\(\int \frac {(c+d x) (A+B x+C x^2)}{x^4 (a+b x^2)^{3/2}} \, dx\) [139]

Optimal result

Integrand size = 30, antiderivative size = 198 \[ \int \frac {(c+d x) \left (A+B x+C x^2\right )}{x^4 \left (a+b x^2\right )^{3/2}} \, dx=\frac {2 a C d-3 b (B c+A d)}{2 a^2 \sqrt {a+b x^2}}-\frac {B c+A d}{2 a x^2 \sqrt {a+b x^2}}+\frac {b (A b c-a (c C+B d)) x}{a^3 \sqrt {a+b x^2}}-\frac {A c \sqrt {a+b x^2}}{3 a^2 x^3}+\frac {(5 A b c-3 a (c C+B d)) \sqrt {a+b x^2}}{3 a^3 x}-\frac {(2 a C d-3 b (B c+A d)) \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right )}{2 a^{5/2}} \] Output:

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

Mathematica [A] (verified)

Time = 1.25 (sec) , antiderivative size = 186, normalized size of antiderivative = 0.94 \[ \int \frac {(c+d x) \left (A+B x+C x^2\right )}{x^4 \left (a+b x^2\right )^{3/2}} \, dx=\frac {16 A b^2 c x^4-a b x^2 (A (-8 c+9 d x)+3 x (3 B c+4 c C x+4 B d x))-a^2 (A (2 c+3 d x)+3 x (2 C x (c-d x)+B (c+2 d x)))}{6 a^3 x^3 \sqrt {a+b x^2}}+\frac {2 C d \text {arctanh}\left (\frac {\sqrt {b} x-\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{3/2}}+\frac {3 b (B c+A d) \text {arctanh}\left (\frac {-\sqrt {b} x+\sqrt {a+b x^2}}{\sqrt {a}}\right )}{a^{5/2}} \] Input:

Integrate[((c + d*x)*(A + B*x + C*x^2))/(x^4*(a + b*x^2)^(3/2)),x]
 

Output:

(16*A*b^2*c*x^4 - a*b*x^2*(A*(-8*c + 9*d*x) + 3*x*(3*B*c + 4*c*C*x + 4*B*d 
*x)) - a^2*(A*(2*c + 3*d*x) + 3*x*(2*C*x*(c - d*x) + B*(c + 2*d*x))))/(6*a 
^3*x^3*Sqrt[a + b*x^2]) + (2*C*d*ArcTanh[(Sqrt[b]*x - Sqrt[a + b*x^2])/Sqr 
t[a]])/a^(3/2) + (3*b*(B*c + A*d)*ArcTanh[(-(Sqrt[b]*x) + Sqrt[a + b*x^2]) 
/Sqrt[a]])/a^(5/2)
 

Rubi [A] (verified)

Time = 1.31 (sec) , antiderivative size = 195, normalized size of antiderivative = 0.98, number of steps used = 11, number of rules used = 10, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2336, 25, 2338, 25, 2338, 27, 534, 243, 73, 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.

\(\displaystyle \int \frac {(c+d x) \left (A+B x+C x^2\right )}{x^4 \left (a+b x^2\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 2336

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2338

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 2338

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

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {-\frac {1}{2} \int \frac {2 (5 A b c-3 a (c C+B d))-3 (2 a C d-3 b (B c+A d)) x}{x^2 \sqrt {b x^2+a}}dx-\frac {3 \sqrt {a+b x^2} (A d+B c)}{2 x^2}}{3 a}-\frac {A c \sqrt {a+b x^2}}{3 a x^3}}{a}-\frac {a (-a C d+A b d+b B c)-b x (A b c-a (B d+c C))}{a^3 \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 534

