\(\int \frac {(c+d x^2)^3}{x^{11/2} (a+b x^2)} \, dx\) [771]

Optimal result

Integrand size = 24, antiderivative size = 237 \[ \int \frac {\left (c+d x^2\right )^3}{x^{11/2} \left (a+b x^2\right )} \, dx=-\frac {2 c^3}{9 a x^{9/2}}+\frac {2 c^2 (b c-3 a d)}{5 a^2 x^{5/2}}-\frac {2 c \left (b^2 c^2-3 a b c d+3 a^2 d^2\right )}{a^3 \sqrt {x}}+\frac {(b c-a d)^3 \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{13/4} b^{3/4}}-\frac {(b c-a d)^3 \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{\sqrt {2} a^{13/4} b^{3/4}}+\frac {(b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{\sqrt {2} a^{13/4} b^{3/4}} \] Output:

-2/9*c^3/a/x^(9/2)+2/5*c^2*(-3*a*d+b*c)/a^2/x^(5/2)-2*c*(3*a^2*d^2-3*a*b*c 
*d+b^2*c^2)/a^3/x^(1/2)+1/2*(-a*d+b*c)^3*arctan(1-2^(1/2)*b^(1/4)*x^(1/2)/ 
a^(1/4))*2^(1/2)/a^(13/4)/b^(3/4)-1/2*(-a*d+b*c)^3*arctan(1+2^(1/2)*b^(1/4 
)*x^(1/2)/a^(1/4))*2^(1/2)/a^(13/4)/b^(3/4)+1/2*(-a*d+b*c)^3*arctanh(2^(1/ 
2)*a^(1/4)*b^(1/4)*x^(1/2)/(a^(1/2)+b^(1/2)*x))*2^(1/2)/a^(13/4)/b^(3/4)
 

Mathematica [A] (verified)

Time = 0.26 (sec) , antiderivative size = 193, normalized size of antiderivative = 0.81 \[ \int \frac {\left (c+d x^2\right )^3}{x^{11/2} \left (a+b x^2\right )} \, dx=\frac {-\frac {4 \sqrt [4]{a} c \left (45 b^2 c^2 x^4-9 a b c x^2 \left (c+15 d x^2\right )+a^2 \left (5 c^2+27 c d x^2+135 d^2 x^4\right )\right )}{x^{9/2}}+\frac {45 \sqrt {2} (b c-a d)^3 \arctan \left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )}{b^{3/4}}+\frac {45 \sqrt {2} (b c-a d)^3 \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{b^{3/4}}}{90 a^{13/4}} \] Input:

Integrate[(c + d*x^2)^3/(x^(11/2)*(a + b*x^2)),x]
 

Output:

((-4*a^(1/4)*c*(45*b^2*c^2*x^4 - 9*a*b*c*x^2*(c + 15*d*x^2) + a^2*(5*c^2 + 
 27*c*d*x^2 + 135*d^2*x^4)))/x^(9/2) + (45*Sqrt[2]*(b*c - a*d)^3*ArcTan[(S 
qrt[a] - Sqrt[b]*x)/(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])])/b^(3/4) + (45*Sqrt 
[2]*(b*c - a*d)^3*ArcTanh[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/(Sqrt[a] + Sqr 
t[b]*x)])/b^(3/4))/(90*a^(13/4))
 

Rubi [A] (verified)

Time = 0.46 (sec) , antiderivative size = 310, normalized size of antiderivative = 1.31, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {368, 961, 2009}

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.

