\(\int \frac {A+B x+C x^2+D x^3}{(c+d x)^3 (a+b x^2)^{3/2}} \, dx\) [114]

Optimal result

Integrand size = 34, antiderivative size = 448 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3 \left (a+b x^2\right )^{3/2}} \, dx=-\frac {a \left (b^2 c^2 (B c-3 A d)-a^2 d^2 (C d-3 c D)+a b \left (3 c^2 C d-3 B c d^2+A d^3-c^3 D\right )\right )-\left (A b^2 c \left (b c^2-3 a d^2\right )-a \left (b^2 c^2 (c C-3 B d)-a^2 d^3 D-a b d \left (3 c C d-B d^2-3 c^2 D\right )\right )\right ) x}{a \left (b c^2+a d^2\right )^3 \sqrt {a+b x^2}}-\frac {\left (c^2 C d-B c d^2+A d^3-c^3 D\right ) \sqrt {a+b x^2}}{2 \left (b c^2+a d^2\right )^2 (c+d x)^2}+\frac {\left (2 a d^2 \left (2 c C d-B d^2-3 c^2 D\right )-b c \left (3 c^2 C d-5 B c d^2+7 A d^3-c^3 D\right )\right ) \sqrt {a+b x^2}}{2 \left (b c^2+a d^2\right )^3 (c+d x)}-\frac {\left (2 b^2 c^2 \left (c^2 C-3 B c d+6 A d^2\right )+2 a^2 d^3 (C d-3 c D)-a b d \left (11 c^2 C d-9 B c d^2+3 A d^3-9 c^3 D\right )\right ) \text {arctanh}\left (\frac {a d-b c x}{\sqrt {b c^2+a d^2} \sqrt {a+b x^2}}\right )}{2 \left (b c^2+a d^2\right )^{7/2}} \] Output:

-(a*(b^2*c^2*(-3*A*d+B*c)-a^2*d^2*(C*d-3*D*c)+a*b*(A*d^3-3*B*c*d^2+3*C*c^2 
*d-D*c^3))-(A*b^2*c*(-3*a*d^2+b*c^2)-a*(b^2*c^2*(-3*B*d+C*c)-a^2*d^3*D-a*b 
*d*(-B*d^2+3*C*c*d-3*D*c^2)))*x)/a/(a*d^2+b*c^2)^3/(b*x^2+a)^(1/2)-1/2*(A* 
d^3-B*c*d^2+C*c^2*d-D*c^3)*(b*x^2+a)^(1/2)/(a*d^2+b*c^2)^2/(d*x+c)^2+1/2*( 
2*a*d^2*(-B*d^2+2*C*c*d-3*D*c^2)-b*c*(7*A*d^3-5*B*c*d^2+3*C*c^2*d-D*c^3))* 
(b*x^2+a)^(1/2)/(a*d^2+b*c^2)^3/(d*x+c)-1/2*(2*b^2*c^2*(6*A*d^2-3*B*c*d+C* 
c^2)+2*a^2*d^3*(C*d-3*D*c)-a*b*d*(3*A*d^3-9*B*c*d^2+11*C*c^2*d-9*D*c^3))*a 
rctanh((-b*c*x+a*d)/(a*d^2+b*c^2)^(1/2)/(b*x^2+a)^(1/2))/(a*d^2+b*c^2)^(7/ 
2)
 

Mathematica [A] (verified)

