\(\int \frac {x^{17/2}}{(b x^2+c x^4)^{3/2}} \, dx\) [267]

Optimal result

Integrand size = 21, antiderivative size = 174 \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}-\frac {15 b \sqrt {b x^2+c x^4}}{7 c^3 \sqrt {x}}+\frac {9 x^{3/2} \sqrt {b x^2+c x^4}}{7 c^2}+\frac {15 b^{7/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{14 c^{13/4} \sqrt {b x^2+c x^4}} \] Output:

-x^(11/2)/c/(c*x^4+b*x^2)^(1/2)-15/7*b*(c*x^4+b*x^2)^(1/2)/c^3/x^(1/2)+9/7 
*x^(3/2)*(c*x^4+b*x^2)^(1/2)/c^2+15/14*b^(7/4)*x*(b^(1/2)+c^(1/2)*x)*((c*x 
^2+b)/(b^(1/2)+c^(1/2)*x)^2)^(1/2)*InverseJacobiAM(2*arctan(c^(1/4)*x^(1/2 
)/b^(1/4)),1/2*2^(1/2))/c^(13/4)/(c*x^4+b*x^2)^(1/2)
 

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.

Time = 10.03 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.49 \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {x^{3/2} \left (-15 b^2-6 b c x^2+2 c^2 x^4+15 b^2 \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {1}{2},\frac {5}{4},-\frac {c x^2}{b}\right )\right )}{7 c^3 \sqrt {x^2 \left (b+c x^2\right )}} \] Input:

Integrate[x^(17/2)/(b*x^2 + c*x^4)^(3/2),x]
 

Output:

(x^(3/2)*(-15*b^2 - 6*b*c*x^2 + 2*c^2*x^4 + 15*b^2*Sqrt[1 + (c*x^2)/b]*Hyp 
ergeometric2F1[1/4, 1/2, 5/4, -((c*x^2)/b)]))/(7*c^3*Sqrt[x^2*(b + c*x^2)] 
)
 

Rubi [A] (verified)

Time = 0.57 (sec) , antiderivative size = 190, normalized size of antiderivative = 1.09, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {1427, 1429, 1429, 1431, 266, 761}

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.

\(\displaystyle \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1427

\(\displaystyle \frac {9 \int \frac {x^{9/2}}{\sqrt {c x^4+b x^2}}dx}{2 c}-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 1429

\(\displaystyle \frac {9 \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \int \frac {x^{5/2}}{\sqrt {c x^4+b x^2}}dx}{7 c}\right )}{2 c}-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 1429

\(\displaystyle \frac {9 \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b \int \frac {\sqrt {x}}{\sqrt {c x^4+b x^2}}dx}{3 c}\right )}{7 c}\right )}{2 c}-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 1431

\(\displaystyle \frac {9 \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {x} \sqrt {c x^2+b}}dx}{3 c \sqrt {b x^2+c x^4}}\right )}{7 c}\right )}{2 c}-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 266

\(\displaystyle \frac {9 \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {2 b x \sqrt {b+c x^2} \int \frac {1}{\sqrt {c x^2+b}}d\sqrt {x}}{3 c \sqrt {b x^2+c x^4}}\right )}{7 c}\right )}{2 c}-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}\)

\(\Big \downarrow \) 761

\(\displaystyle \frac {9 \left (\frac {2 x^{3/2} \sqrt {b x^2+c x^4}}{7 c}-\frac {5 b \left (\frac {2 \sqrt {b x^2+c x^4}}{3 c \sqrt {x}}-\frac {b^{3/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{3 c^{5/4} \sqrt {b x^2+c x^4}}\right )}{7 c}\right )}{2 c}-\frac {x^{11/2}}{c \sqrt {b x^2+c x^4}}\)

Input:

Int[x^(17/2)/(b*x^2 + c*x^4)^(3/2),x]
 

Output:

-(x^(11/2)/(c*Sqrt[b*x^2 + c*x^4])) + (9*((2*x^(3/2)*Sqrt[b*x^2 + c*x^4])/ 
(7*c) - (5*b*((2*Sqrt[b*x^2 + c*x^4])/(3*c*Sqrt[x]) - (b^(3/4)*x*(Sqrt[b] 
+ Sqrt[c]*x)*Sqrt[(b + c*x^2)/(Sqrt[b] + Sqrt[c]*x)^2]*EllipticF[2*ArcTan[ 
(c^(1/4)*Sqrt[x])/b^(1/4)], 1/2])/(3*c^(5/4)*Sqrt[b*x^2 + c*x^4])))/(7*c)) 
)/(2*c)
 

Defintions of rubi rules used

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 761

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[( 
1 + q^2*x^2)*(Sqrt[(a + b*x^4)/(a*(1 + q^2*x^2)^2)]/(2*q*Sqrt[a + b*x^4]))* 
EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]
 

rule 1427

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[d^3*(d*x)^(m - 3)*((b*x^2 + c*x^4)^(p + 1)/(2*c*(p + 1))), x] - Simp[d^4*( 
(m + 2*p - 1)/(2*c*(p + 1)))   Int[(d*x)^(m - 4)*(b*x^2 + c*x^4)^(p + 1), x 
], x] /; FreeQ[{b, c, d, m, p}, x] &&  !IntegerQ[p] && LtQ[p, -1] && GtQ[m 
+ 2*p + 1, 2]
 

