\(\int \frac {1}{(c x)^{11/2} (a+b x^2)^{5/4}} \, dx\) [1061]

Optimal result

Integrand size = 19, antiderivative size = 157 \[ \int \frac {1}{(c x)^{11/2} \left (a+b x^2\right )^{5/4}} \, dx=-\frac {2}{9 a c (c x)^{9/2} \sqrt [4]{a+b x^2}}+\frac {4 b}{9 a^2 c^3 (c x)^{5/2} \sqrt [4]{a+b x^2}}-\frac {8 b^2}{3 a^3 c^5 \sqrt {c x} \sqrt [4]{a+b x^2}}+\frac {16 b^{5/2} \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {c x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{3 a^{7/2} c^6 \sqrt [4]{a+b x^2}} \] Output:

-2/9/a/c/(c*x)^(9/2)/(b*x^2+a)^(1/4)+4/9*b/a^2/c^3/(c*x)^(5/2)/(b*x^2+a)^( 
1/4)-8/3*b^2/a^3/c^5/(c*x)^(1/2)/(b*x^2+a)^(1/4)+16/3*b^(5/2)*(1+a/b/x^2)^ 
(1/4)*(c*x)^(1/2)*EllipticE(sin(1/2*arccot(b^(1/2)*x/a^(1/2))),2^(1/2))/a^ 
(7/2)/c^6/(b*x^2+a)^(1/4)
 

Mathematica [C] (verified)

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

Time = 10.01 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.38 \[ \int \frac {1}{(c x)^{11/2} \left (a+b x^2\right )^{5/4}} \, dx=-\frac {2 x \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {9}{4},\frac {5}{4},-\frac {5}{4},-\frac {b x^2}{a}\right )}{9 a (c x)^{11/2} \sqrt [4]{a+b x^2}} \] Input:

Integrate[1/((c*x)^(11/2)*(a + b*x^2)^(5/4)),x]
 

Output:

(-2*x*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[-9/4, 5/4, -5/4, -((b*x^2)/a 
)])/(9*a*(c*x)^(11/2)*(a + b*x^2)^(1/4))
 

Rubi [A] (verified)

Time = 0.28 (sec) , antiderivative size = 175, normalized size of antiderivative = 1.11, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.316, Rules used = {251, 251, 251, 249, 858, 212}

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.

\(\displaystyle \int \frac {1}{(c x)^{11/2} \left (a+b x^2\right )^{5/4}} \, dx\)

\(\Big \downarrow \) 251

\(\displaystyle -\frac {10 b \int \frac {1}{(c x)^{7/2} \left (b x^2+a\right )^{5/4}}dx}{9 a c^2}-\frac {2}{9 a c (c x)^{9/2} \sqrt [4]{a+b x^2}}\)

\(\Big \downarrow \) 251

\(\displaystyle -\frac {10 b \left (-\frac {6 b \int \frac {1}{(c x)^{3/2} \left (b x^2+a\right )^{5/4}}dx}{5 a c^2}-\frac {2}{5 a c (c x)^{5/2} \sqrt [4]{a+b x^2}}\right )}{9 a c^2}-\frac {2}{9 a c (c x)^{9/2} \sqrt [4]{a+b x^2}}\)

\(\Big \downarrow \) 251

\(\displaystyle -\frac {10 b \left (-\frac {6 b \left (-\frac {2 b \int \frac {\sqrt {c x}}{\left (b x^2+a\right )^{5/4}}dx}{a c^2}-\frac {2}{a c \sqrt {c x} \sqrt [4]{a+b x^2}}\right )}{5 a c^2}-\frac {2}{5 a c (c x)^{5/2} \sqrt [4]{a+b x^2}}\right )}{9 a c^2}-\frac {2}{9 a c (c x)^{9/2} \sqrt [4]{a+b x^2}}\)

\(\Big \downarrow \) 249

\(\displaystyle -\frac {10 b \left (-\frac {6 b \left (-\frac {2 \sqrt {c x} \sqrt [4]{\frac {a}{b x^2}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4} x^2}dx}{a c^2 \sqrt [4]{a+b x^2}}-\frac {2}{a c \sqrt {c x} \sqrt [4]{a+b x^2}}\right )}{5 a c^2}-\frac {2}{5 a c (c x)^{5/2} \sqrt [4]{a+b x^2}}\right )}{9 a c^2}-\frac {2}{9 a c (c x)^{9/2} \sqrt [4]{a+b x^2}}\)

\(\Big \downarrow \) 858

\(\displaystyle -\frac {10 b \left (-\frac {6 b \left (\frac {2 \sqrt {c x} \sqrt [4]{\frac {a}{b x^2}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4}}d\frac {1}{x}}{a c^2 \sqrt [4]{a+b x^2}}-\frac {2}{a c \sqrt {c x} \sqrt [4]{a+b x^2}}\right )}{5 a c^2}-\frac {2}{5 a c (c x)^{5/2} \sqrt [4]{a+b x^2}}\right )}{9 a c^2}-\frac {2}{9 a c (c x)^{9/2} \sqrt [4]{a+b x^2}}\)

\(\Big \downarrow \) 212

\(\displaystyle -\frac {10 b \left (-\frac {6 b \left (\frac {4 \sqrt {b} \sqrt {c x} \sqrt [4]{\frac {a}{b x^2}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x}\right )\right |2\right )}{a^{3/2} c^2 \sqrt [4]{a+b x^2}}-\frac {2}{a c \sqrt {c x} \sqrt [4]{a+b x^2}}\right )}{5 a c^2}-\frac {2}{5 a c (c x)^{5/2} \sqrt [4]{a+b x^2}}\right )}{9 a c^2}-\frac {2}{9 a c (c x)^{9/2} \sqrt [4]{a+b x^2}}\)