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

\(\Big \downarrow \) 243

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

\(\Big \downarrow \) 73

\(\displaystyle \frac {\frac {\frac {1}{2} \left (\frac {3 (2 a C d-3 b (A d+B c)) \int \frac {1}{\frac {x^4}{b}-\frac {a}{b}}d\sqrt {b x^2+a}}{b}+\frac {2 \sqrt {a+b x^2} (5 A b c-3 a (B d+c C))}{a x}\right )-\frac {3 \sqrt {a+b x^2} (A d+B c)}{2 x^2}}{3 a}-\frac {A c \sqrt {a+b x^2}}{3 a x^3}}{a}-\frac {a (-a C d+A b d+b B c)-b x (A b c-a (B d+c C))}{a^3 \sqrt {a+b x^2}}\)

\(\Big \downarrow \) 221

\(\displaystyle \frac {\frac {\frac {1}{2} \left (\frac {2 \sqrt {a+b x^2} (5 A b c-3 a (B d+c C))}{a x}-\frac {3 \text {arctanh}\left (\frac {\sqrt {a+b x^2}}{\sqrt {a}}\right ) (2 a C d-3 b (A d+B c))}{\sqrt {a}}\right )-\frac {3 \sqrt {a+b x^2} (A d+B c)}{2 x^2}}{3 a}-\frac {A c \sqrt {a+b x^2}}{3 a x^3}}{a}-\frac {a (-a C d+A b d+b B c)-b x (A b c-a (B d+c C))}{a^3 \sqrt {a+b x^2}}\)

Input:

Int[((c + d*x)*(A + B*x + C*x^2))/(x^4*(a + b*x^2)^(3/2)),x]
 

Output:

-((a*(b*B*c + A*b*d - a*C*d) - b*(A*b*c - a*(c*C + B*d))*x)/(a^3*Sqrt[a + 
b*x^2])) + (-1/3*(A*c*Sqrt[a + b*x^2])/(a*x^3) + ((-3*(B*c + A*d)*Sqrt[a + 
 b*x^2])/(2*x^2) + ((2*(5*A*b*c - 3*a*(c*C + B*d))*Sqrt[a + b*x^2])/(a*x) 
- (3*(2*a*C*d - 3*b*(B*c + A*d))*ArcTanh[Sqrt[a + b*x^2]/Sqrt[a]])/Sqrt[a] 
)/2)/(3*a))/a
 

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 73

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[ 
{p = Denominator[m]}, Simp[p/b   Subst[Int[x^(p*(m + 1) - 1)*(c - a*(d/b) + 
 d*(x^p/b))^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] && Lt 
Q[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntL 
inearQ[a, b, c, d, m, n, x]
 

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]
 

rule 243

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[1/2   Subst[In 
t[x^((m - 1)/2)*(a + b*x)^p, x], x, x^2], x] /; FreeQ[{a, b, m, p}, x] && I 
ntegerQ[(m - 1)/2]
 

rule 534

Int[(x_)^(m_)*((c_) + (d_.)*(x_))*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> 
Simp[(-c)*x^(m + 1)*((a + b*x^2)^(p + 1)/(2*a*(p + 1))), x] + Simp[d   Int[ 
x^(m + 1)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, m, p}, x] && ILtQ[m, 
0] && GtQ[p, -1] && EqQ[m + 2*p + 3, 0]
 

rule 2336

Int[(Pq_)*((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[ 
{Q = PolynomialQuotient[(c*x)^m*Pq, a + b*x^2, x], f = Coeff[PolynomialRema 
inder[(c*x)^m*Pq, a + b*x^2, x], x, 0], g = Coeff[PolynomialRemainder[(c*x) 
^m*Pq, a + b*x^2, x], x, 1]}, Simp[(a*g - b*f*x)*((a + b*x^2)^(p + 1)/(2*a* 
b*(p + 1))), x] + Simp[1/(2*a*(p + 1))   Int[(c*x)^m*(a + b*x^2)^(p + 1)*Ex 
pandToSum[(2*a*(p + 1)*Q)/(c*x)^m + (f*(2*p + 3))/(c*x)^m, x], x], x]] /; F 
reeQ[{a, b, c}, x] && PolyQ[Pq, x] && LtQ[p, -1] && ILtQ[m, 0]
 