Time = 11.54 (sec) , antiderivative size = 478, normalized size of antiderivative = 1.07 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3 \left (a+b x^2\right )^{3/2}} \, dx=\frac {1}{2} \left (\frac {\sqrt {a+b x^2} \left (\frac {\left (b c^2+a d^2\right ) \left (-c^2 C d+B c d^2-A d^3+c^3 D\right )}{(c+d x)^2}+\frac {-2 a d^2 \left (-2 c C d+B d^2+3 c^2 D\right )+b c \left (-3 c^2 C d+5 B c d^2-7 A d^3+c^3 D\right )}{c+d x}+\frac {2 \left (A b^3 c^3 x+a^3 d^2 (C d-3 c D+d D x)-a b^2 c \left (c^2 C x+B c (c-3 d x)+3 A d (-c+d x)\right )+a^2 b \left (c^3 D-d^3 (A+B x)+3 c d^2 (B+C x)-3 c^2 d (C+D x)\right )\right )}{a \left (a+b x^2\right )}\right )}{\left (b c^2+a d^2\right )^3}+\frac {\left (2 b^2 c^2 \left (c^2 C-3 B c d+6 A d^2\right )+2 a^2 d^3 (C d-3 c D)+a b d \left (-11 c^2 C d+9 B c d^2-3 A d^3+9 c^3 D\right )\right ) \log (c+d x)}{\left (b c^2+a d^2\right )^{7/2}}-\frac {\left (2 b^2 c^2 \left (c^2 C-3 B c d+6 A d^2\right )+2 a^2 d^3 (C d-3 c D)+a b d \left (-11 c^2 C d+9 B c d^2-3 A d^3+9 c^3 D\right )\right ) \log \left (a d-b c x+\sqrt {b c^2+a d^2} \sqrt {a+b x^2}\right )}{\left (b c^2+a d^2\right )^{7/2}}\right ) \] Input:

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

Output:

((Sqrt[a + b*x^2]*(((b*c^2 + a*d^2)*(-(c^2*C*d) + B*c*d^2 - A*d^3 + c^3*D) 
)/(c + d*x)^2 + (-2*a*d^2*(-2*c*C*d + B*d^2 + 3*c^2*D) + b*c*(-3*c^2*C*d + 
 5*B*c*d^2 - 7*A*d^3 + c^3*D))/(c + d*x) + (2*(A*b^3*c^3*x + a^3*d^2*(C*d 
- 3*c*D + d*D*x) - a*b^2*c*(c^2*C*x + B*c*(c - 3*d*x) + 3*A*d*(-c + d*x)) 
+ a^2*b*(c^3*D - d^3*(A + B*x) + 3*c*d^2*(B + C*x) - 3*c^2*d*(C + D*x))))/ 
(a*(a + b*x^2))))/(b*c^2 + a*d^2)^3 + ((2*b^2*c^2*(c^2*C - 3*B*c*d + 6*A*d 
^2) + 2*a^2*d^3*(C*d - 3*c*D) + a*b*d*(-11*c^2*C*d + 9*B*c*d^2 - 3*A*d^3 + 
 9*c^3*D))*Log[c + d*x])/(b*c^2 + a*d^2)^(7/2) - ((2*b^2*c^2*(c^2*C - 3*B* 
c*d + 6*A*d^2) + 2*a^2*d^3*(C*d - 3*c*D) + a*b*d*(-11*c^2*C*d + 9*B*c*d^2 
- 3*A*d^3 + 9*c^3*D))*Log[a*d - b*c*x + Sqrt[b*c^2 + a*d^2]*Sqrt[a + b*x^2 
]])/(b*c^2 + a*d^2)^(7/2))/2
 

Rubi [A] (verified)

Time = 1.61 (sec) , antiderivative size = 460, normalized size of antiderivative = 1.03, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.206, Rules used = {2178, 25, 2182, 27, 679, 488, 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.

