\(\int \frac {(c+d x^2)^3}{x^{13/2} (a+b x^2)^2} \, dx\) [784]

Optimal result

Integrand size = 24, antiderivative size = 292 \[ \int \frac {\left (c+d x^2\right )^3}{x^{13/2} \left (a+b x^2\right )^2} \, dx=-\frac {2 c^3}{11 a^2 x^{11/2}}+\frac {2 c^2 (2 b c-3 a d)}{7 a^3 x^{7/2}}-\frac {2 c (b c-a d)^2}{a^4 x^{3/2}}-\frac {(b c-a d)^3 \sqrt {x}}{2 a^4 \left (a+b x^2\right )}+\frac {3 (b c-a d)^2 (5 b c-a d) \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{19/4} \sqrt [4]{b}}-\frac {3 (b c-a d)^2 (5 b c-a d) \arctan \left (1+\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )}{4 \sqrt {2} a^{19/4} \sqrt [4]{b}}-\frac {3 (b c-a d)^2 (5 b c-a d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{4 \sqrt {2} a^{19/4} \sqrt [4]{b}} \] Output:

-2/11*c^3/a^2/x^(11/2)+2/7*c^2*(-3*a*d+2*b*c)/a^3/x^(7/2)-2*c*(-a*d+b*c)^2 
/a^4/x^(3/2)-1/2*(-a*d+b*c)^3*x^(1/2)/a^4/(b*x^2+a)+3/8*(-a*d+b*c)^2*(-a*d 
+5*b*c)*arctan(1-2^(1/2)*b^(1/4)*x^(1/2)/a^(1/4))*2^(1/2)/a^(19/4)/b^(1/4) 
-3/8*(-a*d+b*c)^2*(-a*d+5*b*c)*arctan(1+2^(1/2)*b^(1/4)*x^(1/2)/a^(1/4))*2 
^(1/2)/a^(19/4)/b^(1/4)-3/8*(-a*d+b*c)^2*(-a*d+5*b*c)*arctanh(2^(1/2)*a^(1 
/4)*b^(1/4)*x^(1/2)/(a^(1/2)+b^(1/2)*x))*2^(1/2)/a^(19/4)/b^(1/4)
 

Mathematica [A] (verified)

Time = 0.52 (sec) , antiderivative size = 265, normalized size of antiderivative = 0.91 \[ \int \frac {\left (c+d x^2\right )^3}{x^{13/2} \left (a+b x^2\right )^2} \, dx=\frac {-\frac {4 a^{3/4} \left (385 b^3 c^3 x^6+11 a b^2 c^2 x^4 \left (20 c-77 d x^2\right )+a^2 b c x^2 \left (-60 c^2-484 c d x^2+539 d^2 x^4\right )+a^3 \left (28 c^3+132 c^2 d x^2+308 c d^2 x^4-77 d^3 x^6\right )\right )}{x^{11/2} \left (a+b x^2\right )}+\frac {231 \sqrt {2} (b c-a d)^2 (5 b c-a d) \arctan \left (\frac {\sqrt {a}-\sqrt {b} x}{\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}\right )}{\sqrt [4]{b}}+\frac {231 \sqrt {2} (b c-a d)^2 (-5 b c+a d) \text {arctanh}\left (\frac {\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}}{\sqrt {a}+\sqrt {b} x}\right )}{\sqrt [4]{b}}}{616 a^{19/4}} \] Input:

Integrate[(c + d*x^2)^3/(x^(13/2)*(a + b*x^2)^2),x]
 

Output:

((-4*a^(3/4)*(385*b^3*c^3*x^6 + 11*a*b^2*c^2*x^4*(20*c - 77*d*x^2) + a^2*b 
*c*x^2*(-60*c^2 - 484*c*d*x^2 + 539*d^2*x^4) + a^3*(28*c^3 + 132*c^2*d*x^2 
 + 308*c*d^2*x^4 - 77*d^3*x^6)))/(x^(11/2)*(a + b*x^2)) + (231*Sqrt[2]*(b* 
c - a*d)^2*(5*b*c - a*d)*ArcTan[(Sqrt[a] - Sqrt[b]*x)/(Sqrt[2]*a^(1/4)*b^( 
1/4)*Sqrt[x])])/b^(1/4) + (231*Sqrt[2]*(b*c - a*d)^2*(-5*b*c + a*d)*ArcTan 
h[(Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x])/(Sqrt[a] + Sqrt[b]*x)])/b^(1/4))/(616* 
a^(19/4))
 

Rubi [A] (verified)

Time = 0.62 (sec) , antiderivative size = 416, normalized size of antiderivative = 1.42, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.208, Rules used = {368, 968, 25, 1040, 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.