rule 2338

Int[(Pq_)*((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{ 
Q = PolynomialQuotient[Pq, c*x, x], R = PolynomialRemainder[Pq, c*x, x]}, S 
imp[R*(c*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*c*(m + 1))), x] + Simp[1/(a*c*( 
m + 1))   Int[(c*x)^(m + 1)*(a + b*x^2)^p*ExpandToSum[a*c*(m + 1)*Q - b*R*( 
m + 2*p + 3)*x, x], x], x]] /; FreeQ[{a, b, c, p}, x] && PolyQ[Pq, x] && Lt 
Q[m, -1] && (IntegerQ[2*p] || NeQ[Expon[Pq, x], 1])
 

Maple [A] (verified)

Time = 0.27 (sec) , antiderivative size = 205, normalized size of antiderivative = 1.04

method	result	size
risch	\(-\frac {\sqrt {b \,x^{2}+a}\, \left (-10 A b c \,x^{2}+6 B a d \,x^{2}+6 C a c \,x^{2}+3 A a d x +3 B a c x +2 A a c \right )}{6 a^{3} x^{3}}-\frac {a \left (3 A b d +3 B b c -2 a C d \right ) \left (\frac {1}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}\right )-\frac {A b d}{\sqrt {b \,x^{2}+a}}-\frac {B b c}{\sqrt {b \,x^{2}+a}}-\frac {2 A \,b^{2} c x}{a \sqrt {b \,x^{2}+a}}+\frac {2 B d b x}{\sqrt {b \,x^{2}+a}}+\frac {2 C c b x}{\sqrt {b \,x^{2}+a}}}{2 a^{2}}\)	\(205\)
default	\(\left (A d +B c \right ) \left (-\frac {1}{2 a \,x^{2} \sqrt {b \,x^{2}+a}}-\frac {3 b \left (\frac {1}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}\right )}{2 a}\right )+\left (B d +C c \right ) \left (-\frac {1}{a x \sqrt {b \,x^{2}+a}}-\frac {2 b x}{a^{2} \sqrt {b \,x^{2}+a}}\right )+A c \left (-\frac {1}{3 a \,x^{3} \sqrt {b \,x^{2}+a}}-\frac {4 b \left (-\frac {1}{a x \sqrt {b \,x^{2}+a}}-\frac {2 b x}{a^{2} \sqrt {b \,x^{2}+a}}\right )}{3 a}\right )+d C \left (\frac {1}{a \sqrt {b \,x^{2}+a}}-\frac {\ln \left (\frac {2 a +2 \sqrt {a}\, \sqrt {b \,x^{2}+a}}{x}\right )}{a^{\frac {3}{2}}}\right )\)	\(224\)

Input:

int((d*x+c)*(C*x^2+B*x+A)/x^4/(b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

-1/6*(b*x^2+a)^(1/2)*(-10*A*b*c*x^2+6*B*a*d*x^2+6*C*a*c*x^2+3*A*a*d*x+3*B* 
a*c*x+2*A*a*c)/a^3/x^3-1/2/a^2*(a*(3*A*b*d+3*B*b*c-2*C*a*d)*(1/a/(b*x^2+a) 
^(1/2)-1/a^(3/2)*ln((2*a+2*a^(1/2)*(b*x^2+a)^(1/2))/x))-A*b*d/(b*x^2+a)^(1 
/2)-B*b*c/(b*x^2+a)^(1/2)-2*A*b^2*c*x/a/(b*x^2+a)^(1/2)+2*B*d/(b*x^2+a)^(1 
/2)*b*x+2*C*c/(b*x^2+a)^(1/2)*b*x)
 

Fricas [A] (verification not implemented)