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

\(\Big \downarrow \) 2178

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2182

\(\displaystyle \frac {-\frac {\int \frac {a b \left (2 \left (a^2 (c C-B d) d^4+2 a b c \left (-2 D c^3+2 C d c^2-B d^2 c-A d^3\right ) d-b^2 c^3 \left (C c^2-3 B d c+6 A d^2\right )\right )-\left (2 a^2 (C d-3 c D) d^4-a b \left (-3 D c^3+7 C d c^2-7 B d^2 c+3 A d^3\right ) d^2-b^2 c^2 \left (-D c^3+C d c^2+B d^2 c-5 A d^3\right )\right ) x\right )}{\left (b c^2+a d^2\right )^2 (c+d x)^2 \sqrt {b x^2+a}}dx}{2 \left (a d^2+b c^2\right )}-\frac {a b \sqrt {a+b x^2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{2 (c+d x)^2 \left (a d^2+b c^2\right )^2}}{a b}-\frac {a \left (-a^2 d^2 (C d-3 c D)+a b \left (A d^3-3 B c d^2+c^3 (-D)+3 c^2 C d\right )+b^2 c^2 (B c-3 A d)\right )-x \left (A b^2 c \left (b c^2-3 a d^2\right )-a \left (-a^2 d^3 D-a b d \left (-B d^2-3 c^2 D+3 c C d\right )+b^2 c^2 (c C-3 B d)\right )\right )}{a \sqrt {a+b x^2} \left (a d^2+b c^2\right )^3}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {-\frac {a b \int \frac {2 \left (a^2 (c C-B d) d^4+2 a b c \left (-2 D c^3+2 C d c^2-B d^2 c-A d^3\right ) d-b^2 c^3 \left (C c^2-3 B d c+6 A d^2\right )\right )-\left (2 a^2 (C d-3 c D) d^4-a b \left (-3 D c^3+7 C d c^2-7 B d^2 c+3 A d^3\right ) d^2-b^2 c^2 \left (-D c^3+C d c^2+B d^2 c-5 A d^3\right )\right ) x}{(c+d x)^2 \sqrt {b x^2+a}}dx}{2 \left (a d^2+b c^2\right )^3}-\frac {a b \sqrt {a+b x^2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{2 (c+d x)^2 \left (a d^2+b c^2\right )^2}}{a b}-\frac {a \left (-a^2 d^2 (C d-3 c D)+a b \left (A d^3-3 B c d^2+c^3 (-D)+3 c^2 C d\right )+b^2 c^2 (B c-3 A d)\right )-x \left (A b^2 c \left (b c^2-3 a d^2\right )-a \left (-a^2 d^3 D-a b d \left (-B d^2-3 c^2 D+3 c C d\right )+b^2 c^2 (c C-3 B d)\right )\right )}{a \sqrt {a+b x^2} \left (a d^2+b c^2\right )^3}\)

\(\Big \downarrow \) 679

\(\displaystyle \frac {-\frac {a b \left (-\left (2 a^2 d^3 (C d-3 c D)-a b d \left (3 A d^3-9 B c d^2-9 c^3 D+11 c^2 C d\right )+2 b^2 c^2 \left (6 A d^2-3 B c d+c^2 C\right )\right ) \int \frac {1}{(c+d x) \sqrt {b x^2+a}}dx-\frac {\sqrt {a+b x^2} \left (2 a d^2 \left (-B d^2-3 c^2 D+2 c C d\right )-b c \left (7 A d^3-5 B c d^2+c^3 (-D)+3 c^2 C d\right )\right )}{c+d x}\right )}{2 \left (a d^2+b c^2\right )^3}-\frac {a b \sqrt {a+b x^2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{2 (c+d x)^2 \left (a d^2+b c^2\right )^2}}{a b}-\frac {a \left (-a^2 d^2 (C d-3 c D)+a b \left (A d^3-3 B c d^2+c^3 (-D)+3 c^2 C d\right )+b^2 c^2 (B c-3 A d)\right )-x \left (A b^2 c \left (b c^2-3 a d^2\right )-a \left (-a^2 d^3 D-a b d \left (-B d^2-3 c^2 D+3 c C d\right )+b^2 c^2 (c C-3 B d)\right )\right )}{a \sqrt {a+b x^2} \left (a d^2+b c^2\right )^3}\)

