\(\int \frac {1}{x^{3/2} (b x^2+c x^4)^{3/2}} \, dx\) [277]

Optimal result

Integrand size = 21, antiderivative size = 173 \[ \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {1}{b x^{5/2} \sqrt {b x^2+c x^4}}-\frac {9 \sqrt {b x^2+c x^4}}{7 b^2 x^{9/2}}+\frac {15 c \sqrt {b x^2+c x^4}}{7 b^3 x^{5/2}}+\frac {15 c^{7/4} x \left (\sqrt {b}+\sqrt {c} x\right ) \sqrt {\frac {b+c x^2}{\left (\sqrt {b}+\sqrt {c} x\right )^2}} \operatorname {EllipticF}\left (2 \arctan \left (\frac {\sqrt [4]{c} \sqrt {x}}{\sqrt [4]{b}}\right ),\frac {1}{2}\right )}{14 b^{13/4} \sqrt {b x^2+c x^4}} \] Output:

1/b/x^(5/2)/(c*x^4+b*x^2)^(1/2)-9/7*(c*x^4+b*x^2)^(1/2)/b^2/x^(9/2)+15/7*c 
*(c*x^4+b*x^2)^(1/2)/b^3/x^(5/2)+15/14*c^(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))/b^(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 = 10.02 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.35 \[ \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )^{3/2}} \, dx=-\frac {2 \sqrt {1+\frac {c x^2}{b}} \operatorname {Hypergeometric2F1}\left (-\frac {7}{4},\frac {3}{2},-\frac {3}{4},-\frac {c x^2}{b}\right )}{7 b x^{5/2} \sqrt {x^2 \left (b+c x^2\right )}} \] Input:

Integrate[1/(x^(3/2)*(b*x^2 + c*x^4)^(3/2)),x]
 

Output:

(-2*Sqrt[1 + (c*x^2)/b]*Hypergeometric2F1[-7/4, 3/2, -3/4, -((c*x^2)/b)])/ 
(7*b*x^(5/2)*Sqrt[x^2*(b + c*x^2)])
 

Rubi [A] (verified)

Time = 0.56 (sec) , antiderivative size = 189, 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 = {1428, 1430, 1430, 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.

\(\displaystyle \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )^{3/2}} \, dx\)

\(\Big \downarrow \) 1428

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

\(\Big \downarrow \) 1430

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

\(\Big \downarrow \) 1430

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

\(\Big \downarrow \) 1431

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

\(\Big \downarrow \) 266

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

\(\Big \downarrow \) 761

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

Input:

Int[1/(x^(3/2)*(b*x^2 + c*x^4)^(3/2)),x]
 

Output:

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

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 1428

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

rule 1430

Int[((d_.)*(x_))^(m_)*((b_.)*(x_)^2 + (c_.)*(x_)^4)^(p_), x_Symbol] :> Simp 
[d*(d*x)^(m - 1)*((b*x^2 + c*x^4)^(p + 1)/(b*(m + 2*p + 1))), x] - Simp[c*( 
(m + 4*p + 3)/(b*d^2*(m + 2*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] && LtQ[m + 2*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 = 1.75 (sec) , antiderivative size = 141, normalized size of antiderivative = 0.82

method	result	size
default	\(\frac {\left (c \,x^{2}+b \right ) \left (15 \sqrt {-b c}\, \sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {2}\, \sqrt {\frac {-c x +\sqrt {-b c}}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {c x +\sqrt {-b c}}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right ) c \,x^{3}+30 c^{2} x^{4}+12 b c \,x^{2}-4 b^{2}\right )}{14 \left (c \,x^{4}+b \,x^{2}\right )^{\frac {3}{2}} \sqrt {x}\, b^{3}}\)	\(141\)
risch	\(-\frac {2 \left (c \,x^{2}+b \right ) \left (-4 c \,x^{2}+b \right )}{7 b^{3} x^{\frac {5}{2}} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}+\frac {c^{2} \left (\frac {4 \sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{c \sqrt {c \,x^{3}+b x}}+7 b \left (\frac {x}{b \sqrt {\left (x^{2}+\frac {b}{c}\right ) c x}}+\frac {\sqrt {-b c}\, \sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}\, \sqrt {-\frac {c x}{\sqrt {-b c}}}\, \operatorname {EllipticF}\left (\sqrt {\frac {\left (x +\frac {\sqrt {-b c}}{c}\right ) c}{\sqrt {-b c}}}, \frac {\sqrt {2}}{2}\right )}{2 b c \sqrt {c \,x^{3}+b x}}\right )\right ) \sqrt {x}\, \sqrt {x \left (c \,x^{2}+b \right )}}{7 b^{3} \sqrt {x^{2} \left (c \,x^{2}+b \right )}}\)	\(316\)

Input:

int(1/x^(3/2)/(c*x^4+b*x^2)^(3/2),x,method=_RETURNVERBOSE)
 

Output:

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

Fricas [A] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.50 \[ \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )^{3/2}} \, dx=\frac {15 \, {\left (c^{2} x^{7} + b c x^{5}\right )} \sqrt {c} {\rm weierstrassPInverse}\left (-\frac {4 \, b}{c}, 0, x\right ) + {\left (15 \, c^{2} x^{4} + 6 \, b c x^{2} - 2 \, b^{2}\right )} \sqrt {c x^{4} + b x^{2}} \sqrt {x}}{7 \, {\left (b^{3} c x^{7} + b^{4} x^{5}\right )}} \] Input:

integrate(1/x^(3/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="fricas")
 

Output:

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

Sympy [F]

\[ \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {1}{x^{\frac {3}{2}} \left (x^{2} \left (b + c x^{2}\right )\right )^{\frac {3}{2}}}\, dx \] Input:

integrate(1/x**(3/2)/(c*x**4+b*x**2)**(3/2),x)
 

Output:

Integral(1/(x**(3/2)*(x**2*(b + c*x**2))**(3/2)), x)
 

Maxima [F]

\[ \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/x^(3/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="maxima")
 

Output:

integrate(1/((c*x^4 + b*x^2)^(3/2)*x^(3/2)), x)
 

Giac [F]

\[ \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )^{3/2}} \, dx=\int { \frac {1}{{\left (c x^{4} + b x^{2}\right )}^{\frac {3}{2}} x^{\frac {3}{2}}} \,d x } \] Input:

integrate(1/x^(3/2)/(c*x^4+b*x^2)^(3/2),x, algorithm="giac")
 

Output:

integrate(1/((c*x^4 + b*x^2)^(3/2)*x^(3/2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {1}{x^{3/2}\,{\left (c\,x^4+b\,x^2\right )}^{3/2}} \,d x \] Input:

int(1/(x^(3/2)*(b*x^2 + c*x^4)^(3/2)),x)
 

Output:

int(1/(x^(3/2)*(b*x^2 + c*x^4)^(3/2)), x)
 

Reduce [F]

\[ \int \frac {1}{x^{3/2} \left (b x^2+c x^4\right )^{3/2}} \, dx=\int \frac {\sqrt {x}\, \sqrt {c \,x^{2}+b}}{c^{2} x^{9}+2 b c \,x^{7}+b^{2} x^{5}}d x \] Input:

int(1/x^(3/2)/(c*x^4+b*x^2)^(3/2),x)
 

Output:

int((sqrt(x)*sqrt(b + c*x**2))/(b**2*x**5 + 2*b*c*x**7 + c**2*x**9),x)