\(\displaystyle \int \frac {\left (c+d x^2\right )^3}{x^{11/2} \left (a+b x^2\right )} \, dx\)

\(\Big \downarrow \) 368

\(\displaystyle 2 \int \frac {\left (d x^2+c\right )^3}{x^5 \left (b x^2+a\right )}d\sqrt {x}\)

\(\Big \downarrow \) 961

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

\(\Big \downarrow \) 2009

\(\displaystyle 2 \left (\frac {\arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ) (b c-a d)^3}{2 \sqrt {2} a^{13/4} b^{3/4}}-\frac {\arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right ) (b c-a d)^3}{2 \sqrt {2} a^{13/4} b^{3/4}}-\frac {(b c-a d)^3 \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{4 \sqrt {2} a^{13/4} b^{3/4}}+\frac {(b c-a d)^3 \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{4 \sqrt {2} a^{13/4} b^{3/4}}+\frac {c^2 (b c-3 a d)}{5 a^2 x^{5/2}}-\frac {c \left (3 a^2 d^2-3 a b c d+b^2 c^2\right )}{a^3 \sqrt {x}}-\frac {c^3}{9 a x^{9/2}}\right )\)

Input:

Int[(c + d*x^2)^3/(x^(11/2)*(a + b*x^2)),x]
 

Output:

2*(-1/9*c^3/(a*x^(9/2)) + (c^2*(b*c - 3*a*d))/(5*a^2*x^(5/2)) - (c*(b^2*c^ 
2 - 3*a*b*c*d + 3*a^2*d^2))/(a^3*Sqrt[x]) + ((b*c - a*d)^3*ArcTan[1 - (Sqr 
t[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(2*Sqrt[2]*a^(13/4)*b^(3/4)) - ((b*c - a*d 
)^3*ArcTan[1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(2*Sqrt[2]*a^(13/4)*b^( 
3/4)) - ((b*c - a*d)^3*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqr 
t[b]*x])/(4*Sqrt[2]*a^(13/4)*b^(3/4)) + ((b*c - a*d)^3*Log[Sqrt[a] + Sqrt[ 
2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(4*Sqrt[2]*a^(13/4)*b^(3/4)))
 

Defintions of rubi rules used

rule 368

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_)*((c_) + (d_.)*(x_)^2)^(q_) 
, x_Symbol] :> With[{k = Denominator[m]}, Simp[k/e   Subst[Int[x^(k*(m + 1) 
 - 1)*(a + b*(x^(k*2)/e^2))^p*(c + d*(x^(k*2)/e^2))^q, x], x, (e*x)^(1/k)], 
 x]] /; FreeQ[{a, b, c, d, e, p, q}, x] && NeQ[b*c - a*d, 0] && FractionQ[m 
] && IntegerQ[p]
 

rule 961

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^( 
n_)), x_Symbol] :> Int[ExpandIntegrand[(e*x)^m*((a + b*x^n)^p/(c + d*x^n)), 
 x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n, 0] 
&& IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

Maple [A] (verified)

Time = 0.51 (sec) , antiderivative size = 208, normalized size of antiderivative = 0.88

method	result	size
derivativedivides	\(-\frac {2 c^{3}}{9 a \,x^{\frac {9}{2}}}-\frac {2 c \left (3 a^{2} d^{2}-3 a b c d +b^{2} c^{2}\right )}{a^{3} \sqrt {x}}-\frac {2 c^{2} \left (3 a d -b c \right )}{5 a^{2} x^{\frac {5}{2}}}+\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{4 a^{3} b \left (\frac {a}{b}\right )^{\frac {1}{4}}}\)	\(208\)
default	\(-\frac {2 c^{3}}{9 a \,x^{\frac {9}{2}}}-\frac {2 c \left (3 a^{2} d^{2}-3 a b c d +b^{2} c^{2}\right )}{a^{3} \sqrt {x}}-\frac {2 c^{2} \left (3 a d -b c \right )}{5 a^{2} x^{\frac {5}{2}}}+\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{4 a^{3} b \left (\frac {a}{b}\right )^{\frac {1}{4}}}\)	\(208\)
risch	\(-\frac {2 \left (135 a^{2} d^{2} x^{4}-135 a b c d \,x^{4}+45 b^{2} c^{2} x^{4}+27 a^{2} c d \,x^{2}-9 a b \,c^{2} x^{2}+5 a^{2} c^{2}\right ) c}{45 a^{3} x^{\frac {9}{2}}}+\frac {\left (a^{3} d^{3}-3 a^{2} b c \,d^{2}+3 a \,b^{2} c^{2} d -b^{3} c^{3}\right ) \sqrt {2}\, \left (\ln \left (\frac {x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{4 a^{3} b \left (\frac {a}{b}\right )^{\frac {1}{4}}}\)	\(215\)