\(\Big \downarrow \) 488

\(\displaystyle \frac {-\frac {a b \left (\left (2 a^2 d^3 (C d-3 c D)-a b d \left (3 A d^3-9 B c d^2-9 c^3 D+11 c^2 C d\right )+2 b^2 c^2 \left (6 A d^2-3 B c d+c^2 C\right )\right ) \int \frac {1}{b c^2+a d^2-\frac {(a d-b c x)^2}{b x^2+a}}d\frac {a d-b c x}{\sqrt {b x^2+a}}-\frac {\sqrt {a+b x^2} \left (2 a d^2 \left (-B d^2-3 c^2 D+2 c C d\right )-b c \left (7 A d^3-5 B c d^2+c^3 (-D)+3 c^2 C d\right )\right )}{c+d x}\right )}{2 \left (a d^2+b c^2\right )^3}-\frac {a b \sqrt {a+b x^2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{2 (c+d x)^2 \left (a d^2+b c^2\right )^2}}{a b}-\frac {a \left (-a^2 d^2 (C d-3 c D)+a b \left (A d^3-3 B c d^2+c^3 (-D)+3 c^2 C d\right )+b^2 c^2 (B c-3 A d)\right )-x \left (A b^2 c \left (b c^2-3 a d^2\right )-a \left (-a^2 d^3 D-a b d \left (-B d^2-3 c^2 D+3 c C d\right )+b^2 c^2 (c C-3 B d)\right )\right )}{a \sqrt {a+b x^2} \left (a d^2+b c^2\right )^3}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {-\frac {a b \left (\frac {\text {arctanh}\left (\frac {a d-b c x}{\sqrt {a+b x^2} \sqrt {a d^2+b c^2}}\right ) \left (2 a^2 d^3 (C d-3 c D)-a b d \left (3 A d^3-9 B c d^2-9 c^3 D+11 c^2 C d\right )+2 b^2 c^2 \left (6 A d^2-3 B c d+c^2 C\right )\right )}{\sqrt {a d^2+b c^2}}-\frac {\sqrt {a+b x^2} \left (2 a d^2 \left (-B d^2-3 c^2 D+2 c C d\right )-b c \left (7 A d^3-5 B c d^2+c^3 (-D)+3 c^2 C d\right )\right )}{c+d x}\right )}{2 \left (a d^2+b c^2\right )^3}-\frac {a b \sqrt {a+b x^2} \left (A d^3-B c d^2+c^3 (-D)+c^2 C d\right )}{2 (c+d x)^2 \left (a d^2+b c^2\right )^2}}{a b}-\frac {a \left (-a^2 d^2 (C d-3 c D)+a b \left (A d^3-3 B c d^2+c^3 (-D)+3 c^2 C d\right )+b^2 c^2 (B c-3 A d)\right )-x \left (A b^2 c \left (b c^2-3 a d^2\right )-a \left (-a^2 d^3 D-a b d \left (-B d^2-3 c^2 D+3 c C d\right )+b^2 c^2 (c C-3 B d)\right )\right )}{a \sqrt {a+b x^2} \left (a d^2+b c^2\right )^3}\)

Input:

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

Output:

-((a*(b^2*c^2*(B*c - 3*A*d) - a^2*d^2*(C*d - 3*c*D) + a*b*(3*c^2*C*d - 3*B 
*c*d^2 + A*d^3 - c^3*D)) - (A*b^2*c*(b*c^2 - 3*a*d^2) - a*(b^2*c^2*(c*C - 
3*B*d) - a^2*d^3*D - a*b*d*(3*c*C*d - B*d^2 - 3*c^2*D)))*x)/(a*(b*c^2 + a* 
d^2)^3*Sqrt[a + b*x^2])) + (-1/2*(a*b*(c^2*C*d - B*c*d^2 + A*d^3 - c^3*D)* 
Sqrt[a + b*x^2])/((b*c^2 + a*d^2)^2*(c + d*x)^2) - (a*b*(-(((2*a*d^2*(2*c* 
C*d - B*d^2 - 3*c^2*D) - b*c*(3*c^2*C*d - 5*B*c*d^2 + 7*A*d^3 - c^3*D))*Sq 
rt[a + b*x^2])/(c + d*x)) + ((2*b^2*c^2*(c^2*C - 3*B*c*d + 6*A*d^2) + 2*a^ 
2*d^3*(C*d - 3*c*D) - a*b*d*(11*c^2*C*d - 9*B*c*d^2 + 3*A*d^3 - 9*c^3*D))* 
ArcTanh[(a*d - b*c*x)/(Sqrt[b*c^2 + a*d^2]*Sqrt[a + b*x^2])])/Sqrt[b*c^2 + 
 a*d^2]))/(2*(b*c^2 + a*d^2)^3))/(a*b)
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