Input:

Int[1/((c*x)^(11/2)*(a + b*x^2)^(5/4)),x]
 

Output:

-2/(9*a*c*(c*x)^(9/2)*(a + b*x^2)^(1/4)) - (10*b*(-2/(5*a*c*(c*x)^(5/2)*(a 
 + b*x^2)^(1/4)) - (6*b*(-2/(a*c*Sqrt[c*x]*(a + b*x^2)^(1/4)) + (4*Sqrt[b] 
*(1 + a/(b*x^2))^(1/4)*Sqrt[c*x]*EllipticE[ArcTan[Sqrt[a]/(Sqrt[b]*x)]/2, 
2])/(a^(3/2)*c^2*(a + b*x^2)^(1/4))))/(5*a*c^2)))/(9*a*c^2)
 

Defintions of rubi rules used

rule 212

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2/(a^(5/4)*Rt[b/a, 2]) 
)*EllipticE[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a 
, 0] && PosQ[b/a]
 

rule 249

Int[Sqrt[(c_.)*(x_)]/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[Sqrt[c* 
x]*((1 + a/(b*x^2))^(1/4)/(b*(a + b*x^2)^(1/4)))   Int[1/(x^2*(1 + a/(b*x^2 
))^(5/4)), x], x] /; FreeQ[{a, b, c}, x] && PosQ[b/a]
 

rule 251

Int[((c_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[(c*x)^ 
(m + 1)/(a*c*(m + 1)*(a + b*x^2)^(1/4)), x] - Simp[b*((2*m + 1)/(2*a*c^2*(m 
 + 1)))   Int[(c*x)^(m + 2)/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a, b, c}, x 
] && PosQ[b/a] && IntegerQ[2*m] && LtQ[m, -1]
 

rule 858

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + 
b/x^n)^p/x^(m + 2), x], x, 1/x] /; FreeQ[{a, b, p}, x] && ILtQ[n, 0] && Int 
egerQ[m]
 

Maple [F]

\[\int \frac {1}{\left (c x \right )^{\frac {11}{2}} \left (b \,x^{2}+a \right )^{\frac {5}{4}}}d x\]

Input:

int(1/(c*x)^(11/2)/(b*x^2+a)^(5/4),x)
 

Output:

int(1/(c*x)^(11/2)/(b*x^2+a)^(5/4),x)
 

Fricas [F]

\[ \int \frac {1}{(c x)^{11/2} \left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} \left (c x\right )^{\frac {11}{2}}} \,d x } \] Input:

integrate(1/(c*x)^(11/2)/(b*x^2+a)^(5/4),x, algorithm="fricas")
 

Output:

integral((b*x^2 + a)^(3/4)*sqrt(c*x)/(b^2*c^6*x^10 + 2*a*b*c^6*x^8 + a^2*c 
^6*x^6), x)
 

Sympy [F(-1)]

Timed out. \[ \int \frac {1}{(c x)^{11/2} \left (a+b x^2\right )^{5/4}} \, dx=\text {Timed out} \] Input:

integrate(1/(c*x)**(11/2)/(b*x**2+a)**(5/4),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \frac {1}{(c x)^{11/2} \left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} \left (c x\right )^{\frac {11}{2}}} \,d x } \] Input:

integrate(1/(c*x)^(11/2)/(b*x^2+a)^(5/4),x, algorithm="maxima")
 

Output:

integrate(1/((b*x^2 + a)^(5/4)*(c*x)^(11/2)), x)
 

Giac [F]

\[ \int \frac {1}{(c x)^{11/2} \left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} \left (c x\right )^{\frac {11}{2}}} \,d x } \] Input:

integrate(1/(c*x)^(11/2)/(b*x^2+a)^(5/4),x, algorithm="giac")
 

Output:

integrate(1/((b*x^2 + a)^(5/4)*(c*x)^(11/2)), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {1}{(c x)^{11/2} \left (a+b x^2\right )^{5/4}} \, dx=\int \frac {1}{{\left (c\,x\right )}^{11/2}\,{\left (b\,x^2+a\right )}^{5/4}} \,d x \] Input:

int(1/((c*x)^(11/2)*(a + b*x^2)^(5/4)),x)
 

Output:

int(1/((c*x)^(11/2)*(a + b*x^2)^(5/4)), x)
 

Reduce [B] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 57, normalized size of antiderivative = 0.36 \[ \int \frac {1}{(c x)^{11/2} \left (a+b x^2\right )^{5/4}} \, dx=\frac {2 \sqrt {c}\, \left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (-32 b^{2} x^{4}+8 a b \,x^{2}-5 a^{2}\right )}{45 \sqrt {x}\, \sqrt {b \,x^{2}+a}\, a^{3} c^{6} x^{4}} \] Input:

int(1/(c*x)^(11/2)/(b*x^2+a)^(5/4),x)
 

Output:

(2*sqrt(c)*(a + b*x**2)**(1/4)*( - 5*a**2 + 8*a*b*x**2 - 32*b**2*x**4))/(4 
5*sqrt(x)*sqrt(a + b*x**2)*a**3*c**6*x**4)