Input:

int((d*x^2+c)^3/x^(11/2)/(b*x^2+a),x,method=_RETURNVERBOSE)
 

Output:

-2/9*c^3/a/x^(9/2)-2*c*(3*a^2*d^2-3*a*b*c*d+b^2*c^2)/a^3/x^(1/2)-2/5*c^2*( 
3*a*d-b*c)/a^2/x^(5/2)+1/4*(a^3*d^3-3*a^2*b*c*d^2+3*a*b^2*c^2*d-b^3*c^3)/a 
^3/b/(a/b)^(1/4)*2^(1/2)*(ln((x-(a/b)^(1/4)*x^(1/2)*2^(1/2)+(a/b)^(1/2))/( 
x+(a/b)^(1/4)*x^(1/2)*2^(1/2)+(a/b)^(1/2)))+2*arctan(2^(1/2)/(a/b)^(1/4)*x 
^(1/2)+1)+2*arctan(2^(1/2)/(a/b)^(1/4)*x^(1/2)-1))
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.11 (sec) , antiderivative size = 2013, normalized size of antiderivative = 8.49 \[ \int \frac {\left (c+d x^2\right )^3}{x^{11/2} \left (a+b x^2\right )} \, dx=\text {Too large to display} \] Input:

integrate((d*x^2+c)^3/x^(11/2)/(b*x^2+a),x, algorithm="fricas")
 

Output:

1/90*(45*a^3*x^5*(-(b^12*c^12 - 12*a*b^11*c^11*d + 66*a^2*b^10*c^10*d^2 - 
220*a^3*b^9*c^9*d^3 + 495*a^4*b^8*c^8*d^4 - 792*a^5*b^7*c^7*d^5 + 924*a^6* 
b^6*c^6*d^6 - 792*a^7*b^5*c^5*d^7 + 495*a^8*b^4*c^4*d^8 - 220*a^9*b^3*c^3* 
d^9 + 66*a^10*b^2*c^2*d^10 - 12*a^11*b*c*d^11 + a^12*d^12)/(a^13*b^3))^(1/ 
4)*log(a^10*b^2*(-(b^12*c^12 - 12*a*b^11*c^11*d + 66*a^2*b^10*c^10*d^2 - 2 
20*a^3*b^9*c^9*d^3 + 495*a^4*b^8*c^8*d^4 - 792*a^5*b^7*c^7*d^5 + 924*a^6*b 
^6*c^6*d^6 - 792*a^7*b^5*c^5*d^7 + 495*a^8*b^4*c^4*d^8 - 220*a^9*b^3*c^3*d 
^9 + 66*a^10*b^2*c^2*d^10 - 12*a^11*b*c*d^11 + a^12*d^12)/(a^13*b^3))^(3/4 
) - (b^9*c^9 - 9*a*b^8*c^8*d + 36*a^2*b^7*c^7*d^2 - 84*a^3*b^6*c^6*d^3 + 1 
26*a^4*b^5*c^5*d^4 - 126*a^5*b^4*c^4*d^5 + 84*a^6*b^3*c^3*d^6 - 36*a^7*b^2 
*c^2*d^7 + 9*a^8*b*c*d^8 - a^9*d^9)*sqrt(x)) - 45*I*a^3*x^5*(-(b^12*c^12 - 
 12*a*b^11*c^11*d + 66*a^2*b^10*c^10*d^2 - 220*a^3*b^9*c^9*d^3 + 495*a^4*b 
^8*c^8*d^4 - 792*a^5*b^7*c^7*d^5 + 924*a^6*b^6*c^6*d^6 - 792*a^7*b^5*c^5*d 
^7 + 495*a^8*b^4*c^4*d^8 - 220*a^9*b^3*c^3*d^9 + 66*a^10*b^2*c^2*d^10 - 12 
*a^11*b*c*d^11 + a^12*d^12)/(a^13*b^3))^(1/4)*log(I*a^10*b^2*(-(b^12*c^12 
- 12*a*b^11*c^11*d + 66*a^2*b^10*c^10*d^2 - 220*a^3*b^9*c^9*d^3 + 495*a^4* 
b^8*c^8*d^4 - 792*a^5*b^7*c^7*d^5 + 924*a^6*b^6*c^6*d^6 - 792*a^7*b^5*c^5* 
d^7 + 495*a^8*b^4*c^4*d^8 - 220*a^9*b^3*c^3*d^9 + 66*a^10*b^2*c^2*d^10 - 1 
2*a^11*b*c*d^11 + a^12*d^12)/(a^13*b^3))^(3/4) - (b^9*c^9 - 9*a*b^8*c^8*d 
+ 36*a^2*b^7*c^7*d^2 - 84*a^3*b^6*c^6*d^3 + 126*a^4*b^5*c^5*d^4 - 126*a...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^2\right )^3}{x^{11/2} \left (a+b x^2\right )} \, dx=\text {Timed out} \] Input:

integrate((d*x**2+c)**3/x**(11/2)/(b*x**2+a),x)
 

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.15 (sec) , antiderivative size = 281, normalized size of antiderivative = 1.19 \[ \int \frac {\left (c+d x^2\right )^3}{x^{11/2} \left (a+b x^2\right )} \, dx=-\frac {{\left (b^{3} c^{3} - 3 \, a b^{2} c^{2} d + 3 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {\sqrt {a} \sqrt {b}} \sqrt {b}} - \frac {\sqrt {2} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}} + \frac {\sqrt {2} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {1}{4}} b^{\frac {3}{4}}}\right )}}{4 \, a^{3}} - \frac {2 \, {\left (5 \, a^{2} c^{3} + 45 \, {\left (b^{2} c^{3} - 3 \, a b c^{2} d + 3 \, a^{2} c d^{2}\right )} x^{4} - 9 \, {\left (a b c^{3} - 3 \, a^{2} c^{2} d\right )} x^{2}\right )}}{45 \, a^{3} x^{\frac {9}{2}}} \] Input:

integrate((d*x^2+c)^3/x^(11/2)/(b*x^2+a),x, algorithm="maxima")
 

Output:

-1/4*(b^3*c^3 - 3*a*b^2*c^2*d + 3*a^2*b*c*d^2 - a^3*d^3)*(2*sqrt(2)*arctan 
(1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) + 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sq 
rt(b)))/(sqrt(sqrt(a)*sqrt(b))*sqrt(b)) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(s 
qrt(2)*a^(1/4)*b^(1/4) - 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(s 
qrt(a)*sqrt(b))*sqrt(b)) - sqrt(2)*log(sqrt(2)*a^(1/4)*b^(1/4)*sqrt(x) + s 
qrt(b)*x + sqrt(a))/(a^(1/4)*b^(3/4)) + sqrt(2)*log(-sqrt(2)*a^(1/4)*b^(1/ 
4)*sqrt(x) + sqrt(b)*x + sqrt(a))/(a^(1/4)*b^(3/4)))/a^3 - 2/45*(5*a^2*c^3 
 + 45*(b^2*c^3 - 3*a*b*c^2*d + 3*a^2*c*d^2)*x^4 - 9*(a*b*c^3 - 3*a^2*c^2*d 
)*x^2)/(a^3*x^(9/2))
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 483 vs. \(2 (183) = 366\).