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 488

Int[1/(((c_) + (d_.)*(x_))*Sqrt[(a_) + (b_.)*(x_)^2]), x_Symbol] :> -Subst[ 
Int[1/(b*c^2 + a*d^2 - x^2), x], x, (a*d - b*c*x)/Sqrt[a + b*x^2]] /; FreeQ 
[{a, b, c, d}, x]
 

rule 679

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

rule 2178

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] : 
> With[{Qx = PolynomialQuotient[(d + e*x)^m*Pq, a + b*x^2, x], R = Coeff[Po 
lynomialRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 0], S = Coeff[Polynomia 
lRemainder[(d + e*x)^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*S - b*R*x)*((a + 
b*x^2)^(p + 1)/(2*a*b*(p + 1))), x] + Simp[1/(2*a*b*(p + 1))   Int[(d + e*x 
)^m*(a + b*x^2)^(p + 1)*ExpandToSum[(2*a*b*(p + 1)*Qx)/(d + e*x)^m + (b*R*( 
2*p + 3))/(d + e*x)^m, x], x], x]] /; FreeQ[{a, b, d, e}, x] && PolyQ[Pq, x 
] && NeQ[b*d^2 + a*e^2, 0] && LtQ[p, -1] && ILtQ[m, 0]
 

rule 2182

Int[(Pq_)*((d_) + (e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
 With[{Qx = PolynomialQuotient[Pq, d + e*x, x], R = PolynomialRemainder[Pq, 
 d + e*x, x]}, Simp[e*R*(d + e*x)^(m + 1)*((a + b*x^2)^(p + 1)/((m + 1)*(b* 
d^2 + a*e^2))), x] + Simp[1/((m + 1)*(b*d^2 + a*e^2))   Int[(d + e*x)^(m + 
1)*(a + b*x^2)^p*ExpandToSum[(m + 1)*(b*d^2 + a*e^2)*Qx + b*d*R*(m + 1) - b 
*e*R*(m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, d, e, p}, x] && PolyQ[Pq, 
 x] && NeQ[b*d^2 + a*e^2, 0] && LtQ[m, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1829\) vs. \(2(430)=860\).

Time = 1.51 (sec) , antiderivative size = 1830, normalized size of antiderivative = 4.08

method	result	size
default	\(\text {Expression too large to display}\)	\(1830\)

Input:

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

Output:

D/d^3/(b*x^2+a)^(1/2)/a*x+1/d^4*(C*d-3*D*c)*(1/(a*d^2+b*c^2)*d^2/(b*(x+c/d 
)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2)+2*b*c*d/(a*d^2+b*c^2)*(2*b*(x 
+c/d)-2*b*c/d)/(4*b*(a*d^2+b*c^2)/d^2-4*b^2*c^2/d^2)/(b*(x+c/d)^2-2*b*c/d* 
(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2)-1/(a*d^2+b*c^2)*d^2/((a*d^2+b*c^2)/d^2)^( 
1/2)*ln((2*(a*d^2+b*c^2)/d^2-2*b*c/d*(x+c/d)+2*((a*d^2+b*c^2)/d^2)^(1/2)*( 
b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2))/(x+c/d)))+1/d^5*(B*d 
^2-2*C*c*d+3*D*c^2)*(-1/(a*d^2+b*c^2)*d^2/(x+c/d)/(b*(x+c/d)^2-2*b*c/d*(x+ 
c/d)+(a*d^2+b*c^2)/d^2)^(1/2)+3*b*c*d/(a*d^2+b*c^2)*(1/(a*d^2+b*c^2)*d^2/( 
b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2)+2*b*c*d/(a*d^2+b*c^2) 
*(2*b*(x+c/d)-2*b*c/d)/(4*b*(a*d^2+b*c^2)/d^2-4*b^2*c^2/d^2)/(b*(x+c/d)^2- 
2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2)-1/(a*d^2+b*c^2)*d^2/((a*d^2+b*c^2 
)/d^2)^(1/2)*ln((2*(a*d^2+b*c^2)/d^2-2*b*c/d*(x+c/d)+2*((a*d^2+b*c^2)/d^2) 
^(1/2)*(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2))/(x+c/d)))-4* 
b/(a*d^2+b*c^2)*d^2*(2*b*(x+c/d)-2*b*c/d)/(4*b*(a*d^2+b*c^2)/d^2-4*b^2*c^2 
/d^2)/(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2))+1/d^6*(A*d^3- 
B*c*d^2+C*c^2*d-D*c^3)*(-1/2/(a*d^2+b*c^2)*d^2/(x+c/d)^2/(b*(x+c/d)^2-2*b* 
c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2)+5/2*b*c*d/(a*d^2+b*c^2)*(-1/(a*d^2+b* 
c^2)*d^2/(x+c/d)/(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^2+b*c^2)/d^2)^(1/2)+3*b 
*c*d/(a*d^2+b*c^2)*(1/(a*d^2+b*c^2)*d^2/(b*(x+c/d)^2-2*b*c/d*(x+c/d)+(a*d^ 
2+b*c^2)/d^2)^(1/2)+2*b*c*d/(a*d^2+b*c^2)*(2*b*(x+c/d)-2*b*c/d)/(4*b*(a...
 