\(\displaystyle \int \frac {\left (c+d x^2\right )^3}{x^{13/2} \left (a+b x^2\right )^2} \, dx\)

\(\Big \downarrow \) 368

\(\displaystyle 2 \int \frac {\left (d x^2+c\right )^3}{x^6 \left (b x^2+a\right )^2}d\sqrt {x}\)

\(\Big \downarrow \) 968

\(\displaystyle 2 \left (\frac {\left (c+d x^2\right )^2 (b c-a d)}{4 a b x^{11/2} \left (a+b x^2\right )}-\frac {\int -\frac {\left (d x^2+c\right ) \left (d (7 b c-3 a d) x^2+c (15 b c-11 a d)\right )}{x^6 \left (b x^2+a\right )}d\sqrt {x}}{4 a b}\right )\)

\(\Big \downarrow \) 25

\(\displaystyle 2 \left (\frac {\int \frac {\left (d x^2+c\right ) \left (d (7 b c-3 a d) x^2+c (15 b c-11 a d)\right )}{x^6 \left (b x^2+a\right )}d\sqrt {x}}{4 a b}+\frac {\left (c+d x^2\right )^2 (b c-a d)}{4 a b x^{11/2} \left (a+b x^2\right )}\right )\)

\(\Big \downarrow \) 1040

\(\displaystyle 2 \left (\frac {\int \left (-\frac {(11 a d-15 b c) c^2}{a x^6}-\frac {\left (15 b^2 c^2-33 a b d c+14 a^2 d^2\right ) c}{a^2 x^4}+\frac {3 b (a d-5 b c) (a d-b c)^2}{a^3 \left (b x^2+a\right )}-\frac {3 (a d-5 b c) (a d-b c)^2}{a^3 x^2}\right )d\sqrt {x}}{4 a b}+\frac {\left (c+d x^2\right )^2 (b c-a d)}{4 a b x^{11/2} \left (a+b x^2\right )}\right )\)

\(\Big \downarrow \) 2009

\(\displaystyle 2 \left (\frac {\frac {3 b^{3/4} \arctan \left (1-\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right ) (b c-a d)^2 (5 b c-a d)}{2 \sqrt {2} a^{15/4}}-\frac {3 b^{3/4} \arctan \left (\frac {\sqrt {2} \sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}+1\right ) (b c-a d)^2 (5 b c-a d)}{2 \sqrt {2} a^{15/4}}+\frac {3 b^{3/4} (b c-a d)^2 (5 b c-a d) \log \left (-\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{4 \sqrt {2} a^{15/4}}-\frac {3 b^{3/4} (b c-a d)^2 (5 b c-a d) \log \left (\sqrt {2} \sqrt [4]{a} \sqrt [4]{b} \sqrt {x}+\sqrt {a}+\sqrt {b} x\right )}{4 \sqrt {2} a^{15/4}}-\frac {(b c-a d)^2 (5 b c-a d)}{a^3 x^{3/2}}+\frac {c \left (14 a^2 d^2-33 a b c d+15 b^2 c^2\right )}{7 a^2 x^{7/2}}-\frac {c^2 (15 b c-11 a d)}{11 a x^{11/2}}}{4 a b}+\frac {\left (c+d x^2\right )^2 (b c-a d)}{4 a b x^{11/2} \left (a+b x^2\right )}\right )\)

Input:

Int[(c + d*x^2)^3/(x^(13/2)*(a + b*x^2)^2),x]
 

Output:

2*(((b*c - a*d)*(c + d*x^2)^2)/(4*a*b*x^(11/2)*(a + b*x^2)) + (-1/11*(c^2* 
(15*b*c - 11*a*d))/(a*x^(11/2)) + (c*(15*b^2*c^2 - 33*a*b*c*d + 14*a^2*d^2 
))/(7*a^2*x^(7/2)) - ((b*c - a*d)^2*(5*b*c - a*d))/(a^3*x^(3/2)) + (3*b^(3 
/4)*(b*c - a*d)^2*(5*b*c - a*d)*ArcTan[1 - (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/ 
4)])/(2*Sqrt[2]*a^(15/4)) - (3*b^(3/4)*(b*c - a*d)^2*(5*b*c - a*d)*ArcTan[ 
1 + (Sqrt[2]*b^(1/4)*Sqrt[x])/a^(1/4)])/(2*Sqrt[2]*a^(15/4)) + (3*b^(3/4)* 
(b*c - a*d)^2*(5*b*c - a*d)*Log[Sqrt[a] - Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] 
+ Sqrt[b]*x])/(4*Sqrt[2]*a^(15/4)) - (3*b^(3/4)*(b*c - a*d)^2*(5*b*c - a*d 
)*Log[Sqrt[a] + Sqrt[2]*a^(1/4)*b^(1/4)*Sqrt[x] + Sqrt[b]*x])/(4*Sqrt[2]*a 
^(15/4)))/(4*a*b))
 

Defintions of rubi rules used

rule 25

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

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 968

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_ 
))^(q_), x_Symbol] :> Simp[(-(c*b - a*d))*(e*x)^(m + 1)*(a + b*x^n)^(p + 1) 
*((c + d*x^n)^(q - 1)/(a*b*e*n*(p + 1))), x] + Simp[1/(a*b*n*(p + 1))   Int 
[(e*x)^m*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 2)*Simp[c*(c*b*n*(p + 1) + (c 
*b - a*d)*(m + 1)) + d*(c*b*n*(p + 1) + (c*b - a*d)*(m + n*(q - 1) + 1))*x^ 
n, x], x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[ 
n, 0] && LtQ[p, -1] && GtQ[q, 1] && IntBinomialQ[a, b, c, d, e, m, n, p, q, 
 x]
 

rule 1040

Int[((g_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n 
_))^(q_.)*((e_) + (f_.)*(x_)^(n_))^(r_.), x_Symbol] :> Int[ExpandIntegrand[ 
(g*x)^m*(a + b*x^n)^p*(c + d*x^n)^q*(e + f*x^n)^r, x], x] /; FreeQ[{a, b, c 
, d, e, f, g, m, n}, x] && IGtQ[p, -2] && IGtQ[q, 0] && IGtQ[r, 0]
 

rule 2009

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

Maple [A] (verified)

Time = 0.48 (sec) , antiderivative size = 235, normalized size of antiderivative = 0.80

method	result	size
risch	\(-\frac {2 c \left (77 a^{2} d^{2} x^{4}-154 a b c d \,x^{4}+77 b^{2} c^{2} x^{4}+33 a^{2} c d \,x^{2}-22 a b \,c^{2} x^{2}+7 a^{2} c^{2}\right )}{77 a^{4} x^{\frac {11}{2}}}-\frac {\left (2 a^{2} d^{2}-4 a b c d +2 b^{2} c^{2}\right ) \left (\frac {\left (-\frac {a d}{4}+\frac {b c}{4}\right ) \sqrt {x}}{b \,x^{2}+a}-\frac {3 \left (a d -5 b c \right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{32 a}\right )}{a^{4}}\)	\(235\)
derivativedivides	\(\frac {\frac {2 \left (\frac {1}{4} a^{3} d^{3}-\frac {3}{4} a^{2} b c \,d^{2}+\frac {3}{4} a \,b^{2} c^{2} d -\frac {1}{4} b^{3} c^{3}\right ) \sqrt {x}}{b \,x^{2}+a}+\frac {3 \left (a^{3} d^{3}-7 a^{2} b c \,d^{2}+11 a \,b^{2} c^{2} d -5 b^{3} c^{3}\right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{16 a}}{a^{4}}-\frac {2 c^{3}}{11 a^{2} x^{\frac {11}{2}}}-\frac {2 c^{2} \left (3 a d -2 b c \right )}{7 a^{3} x^{\frac {7}{2}}}-\frac {2 c \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{a^{4} x^{\frac {3}{2}}}\)	\(260\)
default	\(\frac {\frac {2 \left (\frac {1}{4} a^{3} d^{3}-\frac {3}{4} a^{2} b c \,d^{2}+\frac {3}{4} a \,b^{2} c^{2} d -\frac {1}{4} b^{3} c^{3}\right ) \sqrt {x}}{b \,x^{2}+a}+\frac {3 \left (a^{3} d^{3}-7 a^{2} b c \,d^{2}+11 a \,b^{2} c^{2} d -5 b^{3} c^{3}\right ) \left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {2}\, \left (\ln \left (\frac {x +\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}{x -\left (\frac {a}{b}\right )^{\frac {1}{4}} \sqrt {x}\, \sqrt {2}+\sqrt {\frac {a}{b}}}\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}+1\right )+2 \arctan \left (\frac {\sqrt {2}\, \sqrt {x}}{\left (\frac {a}{b}\right )^{\frac {1}{4}}}-1\right )\right )}{16 a}}{a^{4}}-\frac {2 c^{3}}{11 a^{2} x^{\frac {11}{2}}}-\frac {2 c^{2} \left (3 a d -2 b c \right )}{7 a^{3} x^{\frac {7}{2}}}-\frac {2 c \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}{a^{4} x^{\frac {3}{2}}}\)	\(260\)