Time = 0.14 (sec) , antiderivative size = 483, normalized size of antiderivative = 2.04 \[ \int \frac {\left (c+d x^2\right )^3}{x^{11/2} \left (a+b x^2\right )} \, dx=-\frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} b^{3} c^{3} - 3 \, \left (a b^{3}\right )^{\frac {3}{4}} a b^{2} c^{2} d + 3 \, \left (a b^{3}\right )^{\frac {3}{4}} a^{2} b c d^{2} - \left (a b^{3}\right )^{\frac {3}{4}} a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} + 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, a^{4} b^{3}} - \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} b^{3} c^{3} - 3 \, \left (a b^{3}\right )^{\frac {3}{4}} a b^{2} c^{2} d + 3 \, \left (a b^{3}\right )^{\frac {3}{4}} a^{2} b c d^{2} - \left (a b^{3}\right )^{\frac {3}{4}} a^{3} d^{3}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \left (\frac {a}{b}\right )^{\frac {1}{4}} - 2 \, \sqrt {x}\right )}}{2 \, \left (\frac {a}{b}\right )^{\frac {1}{4}}}\right )}{2 \, a^{4} b^{3}} + \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} b^{3} c^{3} - 3 \, \left (a b^{3}\right )^{\frac {3}{4}} a b^{2} c^{2} d + 3 \, \left (a b^{3}\right )^{\frac {3}{4}} a^{2} b c d^{2} - \left (a b^{3}\right )^{\frac {3}{4}} a^{3} d^{3}\right )} \log \left (\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{4 \, a^{4} b^{3}} - \frac {\sqrt {2} {\left (\left (a b^{3}\right )^{\frac {3}{4}} b^{3} c^{3} - 3 \, \left (a b^{3}\right )^{\frac {3}{4}} a b^{2} c^{2} d + 3 \, \left (a b^{3}\right )^{\frac {3}{4}} a^{2} b c d^{2} - \left (a b^{3}\right )^{\frac {3}{4}} a^{3} d^{3}\right )} \log \left (-\sqrt {2} \sqrt {x} \left (\frac {a}{b}\right )^{\frac {1}{4}} + x + \sqrt {\frac {a}{b}}\right )}{4 \, a^{4} b^{3}} - \frac {2 \, {\left (45 \, b^{2} c^{3} x^{4} - 135 \, a b c^{2} d x^{4} + 135 \, a^{2} c d^{2} x^{4} - 9 \, a b c^{3} x^{2} + 27 \, a^{2} c^{2} d x^{2} + 5 \, a^{2} c^{3}\right )}}{45 \, a^{3} x^{\frac {9}{2}}} \] Input:

integrate((d*x^2+c)^3/x^(11/2)/(b*x^2+a),x, algorithm="giac")
 

Output:

-1/2*sqrt(2)*((a*b^3)^(3/4)*b^3*c^3 - 3*(a*b^3)^(3/4)*a*b^2*c^2*d + 3*(a*b 
^3)^(3/4)*a^2*b*c*d^2 - (a*b^3)^(3/4)*a^3*d^3)*arctan(1/2*sqrt(2)*(sqrt(2) 
*(a/b)^(1/4) + 2*sqrt(x))/(a/b)^(1/4))/(a^4*b^3) - 1/2*sqrt(2)*((a*b^3)^(3 
/4)*b^3*c^3 - 3*(a*b^3)^(3/4)*a*b^2*c^2*d + 3*(a*b^3)^(3/4)*a^2*b*c*d^2 - 
(a*b^3)^(3/4)*a^3*d^3)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) - 2*sqrt(x 
))/(a/b)^(1/4))/(a^4*b^3) + 1/4*sqrt(2)*((a*b^3)^(3/4)*b^3*c^3 - 3*(a*b^3) 
^(3/4)*a*b^2*c^2*d + 3*(a*b^3)^(3/4)*a^2*b*c*d^2 - (a*b^3)^(3/4)*a^3*d^3)* 
log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^4*b^3) - 1/4*sqrt(2)*( 
(a*b^3)^(3/4)*b^3*c^3 - 3*(a*b^3)^(3/4)*a*b^2*c^2*d + 3*(a*b^3)^(3/4)*a^2* 
b*c*d^2 - (a*b^3)^(3/4)*a^3*d^3)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sq 
rt(a/b))/(a^4*b^3) - 2/45*(45*b^2*c^3*x^4 - 135*a*b*c^2*d*x^4 + 135*a^2*c* 
d^2*x^4 - 9*a*b*c^3*x^2 + 27*a^2*c^2*d*x^2 + 5*a^2*c^3)/(a^3*x^(9/2))
 

Mupad [B] (verification not implemented)