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1654 vs. \(2 (429) = 858\).

Time = 25.04 (sec) , antiderivative size = 3335, normalized size of antiderivative = 7.44 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3 \left (a+b x^2\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

Too large to include
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 3156 vs. \(2 (429) = 858\).

Time = 0.25 (sec) , antiderivative size = 3156, normalized size of antiderivative = 7.04 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3 \left (a+b x^2\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

-15/2*D*b^3*c^6*x/(sqrt(b*x^2 + a)*a*b^3*c^6*d^3 + 3*sqrt(b*x^2 + a)*a^2*b 
^2*c^4*d^5 + 3*sqrt(b*x^2 + a)*a^3*b*c^2*d^7 + sqrt(b*x^2 + a)*a^4*d^9) + 
15/2*C*b^3*c^5*x/(sqrt(b*x^2 + a)*a*b^3*c^6*d^2 + 3*sqrt(b*x^2 + a)*a^2*b^ 
2*c^4*d^4 + 3*sqrt(b*x^2 + a)*a^3*b*c^2*d^6 + sqrt(b*x^2 + a)*a^4*d^8) - 1 
5/2*B*b^3*c^4*x/(sqrt(b*x^2 + a)*a*b^3*c^6*d + 3*sqrt(b*x^2 + a)*a^2*b^2*c 
^4*d^3 + 3*sqrt(b*x^2 + a)*a^3*b*c^2*d^5 + sqrt(b*x^2 + a)*a^4*d^7) - 15/2 
*D*b^2*c^5/(sqrt(b*x^2 + a)*b^3*c^6*d^2 + 3*sqrt(b*x^2 + a)*a*b^2*c^4*d^4 
+ 3*sqrt(b*x^2 + a)*a^2*b*c^2*d^6 + sqrt(b*x^2 + a)*a^3*d^8) + 15/2*A*b^3* 
c^3*x/(sqrt(b*x^2 + a)*a*b^3*c^6 + 3*sqrt(b*x^2 + a)*a^2*b^2*c^4*d^2 + 3*s 
qrt(b*x^2 + a)*a^3*b*c^2*d^4 + sqrt(b*x^2 + a)*a^4*d^6) + 31/2*D*b^2*c^4*x 
/(sqrt(b*x^2 + a)*a*b^2*c^4*d^3 + 2*sqrt(b*x^2 + a)*a^2*b*c^2*d^5 + sqrt(b 
*x^2 + a)*a^3*d^7) + 15/2*C*b^2*c^4/(sqrt(b*x^2 + a)*b^3*c^6*d + 3*sqrt(b* 
x^2 + a)*a*b^2*c^4*d^3 + 3*sqrt(b*x^2 + a)*a^2*b*c^2*d^5 + sqrt(b*x^2 + a) 
*a^3*d^7) - 25/2*C*b^2*c^3*x/(sqrt(b*x^2 + a)*a*b^2*c^4*d^2 + 2*sqrt(b*x^2 
 + a)*a^2*b*c^2*d^4 + sqrt(b*x^2 + a)*a^3*d^6) - 15/2*B*b^2*c^3/(sqrt(b*x^ 
2 + a)*b^3*c^6 + 3*sqrt(b*x^2 + a)*a*b^2*c^4*d^2 + 3*sqrt(b*x^2 + a)*a^2*b 
*c^2*d^4 + sqrt(b*x^2 + a)*a^3*d^6) + 5/2*D*b*c^4/(sqrt(b*x^2 + a)*b^2*c^4 
*d^3*x + 2*sqrt(b*x^2 + a)*a*b*c^2*d^5*x + sqrt(b*x^2 + a)*a^2*d^7*x + sqr 
t(b*x^2 + a)*b^2*c^5*d^2 + 2*sqrt(b*x^2 + a)*a*b*c^3*d^4 + sqrt(b*x^2 + a) 
*a^2*c*d^6) + 19/2*B*b^2*c^2*x/(sqrt(b*x^2 + a)*a*b^2*c^4*d + 2*sqrt(b*...
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1762 vs. \(2 (429) = 858\).