Input:

int((d*x^2+c)^3/x^(13/2)/(b*x^2+a)^2,x,method=_RETURNVERBOSE)
 

Output:

-2/77*c*(77*a^2*d^2*x^4-154*a*b*c*d*x^4+77*b^2*c^2*x^4+33*a^2*c*d*x^2-22*a 
*b*c^2*x^2+7*a^2*c^2)/a^4/x^(11/2)-1/a^4*(2*a^2*d^2-4*a*b*c*d+2*b^2*c^2)*( 
(-1/4*a*d+1/4*b*c)*x^(1/2)/(b*x^2+a)-3/32*(a*d-5*b*c)*(a/b)^(1/4)/a*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 = 1783, normalized size of antiderivative = 6.11 \[ \int \frac {\left (c+d x^2\right )^3}{x^{13/2} \left (a+b x^2\right )^2} \, dx=\text {Too large to display} \] Input:

integrate((d*x^2+c)^3/x^(13/2)/(b*x^2+a)^2,x, algorithm="fricas")
 

Output:

1/616*(231*(a^4*b*x^8 + a^5*x^6)*(-(625*b^12*c^12 - 5500*a*b^11*c^11*d + 2 
1650*a^2*b^10*c^10*d^2 - 50220*a^3*b^9*c^9*d^3 + 76111*a^4*b^8*c^8*d^4 - 7 
8968*a^5*b^7*c^7*d^5 + 57148*a^6*b^6*c^6*d^6 - 28856*a^7*b^5*c^5*d^7 + 100 
15*a^8*b^4*c^4*d^8 - 2316*a^9*b^3*c^3*d^9 + 338*a^10*b^2*c^2*d^10 - 28*a^1 
1*b*c*d^11 + a^12*d^12)/(a^19*b))^(1/4)*log(3*a^5*(-(625*b^12*c^12 - 5500* 
a*b^11*c^11*d + 21650*a^2*b^10*c^10*d^2 - 50220*a^3*b^9*c^9*d^3 + 76111*a^ 
4*b^8*c^8*d^4 - 78968*a^5*b^7*c^7*d^5 + 57148*a^6*b^6*c^6*d^6 - 28856*a^7* 
b^5*c^5*d^7 + 10015*a^8*b^4*c^4*d^8 - 2316*a^9*b^3*c^3*d^9 + 338*a^10*b^2* 
c^2*d^10 - 28*a^11*b*c*d^11 + a^12*d^12)/(a^19*b))^(1/4) - 3*(5*b^3*c^3 - 
11*a*b^2*c^2*d + 7*a^2*b*c*d^2 - a^3*d^3)*sqrt(x)) - 231*(-I*a^4*b*x^8 - I 
*a^5*x^6)*(-(625*b^12*c^12 - 5500*a*b^11*c^11*d + 21650*a^2*b^10*c^10*d^2 
- 50220*a^3*b^9*c^9*d^3 + 76111*a^4*b^8*c^8*d^4 - 78968*a^5*b^7*c^7*d^5 + 
57148*a^6*b^6*c^6*d^6 - 28856*a^7*b^5*c^5*d^7 + 10015*a^8*b^4*c^4*d^8 - 23 
16*a^9*b^3*c^3*d^9 + 338*a^10*b^2*c^2*d^10 - 28*a^11*b*c*d^11 + a^12*d^12) 
/(a^19*b))^(1/4)*log(3*I*a^5*(-(625*b^12*c^12 - 5500*a*b^11*c^11*d + 21650 
*a^2*b^10*c^10*d^2 - 50220*a^3*b^9*c^9*d^3 + 76111*a^4*b^8*c^8*d^4 - 78968 
*a^5*b^7*c^7*d^5 + 57148*a^6*b^6*c^6*d^6 - 28856*a^7*b^5*c^5*d^7 + 10015*a 
^8*b^4*c^4*d^8 - 2316*a^9*b^3*c^3*d^9 + 338*a^10*b^2*c^2*d^10 - 28*a^11*b* 
c*d^11 + a^12*d^12)/(a^19*b))^(1/4) - 3*(5*b^3*c^3 - 11*a*b^2*c^2*d + 7*a^ 
2*b*c*d^2 - a^3*d^3)*sqrt(x)) - 231*(I*a^4*b*x^8 + I*a^5*x^6)*(-(625*b^...
 