Time = 0.29 (sec) , antiderivative size = 440, normalized size of antiderivative = 2.22 \[ \int \frac {(c+d x) \left (A+B x+C x^2\right )}{x^4 \left (a+b x^2\right )^{3/2}} \, dx=\left [\frac {3 \, {\left ({\left (3 \, B b^{2} c - {\left (2 \, C a b - 3 \, A b^{2}\right )} d\right )} x^{5} + {\left (3 \, B a b c - {\left (2 \, C a^{2} - 3 \, A a b\right )} d\right )} x^{3}\right )} \sqrt {a} \log \left (-\frac {b x^{2} + 2 \, \sqrt {b x^{2} + a} \sqrt {a} + 2 \, a}{x^{2}}\right ) - 2 \, {\left (4 \, {\left (3 \, B a b d + {\left (3 \, C a b - 4 \, A b^{2}\right )} c\right )} x^{4} + 2 \, A a^{2} c + 3 \, {\left (3 \, B a b c - {\left (2 \, C a^{2} - 3 \, A a b\right )} d\right )} x^{3} + 2 \, {\left (3 \, B a^{2} d + {\left (3 \, C a^{2} - 4 \, A a b\right )} c\right )} x^{2} + 3 \, {\left (B a^{2} c + A a^{2} d\right )} x\right )} \sqrt {b x^{2} + a}}{12 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}}, -\frac {3 \, {\left ({\left (3 \, B b^{2} c - {\left (2 \, C a b - 3 \, A b^{2}\right )} d\right )} x^{5} + {\left (3 \, B a b c - {\left (2 \, C a^{2} - 3 \, A a b\right )} d\right )} x^{3}\right )} \sqrt {-a} \arctan \left (\frac {\sqrt {b x^{2} + a} \sqrt {-a}}{a}\right ) + {\left (4 \, {\left (3 \, B a b d + {\left (3 \, C a b - 4 \, A b^{2}\right )} c\right )} x^{4} + 2 \, A a^{2} c + 3 \, {\left (3 \, B a b c - {\left (2 \, C a^{2} - 3 \, A a b\right )} d\right )} x^{3} + 2 \, {\left (3 \, B a^{2} d + {\left (3 \, C a^{2} - 4 \, A a b\right )} c\right )} x^{2} + 3 \, {\left (B a^{2} c + A a^{2} d\right )} x\right )} \sqrt {b x^{2} + a}}{6 \, {\left (a^{3} b x^{5} + a^{4} x^{3}\right )}}\right ] \] Input:

integrate((d*x+c)*(C*x^2+B*x+A)/x^4/(b*x^2+a)^(3/2),x, algorithm="fricas")
 

Output:

[1/12*(3*((3*B*b^2*c - (2*C*a*b - 3*A*b^2)*d)*x^5 + (3*B*a*b*c - (2*C*a^2 
- 3*A*a*b)*d)*x^3)*sqrt(a)*log(-(b*x^2 + 2*sqrt(b*x^2 + a)*sqrt(a) + 2*a)/ 
x^2) - 2*(4*(3*B*a*b*d + (3*C*a*b - 4*A*b^2)*c)*x^4 + 2*A*a^2*c + 3*(3*B*a 
*b*c - (2*C*a^2 - 3*A*a*b)*d)*x^3 + 2*(3*B*a^2*d + (3*C*a^2 - 4*A*a*b)*c)* 
x^2 + 3*(B*a^2*c + A*a^2*d)*x)*sqrt(b*x^2 + a))/(a^3*b*x^5 + a^4*x^3), -1/ 
6*(3*((3*B*b^2*c - (2*C*a*b - 3*A*b^2)*d)*x^5 + (3*B*a*b*c - (2*C*a^2 - 3* 
A*a*b)*d)*x^3)*sqrt(-a)*arctan(sqrt(b*x^2 + a)*sqrt(-a)/a) + (4*(3*B*a*b*d 
 + (3*C*a*b - 4*A*b^2)*c)*x^4 + 2*A*a^2*c + 3*(3*B*a*b*c - (2*C*a^2 - 3*A* 
a*b)*d)*x^3 + 2*(3*B*a^2*d + (3*C*a^2 - 4*A*a*b)*c)*x^2 + 3*(B*a^2*c + A*a 
^2*d)*x)*sqrt(b*x^2 + a))/(a^3*b*x^5 + a^4*x^3)]
 

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 683 vs. \(2 (178) = 356\).