Time = 0.29 (sec) , antiderivative size = 1762, normalized size of antiderivative = 3.93 \[ \int \frac {A+B x+C x^2+D x^3}{(c+d x)^3 \left (a+b x^2\right )^{3/2}} \, dx=\text {Too large to display} \] Input:

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

Output:

-((C*a*b^5*c^9 - A*b^6*c^9 + 3*D*a^2*b^4*c^8*d - 3*B*a*b^5*c^8*d + 8*D*a^3 
*b^3*c^6*d^3 - 8*B*a^2*b^4*c^6*d^3 - 6*C*a^3*b^3*c^5*d^4 + 6*A*a^2*b^4*c^5 
*d^4 + 6*D*a^4*b^2*c^4*d^5 - 6*B*a^3*b^3*c^4*d^5 - 8*C*a^4*b^2*c^3*d^6 + 8 
*A*a^3*b^3*c^3*d^6 - 3*C*a^5*b*c*d^8 + 3*A*a^4*b^2*c*d^8 - D*a^6*d^9 + B*a 
^5*b*d^9)*x/(a*b^6*c^12 + 6*a^2*b^5*c^10*d^2 + 15*a^3*b^4*c^8*d^4 + 20*a^4 
*b^3*c^6*d^6 + 15*a^5*b^2*c^4*d^8 + 6*a^6*b*c^2*d^10 + a^7*d^12) - (D*a^2* 
b^4*c^9 - B*a*b^5*c^9 - 3*C*a^2*b^4*c^8*d + 3*A*a*b^5*c^8*d - 8*C*a^3*b^3* 
c^6*d^3 + 8*A*a^2*b^4*c^6*d^3 - 6*D*a^4*b^2*c^5*d^4 + 6*B*a^3*b^3*c^5*d^4 
- 6*C*a^4*b^2*c^4*d^5 + 6*A*a^3*b^3*c^4*d^5 - 8*D*a^5*b*c^3*d^6 + 8*B*a^4* 
b^2*c^3*d^6 - 3*D*a^6*c*d^8 + 3*B*a^5*b*c*d^8 + C*a^6*d^9 - A*a^5*b*d^9)/( 
a*b^6*c^12 + 6*a^2*b^5*c^10*d^2 + 15*a^3*b^4*c^8*d^4 + 20*a^4*b^3*c^6*d^6 
+ 15*a^5*b^2*c^4*d^8 + 6*a^6*b*c^2*d^10 + a^7*d^12))/sqrt(b*x^2 + a) - (2* 
C*b^2*c^4 + 9*D*a*b*c^3*d - 6*B*b^2*c^3*d - 11*C*a*b*c^2*d^2 + 12*A*b^2*c^ 
2*d^2 - 6*D*a^2*c*d^3 + 9*B*a*b*c*d^3 + 2*C*a^2*d^4 - 3*A*a*b*d^4)*arctan( 
((sqrt(b)*x - sqrt(b*x^2 + a))*d + sqrt(b)*c)/sqrt(-b*c^2 - a*d^2))/((b^3* 
c^6 + 3*a*b^2*c^4*d^2 + 3*a^2*b*c^2*d^4 + a^3*d^6)*sqrt(-b*c^2 - a*d^2)) - 
 (2*(sqrt(b)*x - sqrt(b*x^2 + a))^3*C*b^2*c^4*d^2 + 7*(sqrt(b)*x - sqrt(b* 
x^2 + a))^3*D*a*b*c^3*d^3 - 4*(sqrt(b)*x - sqrt(b*x^2 + a))^3*B*b^2*c^3*d^ 
3 - 5*(sqrt(b)*x - sqrt(b*x^2 + a))^3*C*a*b*c^2*d^4 + 6*(sqrt(b)*x - sqrt( 
b*x^2 + a))^3*A*b^2*c^2*d^4 + 3*(sqrt(b)*x - sqrt(b*x^2 + a))^3*B*a*b*c...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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