Sympy [F(-1)]

Timed out. \[ \int \frac {\left (c+d x^2\right )^3}{x^{13/2} \left (a+b x^2\right )^2} \, dx=\text {Timed out} \] Input:

integrate((d*x**2+c)**3/x**(13/2)/(b*x**2+a)**2,x)
 

Output:

Timed out
 

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 454 vs. \(2 (226) = 452\).

Time = 0.13 (sec) , antiderivative size = 454, normalized size of antiderivative = 1.55 \[ \int \frac {\left (c+d x^2\right )^3}{x^{13/2} \left (a+b x^2\right )^2} \, dx=-\frac {77 \, {\left (5 \, b^{3} c^{3} - 11 \, a b^{2} c^{2} d + 7 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} x^{6} + 28 \, a^{3} c^{3} + 44 \, {\left (5 \, a b^{2} c^{3} - 11 \, a^{2} b c^{2} d + 7 \, a^{3} c d^{2}\right )} x^{4} - 12 \, {\left (5 \, a^{2} b c^{3} - 11 \, a^{3} c^{2} d\right )} x^{2}}{154 \, {\left (a^{4} b x^{\frac {15}{2}} + a^{5} x^{\frac {11}{2}}\right )}} - \frac {3 \, {\left (\frac {2 \, \sqrt {2} {\left (5 \, b^{3} c^{3} - 11 \, a b^{2} c^{2} d + 7 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} + 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {2 \, \sqrt {2} {\left (5 \, b^{3} c^{3} - 11 \, a b^{2} c^{2} d + 7 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} - 2 \, \sqrt {b} \sqrt {x}\right )}}{2 \, \sqrt {\sqrt {a} \sqrt {b}}}\right )}{\sqrt {a} \sqrt {\sqrt {a} \sqrt {b}}} + \frac {\sqrt {2} {\left (5 \, b^{3} c^{3} - 11 \, a b^{2} c^{2} d + 7 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}} - \frac {\sqrt {2} {\left (5 \, b^{3} c^{3} - 11 \, a b^{2} c^{2} d + 7 \, a^{2} b c d^{2} - a^{3} d^{3}\right )} \log \left (-\sqrt {2} a^{\frac {1}{4}} b^{\frac {1}{4}} \sqrt {x} + \sqrt {b} x + \sqrt {a}\right )}{a^{\frac {3}{4}} b^{\frac {1}{4}}}\right )}}{16 \, a^{4}} \] Input:

integrate((d*x^2+c)^3/x^(13/2)/(b*x^2+a)^2,x, algorithm="maxima")
 

Output:

-1/154*(77*(5*b^3*c^3 - 11*a*b^2*c^2*d + 7*a^2*b*c*d^2 - a^3*d^3)*x^6 + 28 
*a^3*c^3 + 44*(5*a*b^2*c^3 - 11*a^2*b*c^2*d + 7*a^3*c*d^2)*x^4 - 12*(5*a^2 
*b*c^3 - 11*a^3*c^2*d)*x^2)/(a^4*b*x^(15/2) + a^5*x^(11/2)) - 3/16*(2*sqrt 
(2)*(5*b^3*c^3 - 11*a*b^2*c^2*d + 7*a^2*b*c*d^2 - a^3*d^3)*arctan(1/2*sqrt 
(2)*(sqrt(2)*a^(1/4)*b^(1/4) + 2*sqrt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/( 
sqrt(a)*sqrt(sqrt(a)*sqrt(b))) + 2*sqrt(2)*(5*b^3*c^3 - 11*a*b^2*c^2*d + 7 
*a^2*b*c*d^2 - a^3*d^3)*arctan(-1/2*sqrt(2)*(sqrt(2)*a^(1/4)*b^(1/4) - 2*s 
qrt(b)*sqrt(x))/sqrt(sqrt(a)*sqrt(b)))/(sqrt(a)*sqrt(sqrt(a)*sqrt(b))) + s 
qrt(2)*(5*b^3*c^3 - 11*a*b^2*c^2*d + 7*a^2*b*c*d^2 - a^3*d^3)*log(sqrt(2)* 
a^(1/4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/(a^(3/4)*b^(1/4)) - sqrt(2) 
*(5*b^3*c^3 - 11*a*b^2*c^2*d + 7*a^2*b*c*d^2 - a^3*d^3)*log(-sqrt(2)*a^(1/ 
4)*b^(1/4)*sqrt(x) + sqrt(b)*x + sqrt(a))/(a^(3/4)*b^(1/4)))/a^4
 

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 549 vs. \(2 (226) = 452\).

Time = 0.14 (sec) , antiderivative size = 549, normalized size of antiderivative = 1.88 \[ \int \frac {\left (c+d x^2\right )^3}{x^{13/2} \left (a+b x^2\right )^2} \, dx=-\frac {3 \, \sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{3} c^{3} - 11 \, \left (a b^{3}\right )^{\frac {1}{4}} a b^{2} c^{2} d + 7 \, \left (a b^{3}\right )^{\frac {1}{4}} a^{2} b c d^{2} - \left (a b^{3}\right )^{\frac {1}{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 )}{8 \, a^{5} b} - \frac {3 \, \sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{3} c^{3} - 11 \, \left (a b^{3}\right )^{\frac {1}{4}} a b^{2} c^{2} d + 7 \, \left (a b^{3}\right )^{\frac {1}{4}} a^{2} b c d^{2} - \left (a b^{3}\right )^{\frac {1}{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 )}{8 \, a^{5} b} - \frac {3 \, \sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{3} c^{3} - 11 \, \left (a b^{3}\right )^{\frac {1}{4}} a b^{2} c^{2} d + 7 \, \left (a b^{3}\right )^{\frac {1}{4}} a^{2} b c d^{2} - \left (a b^{3}\right )^{\frac {1}{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 )}{16 \, a^{5} b} + \frac {3 \, \sqrt {2} {\left (5 \, \left (a b^{3}\right )^{\frac {1}{4}} b^{3} c^{3} - 11 \, \left (a b^{3}\right )^{\frac {1}{4}} a b^{2} c^{2} d + 7 \, \left (a b^{3}\right )^{\frac {1}{4}} a^{2} b c d^{2} - \left (a b^{3}\right )^{\frac {1}{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 )}{16 \, a^{5} b} - \frac {b^{3} c^{3} \sqrt {x} - 3 \, a b^{2} c^{2} d \sqrt {x} + 3 \, a^{2} b c d^{2} \sqrt {x} - a^{3} d^{3} \sqrt {x}}{2 \, {\left (b x^{2} + a\right )} a^{4}} - \frac {2 \, {\left (77 \, b^{2} c^{3} x^{4} - 154 \, a b c^{2} d x^{4} + 77 \, a^{2} c d^{2} x^{4} - 22 \, a b c^{3} x^{2} + 33 \, a^{2} c^{2} d x^{2} + 7 \, a^{2} c^{3}\right )}}{77 \, a^{4} x^{\frac {11}{2}}} \] Input:

integrate((d*x^2+c)^3/x^(13/2)/(b*x^2+a)^2,x, algorithm="giac")
 