rule 1429

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[d^3*(d*x)^(m - 3)*((b*x^2 + c*x^4)^(p + 1)/(c*(m + 4*p + 1))), x] - Simp[b 
*d^2*((m + 2*p - 1)/(c*(m + 4*p + 1)))   Int[(d*x)^(m - 2)*(b*x^2 + c*x^4)^ 
p, x], x] /; FreeQ[{b, c, d, m, p}, x] &&  !IntegerQ[p] && GtQ[m + 2*p - 1, 
 0] && NeQ[m + 4*p + 1, 0]
 

rule 1431

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[(b*x^2 + c*x^4)^p/((d*x)^(2*p)*(b + c*x^2)^p)   Int[(d*x)^(m + 2*p)*(b + c 
*x^2)^p, x], x] /; FreeQ[{b, c, d, m, p}, x] &&  !IntegerQ[p]
 

Maple [A] (verified)

Time = 1.88 (sec) , antiderivative size = 144, normalized size of antiderivative = 0.83

method	result	size
default	\(\frac {x^{\frac {5}{2}} \left (c \,x^{2}+b \right ) \left (15 b^{2} \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )+4 c^{3} x^{5}-12 b \,c^{2} x^{3}-30 b^{2} c x \right )}{14 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} c^{4}}\)	\(144\)
risch	\(-\frac {2 \left (-c \,x^{2}+4 b \right ) x^{\frac {3}{2}} \left (c \,x^{2}+b \right )}{7 c^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {b^{2} \left (\frac {11 \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c \sqrt {c \,x^{3}+b x}}-7 b \left (\frac {x}{b \sqrt {\left (x^{2}+\frac {b}{c}\right ) c x}}+\frac {\sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{2 b c \sqrt {c \,x^{3}+b x}}\right )\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{7 c^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(318\)

Input:

int(x^(17/2)/(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

1/14/(c*x^4+b*x^2)^(3/2)*x^(5/2)*(c*x^2+b)*(15*b^2*(-b*c)^(1/2)*((c*x+(-b* 
c)^(1/2))/(-b*c)^(1/2))^(1/2)*2^(1/2)*((-c*x+(-b*c)^(1/2))/(-b*c)^(1/2))^( 
1/2)*(-c/(-b*c)^(1/2)*x)^(1/2)*EllipticF(((c*x+(-b*c)^(1/2))/(-b*c)^(1/2)) 
^(1/2),1/2*2^(1/2))+4*c^3*x^5-12*b*c^2*x^3-30*b^2*c*x)/c^4
 

Fricas [A] (verification not implemented)

Time = 0.08 (sec) , antiderivative size = 88, normalized size of antiderivative = 0.51 \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {15 \, {\left (b^{2} c x^{3} + b^{3} x\right )} \sqrt {c} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + {\left (2 \, c^{3} x^{4} - 6 \, b c^{2} x^{2} - 15 \, b^{2} c\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}}{7 \, {\left (c^{5} x^{3} + b c^{4} x\right )}} \] Input:

integrate(x^(17/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")
 

Output:

1/7*(15*(b^2*c*x^3 + b^3*x)*sqrt(c)*weierstrassPInverse(-4*b/c, 0, x) + (2 
*c^3*x^4 - 6*b*c^2*x^2 - 15*b^2*c)*sqrt(c*x^4 + b*x^2)*sqrt(x))/(c^5*x^3 + 
 b*c^4*x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\text {Timed out} \] Input:

integrate(x**(17/2)/(c*x**4+b*x**2)**(3/2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {x^{\frac {17}{2}}}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(x^(17/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate(x^(17/2)/(c*x^4 + b*x^2)^(3/2), x)
 

Giac [F]

\[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {x^{\frac {17}{2}}}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}}} \,d x } \] Input:

integrate(x^(17/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
 

Output:

integrate(x^(17/2)/(c*x^4 + b*x^2)^(3/2), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {x^{17/2}}{{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \] Input:

int(x^(17/2)/(b*x^2 + c*x^4)^(3/2),x)
 

Output:

int(x^(17/2)/(b*x^2 + c*x^4)^(3/2), x)
 

Reduce [F]

\[ \int \frac {x^{17/2}}{\left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {-30 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b^{2}-6 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, b c \,x^{2}+2 \sqrt {x}\, \sqrt {c \,x^{2}+b}\, c^{2} x^{4}+15 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c^{2} x^{5}+2 b c \,x^{3}+b^{2} x}d x \right ) b^{4}+15 \left (\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c^{2} x^{5}+2 b c \,x^{3}+b^{2} x}d x \right ) b^{3} c \,x^{2}}{7 c^{3} \left (c \,x^{2}+b \right )} \] Input:

int(x^(17/2)/(c*x^4+b*x^2)^(3/2),x)
 

Output:

( - 30*sqrt(x)*sqrt(b + c*x**2)*b**2 - 6*sqrt(x)*sqrt(b + c*x**2)*b*c*x**2 
 + 2*sqrt(x)*sqrt(b + c*x**2)*c**2*x**4 + 15*int((sqrt(x)*sqrt(b + c*x**2) 
)/(b**2*x + 2*b*c*x**3 + c**2*x**5),x)*b**4 + 15*int((sqrt(x)*sqrt(b + c*x 
**2))/(b**2*x + 2*b*c*x**3 + c**2*x**5),x)*b**3*c*x**2)/(7*c**3*(b + c*x** 
2))