\(\int \frac {(A+B x) (c+d x)^2}{(a+b x^2)^{3/2}} \, dx\) [180]

Optimal result

Integrand size = 24, antiderivative size = 112 \[ \int \frac {(A+B x) (c+d x)^2}{\left (a+b x^2\right )^{3/2}} \, dx=-\frac {(c+d x) (a (B c+A d)-(A b c-a B d) x)}{a b \sqrt {a+b x^2}}-\frac {d (A b c-2 a B d) \sqrt {a+b x^2}}{a b^2}+\frac {d (2 B c+A d) \text {arctanh}\left (\frac {\sqrt {b} x}{\sqrt {a+b x^2}}\right )}{b^{3/2}} \] Output:

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

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 109, normalized size of antiderivative = 0.97 \[ \int \frac {(A+B x) (c+d x)^2}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {2 a^2 B d^2+A b^2 c^2 x-a b \left (A d (2 c+d x)+B \left (c^2+2 c d x-d^2 x^2\right )\right )}{a b^2 \sqrt {a+b x^2}}-\frac {d (2 B c+A d) \log \left (-\sqrt {b} x+\sqrt {a+b x^2}\right )}{b^{3/2}} \] Input:

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

Output:

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

Rubi [A] (verified)

Time = 0.25 (sec) , antiderivative size = 117, normalized size of antiderivative = 1.04, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {684, 27, 455, 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.

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

\(\Big \downarrow \) 684

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 455

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

-(((c + d*x)*(a*(B*c + A*d) - (A*b*c - a*B*d)*x))/(a*b*Sqrt[a + b*x^2])) + 
 (d*(-(((A*b*c - 2*a*B*d)*Sqrt[a + b*x^2])/b) + (a*(2*B*c + A*d)*ArcTanh[( 
Sqrt[b]*x)/Sqrt[a + b*x^2]])/Sqrt[b]))/(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 455

Int[((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_.), x_Symbol] :> Simp[d*(( 
a + b*x^2)^(p + 1)/(2*b*(p + 1))), x] + Simp[c   Int[(a + b*x^2)^p, x], x] 
/; FreeQ[{a, b, c, d, 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 = 1.05 (sec) , antiderivative size = 124, normalized size of antiderivative = 1.11

method	result	size
risch	\(\frac {d^{2} B \sqrt {b \,x^{2}+a}}{b^{2}}+\frac {b d \left (A d +2 B c \right ) \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )-\frac {2 A b c d -a B \,d^{2}+B b \,c^{2}}{b \sqrt {b \,x^{2}+a}}+\frac {A b \,c^{2} x}{a \sqrt {b \,x^{2}+a}}}{b}\)	\(124\)
default	\(\frac {A \,c^{2} x}{\sqrt {b \,x^{2}+a}\, a}+d \left (A d +2 B c \right ) \left (-\frac {x}{b \sqrt {b \,x^{2}+a}}+\frac {\ln \left (\sqrt {b}\, x +\sqrt {b \,x^{2}+a}\right )}{b^{\frac {3}{2}}}\right )-\frac {c \left (2 A d +B c \right )}{b \sqrt {b \,x^{2}+a}}+B \,d^{2} \left (\frac {x^{2}}{b \sqrt {b \,x^{2}+a}}+\frac {2 a}{b^{2} \sqrt {b \,x^{2}+a}}\right )\)	\(126\)

Input:

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

Output:

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

Fricas [A] (verification not implemented)

Time = 0.10 (sec) , antiderivative size = 312, normalized size of antiderivative = 2.79 \[ \int \frac {(A+B x) (c+d x)^2}{\left (a+b x^2\right )^{3/2}} \, dx=\left [\frac {{\left (2 \, B a^{2} c d + A a^{2} d^{2} + {\left (2 \, B a b c d + A a b d^{2}\right )} x^{2}\right )} \sqrt {b} \log \left (-2 \, b x^{2} - 2 \, \sqrt {b x^{2} + a} \sqrt {b} x - a\right ) + 2 \, {\left (B a b d^{2} x^{2} - B a b c^{2} - 2 \, A a b c d + 2 \, B a^{2} d^{2} + {\left (A b^{2} c^{2} - 2 \, B a b c d - A a b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{2 \, {\left (a b^{3} x^{2} + a^{2} b^{2}\right )}}, -\frac {{\left (2 \, B a^{2} c d + A a^{2} d^{2} + {\left (2 \, B a b c d + A a b d^{2}\right )} x^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} x}{\sqrt {b x^{2} + a}}\right ) - {\left (B a b d^{2} x^{2} - B a b c^{2} - 2 \, A a b c d + 2 \, B a^{2} d^{2} + {\left (A b^{2} c^{2} - 2 \, B a b c d - A a b d^{2}\right )} x\right )} \sqrt {b x^{2} + a}}{a b^{3} x^{2} + a^{2} b^{2}}\right ] \] Input:

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

Output:

[1/2*((2*B*a^2*c*d + A*a^2*d^2 + (2*B*a*b*c*d + A*a*b*d^2)*x^2)*sqrt(b)*lo 
g(-2*b*x^2 - 2*sqrt(b*x^2 + a)*sqrt(b)*x - a) + 2*(B*a*b*d^2*x^2 - B*a*b*c 
^2 - 2*A*a*b*c*d + 2*B*a^2*d^2 + (A*b^2*c^2 - 2*B*a*b*c*d - A*a*b*d^2)*x)* 
sqrt(b*x^2 + a))/(a*b^3*x^2 + a^2*b^2), -((2*B*a^2*c*d + A*a^2*d^2 + (2*B* 
a*b*c*d + A*a*b*d^2)*x^2)*sqrt(-b)*arctan(sqrt(-b)*x/sqrt(b*x^2 + a)) - (B 
*a*b*d^2*x^2 - B*a*b*c^2 - 2*A*a*b*c*d + 2*B*a^2*d^2 + (A*b^2*c^2 - 2*B*a* 
b*c*d - A*a*b*d^2)*x)*sqrt(b*x^2 + a))/(a*b^3*x^2 + a^2*b^2)]
 

Sympy [F]

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

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

Output:

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

Maxima [A] (verification not implemented)

Time = 0.06 (sec) , antiderivative size = 143, normalized size of antiderivative = 1.28 \[ \int \frac {(A+B x) (c+d x)^2}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {B d^{2} x^{2}}{\sqrt {b x^{2} + a} b} + \frac {A c^{2} x}{\sqrt {b x^{2} + a} a} - \frac {B c^{2}}{\sqrt {b x^{2} + a} b} - \frac {2 \, A c d}{\sqrt {b x^{2} + a} b} + \frac {2 \, B a d^{2}}{\sqrt {b x^{2} + a} b^{2}} - \frac {{\left (2 \, B c d + A d^{2}\right )} x}{\sqrt {b x^{2} + a} b} + \frac {{\left (2 \, B c d + A d^{2}\right )} \operatorname {arsinh}\left (\frac {b x}{\sqrt {a b}}\right )}{b^{\frac {3}{2}}} \] Input:

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

Output:

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

Giac [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 130, normalized size of antiderivative = 1.16 \[ \int \frac {(A+B x) (c+d x)^2}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {{\left (\frac {B d^{2} x}{b} + \frac {A b^{3} c^{2} - 2 \, B a b^{2} c d - A a b^{2} d^{2}}{a b^{3}}\right )} x - \frac {B a b^{2} c^{2} + 2 \, A a b^{2} c d - 2 \, B a^{2} b d^{2}}{a b^{3}}}{\sqrt {b x^{2} + a}} - \frac {{\left (2 \, B c d + A d^{2}\right )} \log \left ({\left | -\sqrt {b} x + \sqrt {b x^{2} + a} \right |}\right )}{b^{\frac {3}{2}}} \] Input:

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [B] (verification not implemented)

Time = 0.32 (sec) , antiderivative size = 303, normalized size of antiderivative = 2.71 \[ \int \frac {(A+B x) (c+d x)^2}{\left (a+b x^2\right )^{3/2}} \, dx=\frac {-2 \sqrt {b \,x^{2}+a}\, a b c d -\sqrt {b \,x^{2}+a}\, a b \,d^{2} x +2 \sqrt {b \,x^{2}+a}\, a b \,d^{2}+\sqrt {b \,x^{2}+a}\, b^{2} c^{2} x -\sqrt {b \,x^{2}+a}\, b^{2} c^{2}-2 \sqrt {b \,x^{2}+a}\, b^{2} c d x +\sqrt {b \,x^{2}+a}\, b^{2} d^{2} x^{2}+\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a^{2} d^{2}+2 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b c d +\sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) a b \,d^{2} x^{2}+2 \sqrt {b}\, \mathrm {log}\left (\frac {\sqrt {b \,x^{2}+a}+\sqrt {b}\, x}{\sqrt {a}}\right ) b^{2} c d \,x^{2}-\sqrt {b}\, a^{2} d^{2}+\sqrt {b}\, a b \,c^{2}-2 \sqrt {b}\, a b c d -\sqrt {b}\, a b \,d^{2} x^{2}+\sqrt {b}\, b^{2} c^{2} x^{2}-2 \sqrt {b}\, b^{2} c d \,x^{2}}{b^{2} \left (b \,x^{2}+a \right )} \] Input:

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

Output:

( - 2*sqrt(a + b*x**2)*a*b*c*d - sqrt(a + b*x**2)*a*b*d**2*x + 2*sqrt(a + 
b*x**2)*a*b*d**2 + sqrt(a + b*x**2)*b**2*c**2*x - sqrt(a + b*x**2)*b**2*c* 
*2 - 2*sqrt(a + b*x**2)*b**2*c*d*x + sqrt(a + b*x**2)*b**2*d**2*x**2 + sqr 
t(b)*log((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a**2*d**2 + 2*sqrt(b)*log 
((sqrt(a + b*x**2) + sqrt(b)*x)/sqrt(a))*a*b*c*d + sqrt(b)*log((sqrt(a + b 
*x**2) + sqrt(b)*x)/sqrt(a))*a*b*d**2*x**2 + 2*sqrt(b)*log((sqrt(a + b*x** 
2) + sqrt(b)*x)/sqrt(a))*b**2*c*d*x**2 - sqrt(b)*a**2*d**2 + sqrt(b)*a*b*c 
**2 - 2*sqrt(b)*a*b*c*d - sqrt(b)*a*b*d**2*x**2 + sqrt(b)*b**2*c**2*x**2 - 
 2*sqrt(b)*b**2*c*d*x**2)/(b**2*(a + b*x**2))