Output:

-3/8*sqrt(2)*(5*(a*b^3)^(1/4)*b^3*c^3 - 11*(a*b^3)^(1/4)*a*b^2*c^2*d + 7*( 
a*b^3)^(1/4)*a^2*b*c*d^2 - (a*b^3)^(1/4)*a^3*d^3)*arctan(1/2*sqrt(2)*(sqrt 
(2)*(a/b)^(1/4) + 2*sqrt(x))/(a/b)^(1/4))/(a^5*b) - 3/8*sqrt(2)*(5*(a*b^3) 
^(1/4)*b^3*c^3 - 11*(a*b^3)^(1/4)*a*b^2*c^2*d + 7*(a*b^3)^(1/4)*a^2*b*c*d^ 
2 - (a*b^3)^(1/4)*a^3*d^3)*arctan(-1/2*sqrt(2)*(sqrt(2)*(a/b)^(1/4) - 2*sq 
rt(x))/(a/b)^(1/4))/(a^5*b) - 3/16*sqrt(2)*(5*(a*b^3)^(1/4)*b^3*c^3 - 11*( 
a*b^3)^(1/4)*a*b^2*c^2*d + 7*(a*b^3)^(1/4)*a^2*b*c*d^2 - (a*b^3)^(1/4)*a^3 
*d^3)*log(sqrt(2)*sqrt(x)*(a/b)^(1/4) + x + sqrt(a/b))/(a^5*b) + 3/16*sqrt 
(2)*(5*(a*b^3)^(1/4)*b^3*c^3 - 11*(a*b^3)^(1/4)*a*b^2*c^2*d + 7*(a*b^3)^(1 
/4)*a^2*b*c*d^2 - (a*b^3)^(1/4)*a^3*d^3)*log(-sqrt(2)*sqrt(x)*(a/b)^(1/4) 
+ x + sqrt(a/b))/(a^5*b) - 1/2*(b^3*c^3*sqrt(x) - 3*a*b^2*c^2*d*sqrt(x) + 
3*a^2*b*c*d^2*sqrt(x) - a^3*d^3*sqrt(x))/((b*x^2 + a)*a^4) - 2/77*(77*b^2* 
c^3*x^4 - 154*a*b*c^2*d*x^4 + 77*a^2*c*d^2*x^4 - 22*a*b*c^3*x^2 + 33*a^2*c 
^2*d*x^2 + 7*a^2*c^3)/(a^4*x^(11/2))
 

Mupad [B] (verification not implemented)

Time = 1.07 (sec) , antiderivative size = 1774, normalized size of antiderivative = 6.08 \[ \int \frac {\left (c+d x^2\right )^3}{x^{13/2} \left (a+b x^2\right )^2} \, dx=\text {Too large to display} \] Input:

int((c + d*x^2)^3/(x^(13/2)*(a + b*x^2)^2),x)
 

Output:

- ((2*c^3)/(11*a) - (x^6*(a^3*d^3 - 5*b^3*c^3 + 11*a*b^2*c^2*d - 7*a^2*b*c 
*d^2))/(2*a^4) + (6*c^2*x^2*(11*a*d - 5*b*c))/(77*a^2) + (2*c*x^4*(7*a^2*d 
^2 + 5*b^2*c^2 - 11*a*b*c*d))/(7*a^3))/(a*x^(11/2) + b*x^(15/2)) - (atan(( 
((x^(1/2)*(7200*a^12*b^9*c^6 + 288*a^18*b^3*d^6 - 31680*a^13*b^8*c^5*d - 4 
032*a^17*b^4*c*d^5 + 55008*a^14*b^7*c^4*d^2 - 47232*a^15*b^6*c^3*d^3 + 204 
48*a^16*b^5*c^2*d^4) - (3*(a*d - b*c)^2*(a*d - 5*b*c)*(3840*a^17*b^6*c^3 - 
 768*a^20*b^3*d^3 - 8448*a^18*b^5*c^2*d + 5376*a^19*b^4*c*d^2))/(8*(-a)^(1 
9/4)*b^(1/4)))*(a*d - b*c)^2*(a*d - 5*b*c)*3i)/(8*(-a)^(19/4)*b^(1/4)) + ( 
(x^(1/2)*(7200*a^12*b^9*c^6 + 288*a^18*b^3*d^6 - 31680*a^13*b^8*c^5*d - 40 
32*a^17*b^4*c*d^5 + 55008*a^14*b^7*c^4*d^2 - 47232*a^15*b^6*c^3*d^3 + 2044 
8*a^16*b^5*c^2*d^4) + (3*(a*d - b*c)^2*(a*d - 5*b*c)*(3840*a^17*b^6*c^3 - 
768*a^20*b^3*d^3 - 8448*a^18*b^5*c^2*d + 5376*a^19*b^4*c*d^2))/(8*(-a)^(19 
/4)*b^(1/4)))*(a*d - b*c)^2*(a*d - 5*b*c)*3i)/(8*(-a)^(19/4)*b^(1/4)))/((3 
*(x^(1/2)*(7200*a^12*b^9*c^6 + 288*a^18*b^3*d^6 - 31680*a^13*b^8*c^5*d - 4 
032*a^17*b^4*c*d^5 + 55008*a^14*b^7*c^4*d^2 - 47232*a^15*b^6*c^3*d^3 + 204 
48*a^16*b^5*c^2*d^4) - (3*(a*d - b*c)^2*(a*d - 5*b*c)*(3840*a^17*b^6*c^3 - 
 768*a^20*b^3*d^3 - 8448*a^18*b^5*c^2*d + 5376*a^19*b^4*c*d^2))/(8*(-a)^(1 
9/4)*b^(1/4)))*(a*d - b*c)^2*(a*d - 5*b*c))/(8*(-a)^(19/4)*b^(1/4)) - (3*( 
x^(1/2)*(7200*a^12*b^9*c^6 + 288*a^18*b^3*d^6 - 31680*a^13*b^8*c^5*d - 403 
2*a^17*b^4*c*d^5 + 55008*a^14*b^7*c^4*d^2 - 47232*a^15*b^6*c^3*d^3 + 20...
 

Reduce [B] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 1509, normalized size of antiderivative = 5.17 \[ \int \frac {\left (c+d x^2\right )^3}{x^{13/2} \left (a+b x^2\right )^2} \, dx =\text {Too large to display} \] Input:

int((d*x^2+c)^3/x^(13/2)/(b*x^2+a)^2,x)
 

Output:

( - 462*sqrt(x)*b**(3/4)*a**(1/4)*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**4*d**3*x**5 + 3234*sq 
rt(x)*b**(3/4)*a**(1/4)*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*b*c*d**2*x**5 - 462*sqrt(x)*b 
**(3/4)*a**(1/4)*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*b*d**3*x**7 - 5082*sqrt(x)*b**(3/4)* 
a**(1/4)*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**2*c**2*d*x**5 + 3234*sqrt(x)*b**(3/4)*a** 
(1/4)*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**2*c*d**2*x**7 + 2310*sqrt(x)*b**(3/4)*a**(1/ 
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**3*c**3*x**5 - 5082*sqrt(x)*b**(3/4)*a**(1/4)*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**3*c**2*d*x**7 + 2310*sqrt(x)*b**(3/4)*a**(1/4)*sqrt(2)*ata 
n((b**(1/4)*a**(1/4)*sqrt(2) - 2*sqrt(x)*sqrt(b))/(b**(1/4)*a**(1/4)*sqrt( 
2)))*b**4*c**3*x**7 + 462*sqrt(x)*b**(3/4)*a**(1/4)*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**4*d 
**3*x**5 - 3234*sqrt(x)*b**(3/4)*a**(1/4)*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*b*c*d**2*x* 
*5 + 462*sqrt(x)*b**(3/4)*a**(1/4)*sqrt(2)*atan((b**(1/4)*a**(1/4)*sqrt...