Time = 0.78 (sec) , antiderivative size = 591, normalized size of antiderivative = 2.49 \[ \int \frac {\left (c+d x^2\right )^3}{x^{11/2} \left (a+b x^2\right )} \, dx=\frac {\mathrm {atan}\left (\frac {\sqrt {x}\,{\left (a\,d-b\,c\right )}^3\,\left (16\,a^{16}\,b^2\,d^6-96\,a^{15}\,b^3\,c\,d^5+240\,a^{14}\,b^4\,c^2\,d^4-320\,a^{13}\,b^5\,c^3\,d^3+240\,a^{12}\,b^6\,c^4\,d^2-96\,a^{11}\,b^7\,c^5\,d+16\,a^{10}\,b^8\,c^6\right )}{{\left (-a\right )}^{13/4}\,b^{3/4}\,\left (16\,a^{16}\,b\,d^9-144\,a^{15}\,b^2\,c\,d^8+576\,a^{14}\,b^3\,c^2\,d^7-1344\,a^{13}\,b^4\,c^3\,d^6+2016\,a^{12}\,b^5\,c^4\,d^5-2016\,a^{11}\,b^6\,c^5\,d^4+1344\,a^{10}\,b^7\,c^6\,d^3-576\,a^9\,b^8\,c^7\,d^2+144\,a^8\,b^9\,c^8\,d-16\,a^7\,b^{10}\,c^9\right )}\right )\,{\left (a\,d-b\,c\right )}^3}{{\left (-a\right )}^{13/4}\,b^{3/4}}-\frac {\frac {2\,c^3}{9\,a}+\frac {2\,c^2\,x^2\,\left (3\,a\,d-b\,c\right )}{5\,a^2}+\frac {2\,c\,x^4\,\left (3\,a^2\,d^2-3\,a\,b\,c\,d+b^2\,c^2\right )}{a^3}}{x^{9/2}}-\frac {\mathrm {atanh}\left (\frac {\sqrt {x}\,{\left (a\,d-b\,c\right )}^3\,\left (16\,a^{16}\,b^2\,d^6-96\,a^{15}\,b^3\,c\,d^5+240\,a^{14}\,b^4\,c^2\,d^4-320\,a^{13}\,b^5\,c^3\,d^3+240\,a^{12}\,b^6\,c^4\,d^2-96\,a^{11}\,b^7\,c^5\,d+16\,a^{10}\,b^8\,c^6\right )}{{\left (-a\right )}^{13/4}\,b^{3/4}\,\left (16\,a^{16}\,b\,d^9-144\,a^{15}\,b^2\,c\,d^8+576\,a^{14}\,b^3\,c^2\,d^7-1344\,a^{13}\,b^4\,c^3\,d^6+2016\,a^{12}\,b^5\,c^4\,d^5-2016\,a^{11}\,b^6\,c^5\,d^4+1344\,a^{10}\,b^7\,c^6\,d^3-576\,a^9\,b^8\,c^7\,d^2+144\,a^8\,b^9\,c^8\,d-16\,a^7\,b^{10}\,c^9\right )}\right )\,{\left (a\,d-b\,c\right )}^3}{{\left (-a\right )}^{13/4}\,b^{3/4}} \] Input:

int((c + d*x^2)^3/(x^(11/2)*(a + b*x^2)),x)
 

Output:

(atan((x^(1/2)*(a*d - b*c)^3*(16*a^10*b^8*c^6 + 16*a^16*b^2*d^6 - 96*a^11* 
b^7*c^5*d - 96*a^15*b^3*c*d^5 + 240*a^12*b^6*c^4*d^2 - 320*a^13*b^5*c^3*d^ 
3 + 240*a^14*b^4*c^2*d^4))/((-a)^(13/4)*b^(3/4)*(16*a^16*b*d^9 - 16*a^7*b^ 
10*c^9 + 144*a^8*b^9*c^8*d - 144*a^15*b^2*c*d^8 - 576*a^9*b^8*c^7*d^2 + 13 
44*a^10*b^7*c^6*d^3 - 2016*a^11*b^6*c^5*d^4 + 2016*a^12*b^5*c^4*d^5 - 1344 
*a^13*b^4*c^3*d^6 + 576*a^14*b^3*c^2*d^7)))*(a*d - b*c)^3)/((-a)^(13/4)*b^ 
(3/4)) - ((2*c^3)/(9*a) + (2*c^2*x^2*(3*a*d - b*c))/(5*a^2) + (2*c*x^4*(3* 
a^2*d^2 + b^2*c^2 - 3*a*b*c*d))/a^3)/x^(9/2) - (atanh((x^(1/2)*(a*d - b*c) 
^3*(16*a^10*b^8*c^6 + 16*a^16*b^2*d^6 - 96*a^11*b^7*c^5*d - 96*a^15*b^3*c* 
d^5 + 240*a^12*b^6*c^4*d^2 - 320*a^13*b^5*c^3*d^3 + 240*a^14*b^4*c^2*d^4)) 
/((-a)^(13/4)*b^(3/4)*(16*a^16*b*d^9 - 16*a^7*b^10*c^9 + 144*a^8*b^9*c^8*d 
 - 144*a^15*b^2*c*d^8 - 576*a^9*b^8*c^7*d^2 + 1344*a^10*b^7*c^6*d^3 - 2016 
*a^11*b^6*c^5*d^4 + 2016*a^12*b^5*c^4*d^5 - 1344*a^13*b^4*c^3*d^6 + 576*a^ 
14*b^3*c^2*d^7)))*(a*d - b*c)^3)/((-a)^(13/4)*b^(3/4))
 

Reduce [B] (verification not implemented)

Time = 0.21 (sec) , antiderivative size = 768, normalized size of antiderivative = 3.24 \[ \int \frac {\left (c+d x^2\right )^3}{x^{11/2} \left (a+b x^2\right )} \, dx =\text {Too large to display} \] Input:

int((d*x^2+c)^3/x^(11/2)/(b*x^2+a),x)
 

Output:

( - 90*sqrt(x)*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 
 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))*a**3*d**3*x**4 + 270*sqrt 
(x)*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(x)* 
sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))*a**2*b*c*d**2*x**4 - 270*sqrt(x)*b** 
(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(b) 
)/(b**(1/4)*a**(1/4)*sqrt(2)))*a*b**2*c**2*d*x**4 + 90*sqrt(x)*b**(1/4)*a* 
*(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(b))/(b**(1 
/4)*a**(1/4)*sqrt(2)))*b**3*c**3*x**4 + 90*sqrt(x)*b**(1/4)*a**(3/4)*sqrt( 
2)*atan((b**(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4) 
*sqrt(2)))*a**3*d**3*x**4 - 270*sqrt(x)*b**(1/4)*a**(3/4)*sqrt(2)*atan((b* 
*(1/4)*a**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))* 
a**2*b*c*d**2*x**4 + 270*sqrt(x)*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)* 
a**(1/4)*sqrt(2) + 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))*a*b**2* 
c**2*d*x**4 - 90*sqrt(x)*b**(1/4)*a**(3/4)*sqrt(2)*atan((b**(1/4)*a**(1/4) 
*sqrt(2) + 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt(2)))*b**3*c**3*x**4 
+ 45*sqrt(x)*b**(1/4)*a**(3/4)*sqrt(2)*log( - sqrt(x)*b**(1/4)*a**(1/4)*sq 
rt(2) + sqrt(a) + sqrt(b)*x)*a**3*d**3*x**4 - 135*sqrt(x)*b**(1/4)*a**(3/4 
)*sqrt(2)*log( - sqrt(x)*b**(1/4)*a**(1/4)*sqrt(2) + sqrt(a) + sqrt(b)*x)* 
a**2*b*c*d**2*x**4 + 135*sqrt(x)*b**(1/4)*a**(3/4)*sqrt(2)*log( - sqrt(x)* 
b**(1/4)*a**(1/4)*sqrt(2) + sqrt(a) + sqrt(b)*x)*a*b**2*c**2*d*x**4 - 4...