Time = 21.73 (sec) , antiderivative size = 683, normalized size of antiderivative = 3.45 \[ \int \frac {(c+d x) \left (A+B x+C x^2\right )}{x^4 \left (a+b x^2\right )^{3/2}} \, dx=A c \left (- \frac {a^{3} b^{\frac {9}{2}} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{5} b^{4} x^{2} + 6 a^{4} b^{5} x^{4} + 3 a^{3} b^{6} x^{6}} + \frac {3 a^{2} b^{\frac {11}{2}} x^{2} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{5} b^{4} x^{2} + 6 a^{4} b^{5} x^{4} + 3 a^{3} b^{6} x^{6}} + \frac {12 a b^{\frac {13}{2}} x^{4} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{5} b^{4} x^{2} + 6 a^{4} b^{5} x^{4} + 3 a^{3} b^{6} x^{6}} + \frac {8 b^{\frac {15}{2}} x^{6} \sqrt {\frac {a}{b x^{2}} + 1}}{3 a^{5} b^{4} x^{2} + 6 a^{4} b^{5} x^{4} + 3 a^{3} b^{6} x^{6}}\right ) + A d \left (- \frac {1}{2 a \sqrt {b} x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {3 \sqrt {b}}{2 a^{2} x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {3 b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2 a^{\frac {5}{2}}}\right ) + B c \left (- \frac {1}{2 a \sqrt {b} x^{3} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {3 \sqrt {b}}{2 a^{2} x \sqrt {\frac {a}{b x^{2}} + 1}} + \frac {3 b \operatorname {asinh}{\left (\frac {\sqrt {a}}{\sqrt {b} x} \right )}}{2 a^{\frac {5}{2}}}\right ) + B d \left (- \frac {1}{a \sqrt {b} x^{2} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {2 \sqrt {b}}{a^{2} \sqrt {\frac {a}{b x^{2}} + 1}}\right ) + C c \left (- \frac {1}{a \sqrt {b} x^{2} \sqrt {\frac {a}{b x^{2}} + 1}} - \frac {2 \sqrt {b}}{a^{2} \sqrt {\frac {a}{b x^{2}} + 1}}\right ) + C d \left (\frac {2 a^{3} \sqrt {1 + \frac {b x^{2}}{a}}}{2 a^{\frac {9}{2}} + 2 a^{\frac {7}{2}} b x^{2}} + \frac {a^{3} \log {\left (\frac {b x^{2}}{a} \right )}}{2 a^{\frac {9}{2}} + 2 a^{\frac {7}{2}} b x^{2}} - \frac {2 a^{3} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{2 a^{\frac {9}{2}} + 2 a^{\frac {7}{2}} b x^{2}} + \frac {a^{2} b x^{2} \log {\left (\frac {b x^{2}}{a} \right )}}{2 a^{\frac {9}{2}} + 2 a^{\frac {7}{2}} b x^{2}} - \frac {2 a^{2} b x^{2} \log {\left (\sqrt {1 + \frac {b x^{2}}{a}} + 1 \right )}}{2 a^{\frac {9}{2}} + 2 a^{\frac {7}{2}} b x^{2}}\right ) \] Input:

integrate((d*x+c)*(C*x**2+B*x+A)/x**4/(b*x**2+a)**(3/2),x)
 

Output:

A*c*(-a**3*b**(9/2)*sqrt(a/(b*x**2) + 1)/(3*a**5*b**4*x**2 + 6*a**4*b**5*x 
**4 + 3*a**3*b**6*x**6) + 3*a**2*b**(11/2)*x**2*sqrt(a/(b*x**2) + 1)/(3*a* 
*5*b**4*x**2 + 6*a**4*b**5*x**4 + 3*a**3*b**6*x**6) + 12*a*b**(13/2)*x**4* 
sqrt(a/(b*x**2) + 1)/(3*a**5*b**4*x**2 + 6*a**4*b**5*x**4 + 3*a**3*b**6*x* 
*6) + 8*b**(15/2)*x**6*sqrt(a/(b*x**2) + 1)/(3*a**5*b**4*x**2 + 6*a**4*b** 
5*x**4 + 3*a**3*b**6*x**6)) + A*d*(-1/(2*a*sqrt(b)*x**3*sqrt(a/(b*x**2) + 
1)) - 3*sqrt(b)/(2*a**2*x*sqrt(a/(b*x**2) + 1)) + 3*b*asinh(sqrt(a)/(sqrt( 
b)*x))/(2*a**(5/2))) + B*c*(-1/(2*a*sqrt(b)*x**3*sqrt(a/(b*x**2) + 1)) - 3 
*sqrt(b)/(2*a**2*x*sqrt(a/(b*x**2) + 1)) + 3*b*asinh(sqrt(a)/(sqrt(b)*x))/ 
(2*a**(5/2))) + B*d*(-1/(a*sqrt(b)*x**2*sqrt(a/(b*x**2) + 1)) - 2*sqrt(b)/ 
(a**2*sqrt(a/(b*x**2) + 1))) + C*c*(-1/(a*sqrt(b)*x**2*sqrt(a/(b*x**2) + 1 
)) - 2*sqrt(b)/(a**2*sqrt(a/(b*x**2) + 1))) + C*d*(2*a**3*sqrt(1 + b*x**2/ 
a)/(2*a**(9/2) + 2*a**(7/2)*b*x**2) + a**3*log(b*x**2/a)/(2*a**(9/2) + 2*a 
**(7/2)*b*x**2) - 2*a**3*log(sqrt(1 + b*x**2/a) + 1)/(2*a**(9/2) + 2*a**(7 
/2)*b*x**2) + a**2*b*x**2*log(b*x**2/a)/(2*a**(9/2) + 2*a**(7/2)*b*x**2) - 
 2*a**2*b*x**2*log(sqrt(1 + b*x**2/a) + 1)/(2*a**(9/2) + 2*a**(7/2)*b*x**2 
))
 

Maxima [A] (verification not implemented)

Time = 0.05 (sec) , antiderivative size = 212, normalized size of antiderivative = 1.07 \[ \int \frac {(c+d x) \left (A+B x+C x^2\right )}{x^4 \left (a+b x^2\right )^{3/2}} \, dx=\frac {8 \, A b^{2} c x}{3 \, \sqrt {b x^{2} + a} a^{3}} - \frac {C d \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{a^{\frac {3}{2}}} + \frac {C d}{\sqrt {b x^{2} + a} a} - \frac {2 \, {\left (C c + B d\right )} b x}{\sqrt {b x^{2} + a} a^{2}} + \frac {3 \, {\left (B c + A d\right )} b \operatorname {arsinh}\left (\frac {a}{\sqrt {a b} {\left | x \right |}}\right )}{2 \, a^{\frac {5}{2}}} - \frac {3 \, {\left (B c + A d\right )} b}{2 \, \sqrt {b x^{2} + a} a^{2}} + \frac {4 \, A b c}{3 \, \sqrt {b x^{2} + a} a^{2} x} - \frac {C c + B d}{\sqrt {b x^{2} + a} a x} - \frac {A c}{3 \, \sqrt {b x^{2} + a} a x^{3}} - \frac {B c + A d}{2 \, \sqrt {b x^{2} + a} a x^{2}} \] Input:

integrate((d*x+c)*(C*x^2+B*x+A)/x^4/(b*x^2+a)^(3/2),x, algorithm="maxima")
 

Output:

8/3*A*b^2*c*x/(sqrt(b*x^2 + a)*a^3) - C*d*arcsinh(a/(sqrt(a*b)*abs(x)))/a^ 
(3/2) + C*d/(sqrt(b*x^2 + a)*a) - 2*(C*c + B*d)*b*x/(sqrt(b*x^2 + a)*a^2) 
+ 3/2*(B*c + A*d)*b*arcsinh(a/(sqrt(a*b)*abs(x)))/a^(5/2) - 3/2*(B*c + A*d 
)*b/(sqrt(b*x^2 + a)*a^2) + 4/3*A*b*c/(sqrt(b*x^2 + a)*a^2*x) - (C*c + B*d 
)/(sqrt(b*x^2 + a)*a*x) - 1/3*A*c/(sqrt(b*x^2 + a)*a*x^3) - 1/2*(B*c + A*d 
)/(sqrt(b*x^2 + a)*a*x^2)
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 444 vs. \(2 (172) = 344\).

Time = 0.13 (sec) , antiderivative size = 444, normalized size of antiderivative = 2.24 \[ \int \frac {(c+d x) \left (A+B x+C x^2\right )}{x^4 \left (a+b x^2\right )^{3/2}} \, dx=-\frac {\frac {{\left (C a^{3} b c - A a^{2} b^{2} c + B a^{3} b d\right )} x}{a^{5}} + \frac {B a^{3} b c - C a^{4} d + A a^{3} b d}{a^{5}}}{\sqrt {b x^{2} + a}} - \frac {{\left (3 \, B b c - 2 \, C a d + 3 \, A b d\right )} \arctan \left (-\frac {\sqrt {b} x - \sqrt {b x^{2} + a}}{\sqrt {-a}}\right )}{\sqrt {-a} a^{2}} + \frac {3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{5} B b c + 3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{5} A b d + 6 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} C a \sqrt {b} c - 6 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} A b^{\frac {3}{2}} c + 6 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{4} B a \sqrt {b} d - 12 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} C a^{2} \sqrt {b} c + 24 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} A a b^{\frac {3}{2}} c - 12 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} B a^{2} \sqrt {b} d - 3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )} B a^{2} b c - 3 \, {\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )} A a^{2} b d + 6 \, C a^{3} \sqrt {b} c - 10 \, A a^{2} b^{\frac {3}{2}} c + 6 \, B a^{3} \sqrt {b} d}{3 \, {\left ({\left (\sqrt {b} x - \sqrt {b x^{2} + a}\right )}^{2} - a\right )}^{3} a^{2}} \] Input:

integrate((d*x+c)*(C*x^2+B*x+A)/x^4/(b*x^2+a)^(3/2),x, algorithm="giac")
 

Output:

-((C*a^3*b*c - A*a^2*b^2*c + B*a^3*b*d)*x/a^5 + (B*a^3*b*c - C*a^4*d + A*a 
^3*b*d)/a^5)/sqrt(b*x^2 + a) - (3*B*b*c - 2*C*a*d + 3*A*b*d)*arctan(-(sqrt 
(b)*x - sqrt(b*x^2 + a))/sqrt(-a))/(sqrt(-a)*a^2) + 1/3*(3*(sqrt(b)*x - sq 
rt(b*x^2 + a))^5*B*b*c + 3*(sqrt(b)*x - sqrt(b*x^2 + a))^5*A*b*d + 6*(sqrt 
(b)*x - sqrt(b*x^2 + a))^4*C*a*sqrt(b)*c - 6*(sqrt(b)*x - sqrt(b*x^2 + a)) 
^4*A*b^(3/2)*c + 6*(sqrt(b)*x - sqrt(b*x^2 + a))^4*B*a*sqrt(b)*d - 12*(sqr 
t(b)*x - sqrt(b*x^2 + a))^2*C*a^2*sqrt(b)*c + 24*(sqrt(b)*x - sqrt(b*x^2 + 
 a))^2*A*a*b^(3/2)*c - 12*(sqrt(b)*x - sqrt(b*x^2 + a))^2*B*a^2*sqrt(b)*d 
- 3*(sqrt(b)*x - sqrt(b*x^2 + a))*B*a^2*b*c - 3*(sqrt(b)*x - sqrt(b*x^2 + 
a))*A*a^2*b*d + 6*C*a^3*sqrt(b)*c - 10*A*a^2*b^(3/2)*c + 6*B*a^3*sqrt(b)*d 
)/(((sqrt(b)*x - sqrt(b*x^2 + a))^2 - a)^3*a^2)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(c+d x) \left (A+B x+C x^2\right )}{x^4 \left (a+b x^2\right )^{3/2}} \, dx=\int \frac {\left (c+d\,x\right )\,\left (C\,x^2+B\,x+A\right )}{x^4\,{\left (b\,x^2+a\right )}^{3/2}} \,d x \] Input:

int(((c + d*x)*(A + B*x + C*x^2))/(x^4*(a + b*x^2)^(3/2)),x)
 

Output:

int(((c + d*x)*(A + B*x + C*x^2))/(x^4*(a + b*x^2)^(3/2)), x)
 

Reduce [B] (verification not implemented)

Time = 0.19 (sec) , antiderivative size = 703, normalized size of antiderivative = 3.55 \[ \int \frac {(c+d x) \left (A+B x+C x^2\right )}{x^4 \left (a+b x^2\right )^{3/2}} \, dx =\text {Too large to display} \] Input:

int((d*x+c)*(C*x^2+B*x+A)/x^4/(b*x^2+a)^(3/2),x)
 

Output:

( - 2*sqrt(a + b*x**2)*a**3*c - 3*sqrt(a + b*x**2)*a**3*d*x + 8*sqrt(a + b 
*x**2)*a**2*b*c*x**2 - 3*sqrt(a + b*x**2)*a**2*b*c*x - 9*sqrt(a + b*x**2)* 
a**2*b*d*x**3 - 6*sqrt(a + b*x**2)*a**2*b*d*x**2 - 6*sqrt(a + b*x**2)*a**2 
*c**2*x**2 + 6*sqrt(a + b*x**2)*a**2*c*d*x**3 + 16*sqrt(a + b*x**2)*a*b**2 
*c*x**4 - 9*sqrt(a + b*x**2)*a*b**2*c*x**3 - 12*sqrt(a + b*x**2)*a*b**2*d* 
x**4 - 12*sqrt(a + b*x**2)*a*b*c**2*x**4 - 9*sqrt(a)*log((sqrt(a + b*x**2) 
 - sqrt(a) + sqrt(b)*x)/sqrt(a))*a**2*b*d*x**3 + 6*sqrt(a)*log((sqrt(a + b 
*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*a**2*c*d*x**3 - 9*sqrt(a)*log((sqrt 
(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b**2*c*x**3 - 9*sqrt(a)*log 
((sqrt(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b**2*d*x**5 + 6*sqrt( 
a)*log((sqrt(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*a*b*c*d*x**5 - 9* 
sqrt(a)*log((sqrt(a + b*x**2) - sqrt(a) + sqrt(b)*x)/sqrt(a))*b**3*c*x**5 
+ 9*sqrt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*a**2*b*d 
*x**3 - 6*sqrt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt(a))*a* 
*2*c*d*x**3 + 9*sqrt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x)/sqrt( 
a))*a*b**2*c*x**3 + 9*sqrt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt(b)*x) 
/sqrt(a))*a*b**2*d*x**5 - 6*sqrt(a)*log((sqrt(a + b*x**2) + sqrt(a) + sqrt 
(b)*x)/sqrt(a))*a*b*c*d*x**5 + 9*sqrt(a)*log((sqrt(a + b*x**2) + sqrt(a) + 
 sqrt(b)*x)/sqrt(a))*b**3*c*x**5 - 16*sqrt(b)*a**2*b*c*x**3 + 12*sqrt(b)*a 
**2*b*d*x**3 + 12*sqrt(b)*a**2*c**2*x**3 - 16*sqrt(b)*a*b**2*c*x**5 + 1...