b \,x^{2}+a}}{2 b^{2}}+\frac {3 b d \left (2 A b c d -a B \,d^{2}+2 B b \,c^{2}\right ) \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )+\frac {2 A \,d^{3} a -6 A b \,c^{2} d +6 a B c \,d^{2}-2 b B \,c^{3}}{\sqrt {b \,x^{2}+a}}+\frac {2 A \,b^{2} c^{3} x}{a \sqrt {b \,x^{2}+a}}-\frac {a B \,d^{3} x}{\sqrt {b \,x^{2}+a}}}{2 b^{2}}\)	\(179\)
default	\(\frac {A \,c^{3} x}{\sqrt {b \,x^{2}+a}\, a}+d^{2} \left (A d +3 B c \right ) \left (\frac {x^{2}}{b \sqrt {b \,x^{2}+a}}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )+3 c d \left (A d +B c \right ) \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )-\frac {c^{2} \left (3 A d +B c \right )}{b \sqrt {b \,x^{2}+a}}+B \,d^{3} \left (\frac {x^{3}}{2 b \sqrt {b \,x^{2}+a}}-\frac {3 a \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )}{2 b}\right )\)	\(201\)

Input:

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

Output:

1/2*d^2*(B*d*x+2*A*d+6*B*c)*(b*x^2+a)^(1/2)/b^2+1/2/b^2*(3*b*d*(2*A*b*c*d- 
B*a*d^2+2*B*b*c^2)*(-x/b/(b*x^2+a)^(1/2)+1/b^(3/2)*ln(b^(1/2)*x+(b*x^2+a)^ 
(1/2)))+2*(A*a*d^3-3*A*b*c^2*d+3*B*a*c*d^2-B*b*c^3)/(b*x^2+a)^(1/2)+2*A*b^ 
2*c^3*x/a/(b*x^2+a)^(1/2)-a*B*d^3*x/(b*x^2+a)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 508, normalized size of antiderivative = 3.26 \[ \int \frac {(A+B x) (c+d x)^3}{\left (a+b x^2\right )^{3/2}} \, dx=\left [-\frac {3 \, {\left (2 \, B a^{2} b c^{2} d + 2 \, A a^{2} b c d^{2} - B a^{3} d^{3} + {\left (2 \, B a b^{2} c^{2} d + 2 \, A a b^{2} c d^{2} - B a^{2} b d^{3}\right )} x^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) - 2 \, {\left (B a b^{2} d^{3} x^{3} - 2 \, B a b^{2} c^{3} - 6 \, A a b^{2} c^{2} d + 12 \, B a^{2} b c d^{2} + 4 \, A a^{2} b d^{3} + 2 \, {\left (3 \, B a b^{2} c d^{2} + A a b^{2} d^{3}\right )} x^{2} + {\left (2 \, A b^{3} c^{3} - 6 \, B a b^{2} c^{2} d - 6 \, A a b^{2} c d^{2} + 3 \, B a^{2} b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{4 \, {\left (a b^{4} x^{2} + a^{2} b^{3}\right )}}, -\frac {3 \, {\left (2 \, B a^{2} b c^{2} d + 2 \, A a^{2} b c d^{2} - B a^{3} d^{3} + {\left (2 \, B a b^{2} c^{2} d + 2 \, A a b^{2} c d^{2} - B a^{2} b d^{3}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (B a b^{2} d^{3} x^{3} - 2 \, B a b^{2} c^{3} - 6 \, A a b^{2} c^{2} d + 12 \, B a^{2} b c d^{2} + 4 \, A a^{2} b d^{3} + 2 \, {\left (3 \, B a b^{2} c d^{2} + A a b^{2} d^{3}\right )} x^{2} + {\left (2 \, A b^{3} c^{3} - 6 \, B a b^{2} c^{2} d - 6 \, A a b^{2} c d^{2} + 3 \, B a^{2} b d^{3}\right )} x\right )} \sqrt {b x^{2} + a}}{2 \, {\left (a b^{4} x^{2} + a^{2} b^{3}\right )}}\right ] \] Input:

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

Output:

[-1/4*(3*(2*B*a^2*b*c^2*d + 2*A*a^2*b*c*d^2 - B*a^3*d^3 + (2*B*a*b^2*c^2*d 
 + 2*A*a*b^2*c*d^2 - B*a^2*b*d^3)*x^2)*sqrt(b)*log(-2*b*x^2 + 2*sqrt(b*x^2 
 + a)*sqrt(b)*x - a) - 2*(B*a*b^2*d^3*x^3 - 2*B*a*b^2*c^3 - 6*A*a*b^2*c^2* 
d + 12*B*a^2*b*c*d^2 + 4*A*a^2*b*d^3 + 2*(3*B*a*b^2*c*d^2 + A*a*b^2*d^3)*x 
^2 + (2*A*b^3*c^3 - 6*B*a*b^2*c^2*d - 6*A*a*b^2*c*d^2 + 3*B*a^2*b*d^3)*x)* 
sqrt(b*x^2 + a))/(a*b^4*x^2 + a^2*b^3), -1/2*(3*(2*B*a^2*b*c^2*d + 2*A*a^2 
*b*c*d^2 - B*a^3*d^3 + (2*B*a*b^2*c^2*d + 2*A*a*b^2*c*d^2 - B*a^2*b*d^3)*x 
^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (B*a*b^2*d^3*x^3 - 2*B*a 
*b^2*c^3 - 6*A*a*b^2*c^2*d + 12*B*a^2*b*c*d^2 + 4*A*a^2*b*d^3 + 2*(3*B*a*b 
^2*c*d^2 + A*a*b^2*d^3)*x^2 + (2*A*b^3*c^3 - 6*B*a*b^2*c^2*d - 6*A*a*b^2*c 
*d^2 + 3*B*a^2*b*d^3)*x)*sqrt(b*x^2 + a))/(a*b^4*x^2 + a^2*b^3)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.04 (sec) , antiderivative size = 228, normalized size of antiderivative = 1.46 \[ \int \frac {(A+B x) (c+d x)^3}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {B d^{3} x^{3}}{2 \, \sqrt {b x^{2} + a} b} + \frac {A c^{3} x}{\sqrt {b x^{2} + a} a} + \frac {3 \, B a d^{3} x}{2 \, \sqrt {b x^{2} + a} b^{2}} - \frac {3 \, B a d^{3} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{2 \, b^{\frac {5}{2}}} - \frac {B c^{3}}{\sqrt {b x^{2} + a} b} - \frac {3 \, A c^{2} d}{\sqrt {b x^{2} + a} b} + \frac {{\left (3 \, B c d^{2} + A d^{3}\right )} x^{2}}{\sqrt {b x^{2} + a} b} - \frac {3 \, {\left (B c^{2} d + A c d^{2}\right )} x}{\sqrt {b x^{2} + a} b} + \frac {3 \, {\left (B c^{2} d + A c d^{2}\right )} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} + \frac {2 \, {\left (3 \, B c d^{2} + A d^{3}\right )} a}{\sqrt {b x^{2} + a} b^{2}} \] Input:

integrate((B*x+A)*(d*x+c)^3/(b*x^2+a)^(3/2),x, algorithm="maxima")
 

Output:

1/2*B*d^3*x^3/(sqrt(b*x^2 + a)*b) + A*c^3*x/(sqrt(b*x^2 + a)*a) + 3/2*B*a* 
d^3*x/(sqrt(b*x^2 + a)*b^2) - 3/2*B*a*d^3*arcsinh(b*x/sqrt(a*b))/b^(5/2) - 
 B*c^3/(sqrt(b*x^2 + a)*b) - 3*A*c^2*d/(sqrt(b*x^2 + a)*b) + (3*B*c*d^2 + 
A*d^3)*x^2/(sqrt(b*x^2 + a)*b) - 3*(B*c^2*d + A*c*d^2)*x/(sqrt(b*x^2 + a)* 
b) + 3*(B*c^2*d + A*c*d^2)*arcsinh(b*x/sqrt(a*b))/b^(3/2) + 2*(3*B*c*d^2 + 
 A*d^3)*a/(sqrt(b*x^2 + a)*b^2)
 

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 209, normalized size of antiderivative = 1.34 \[ \int \frac {(A+B x) (c+d x)^3}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left ({\left (\frac {B d^{3} x}{b} + \frac {2 \, {\left (3 \, B a b^{3} c d^{2} + A a b^{3} d^{3}\right )}}{a b^{4}}\right )} x + \frac {2 \, A b^{4} c^{3} - 6 \, B a b^{3} c^{2} d - 6 \, A a b^{3} c d^{2} + 3 \, B a^{2} b^{2} d^{3}}{a b^{4}}\right )} x - \frac {2 \, {\left (B a b^{3} c^{3} + 3 \, A a b^{3} c^{2} d - 6 \, B a^{2} b^{2} c d^{2} - 2 \, A a^{2} b^{2} d^{3}\right )}}{a b^{4}}}{2 \, \sqrt {b x^{2} + a}} - \frac {3 \, {\left (2 \, B b c^{2} d + 2 \, A b c d^{2} - B a d^{3}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{2 \, b^{\frac {5}{2}}} \] Input:

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

Output:

1/2*(((B*d^3*x/b + 2*(3*B*a*b^3*c*d^2 + A*a*b^3*d^3)/(a*b^4))*x + (2*A*b^4 
*c^3 - 6*B*a*b^3*c^2*d - 6*A*a*b^3*c*d^2 + 3*B*a^2*b^2*d^3)/(a*b^4))*x - 2 
*(B*a*b^3*c^3 + 3*A*a*b^3*c^2*d - 6*B*a^2*b^2*c*d^2 - 2*A*a^2*b^2*d^3)/(a* 
b^4))/sqrt(b*x^2 + a) - 3/2*(2*B*b*c^2*d + 2*A*b*c*d^2 - B*a*d^3)*log(abs( 
-sqrt(b)*x + sqrt(b*x^2 + a)))/b^(5/2)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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