Integrand size = 19, antiderivative size = 126 \[ \int \frac {1}{(c x)^{7/2} \left (a+b x^2\right )^{5/4}} \, dx=-\frac {2}{5 a c (c x)^{5/2} \sqrt [4]{a+b x^2}}+\frac {12 b}{5 a^2 c^3 \sqrt {c x} \sqrt [4]{a+b x^2}}-\frac {24 b^{3/2} \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {c x} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{5 a^{5/2} c^4 \sqrt [4]{a+b x^2}} \] Output:
-2/5/a/c/(c*x)^(5/2)/(b*x^2+a)^(1/4)+12/5*b/a^2/c^3/(c*x)^(1/2)/(b*x^2+a)^ (1/4)-24/5*b^(3/2)*(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^(5/2)/c^4/(b*x^2+a)^(1/4)
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 = 0.47 \[ \int \frac {1}{(c x)^{7/2} \left (a+b x^2\right )^{5/4}} \, dx=-\frac {2 x \sqrt [4]{1+\frac {b x^2}{a}} \operatorname {Hypergeometric2F1}\left (-\frac {5}{4},\frac {5}{4},-\frac {1}{4},-\frac {b x^2}{a}\right )}{5 a (c x)^{7/2} \sqrt [4]{a+b x^2}} \] Input:
Integrate[1/((c*x)^(7/2)*(a + b*x^2)^(5/4)),x]
Output:
(-2*x*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[-5/4, 5/4, -1/4, -((b*x^2)/a )])/(5*a*(c*x)^(7/2)*(a + b*x^2)^(1/4))
Time = 0.25 (sec) , antiderivative size = 135, normalized size of antiderivative = 1.07, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {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.
\(\displaystyle \int \frac {1}{(c x)^{7/2} \left (a+b x^2\right )^{5/4}} \, dx\) |
\(\Big \downarrow \) 251 |
\(\displaystyle -\frac {6 b \int \frac {1}{(c x)^{3/2} \left (b x^2+a\right )^{5/4}}dx}{5 a c^2}-\frac {2}{5 a c (c x)^{5/2} \sqrt [4]{a+b x^2}}\) |
\(\Big \downarrow \) 251 |
\(\displaystyle -\frac {6 b \left (-\frac {2 b \int \frac {\sqrt {c x}}{\left (b x^2+a\right )^{5/4}}dx}{a c^2}-\frac {2}{a c \sqrt {c x} \sqrt [4]{a+b x^2}}\right )}{5 a c^2}-\frac {2}{5 a c (c x)^{5/2} \sqrt [4]{a+b x^2}}\) |
\(\Big \downarrow \) 249 |
\(\displaystyle -\frac {6 b \left (-\frac {2 \sqrt {c x} \sqrt [4]{\frac {a}{b x^2}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4} x^2}dx}{a c^2 \sqrt [4]{a+b x^2}}-\frac {2}{a c \sqrt {c x} \sqrt [4]{a+b x^2}}\right )}{5 a c^2}-\frac {2}{5 a c (c x)^{5/2} \sqrt [4]{a+b x^2}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\frac {6 b \left (\frac {2 \sqrt {c x} \sqrt [4]{\frac {a}{b x^2}+1} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{5/4}}d\frac {1}{x}}{a c^2 \sqrt [4]{a+b x^2}}-\frac {2}{a c \sqrt {c x} \sqrt [4]{a+b x^2}}\right )}{5 a c^2}-\frac {2}{5 a c (c x)^{5/2} \sqrt [4]{a+b x^2}}\) |
\(\Big \downarrow \) 212 |
\(\displaystyle -\frac {6 b \left (\frac {4 \sqrt {b} \sqrt {c x} \sqrt [4]{\frac {a}{b x^2}+1} E\left (\left .\frac {1}{2} \arctan \left (\frac {\sqrt {a}}{\sqrt {b} x}\right )\right |2\right )}{a^{3/2} c^2 \sqrt [4]{a+b x^2}}-\frac {2}{a c \sqrt {c x} \sqrt [4]{a+b x^2}}\right )}{5 a c^2}-\frac {2}{5 a c (c x)^{5/2} \sqrt [4]{a+b x^2}}\) |
Input:
Int[1/((c*x)^(7/2)*(a + b*x^2)^(5/4)),x]
Output:
-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)
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]
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]
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]
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]
\[\int \frac {1}{\left (c x \right )^{\frac {7}{2}} \left (b \,x^{2}+a \right )^{\frac {5}{4}}}d x\]
Input:
int(1/(c*x)^(7/2)/(b*x^2+a)^(5/4),x)
Output:
int(1/(c*x)^(7/2)/(b*x^2+a)^(5/4),x)
\[ \int \frac {1}{(c x)^{7/2} \left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} \left (c x\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate(1/(c*x)^(7/2)/(b*x^2+a)^(5/4),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(3/4)*sqrt(c*x)/(b^2*c^4*x^8 + 2*a*b*c^4*x^6 + a^2*c^ 4*x^4), x)
Result contains complex when optimal does not.
Time = 22.97 (sec) , antiderivative size = 34, normalized size of antiderivative = 0.27 \[ \int \frac {1}{(c x)^{7/2} \left (a+b x^2\right )^{5/4}} \, dx=- \frac {{{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {5}{2} \\ \frac {7}{2} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{5 b^{\frac {5}{4}} c^{\frac {7}{2}} x^{5}} \] Input:
integrate(1/(c*x)**(7/2)/(b*x**2+a)**(5/4),x)
Output:
-hyper((5/4, 5/2), (7/2,), a*exp_polar(I*pi)/(b*x**2))/(5*b**(5/4)*c**(7/2 )*x**5)
\[ \int \frac {1}{(c x)^{7/2} \left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} \left (c x\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate(1/(c*x)^(7/2)/(b*x^2+a)^(5/4),x, algorithm="maxima")
Output:
integrate(1/((b*x^2 + a)^(5/4)*(c*x)^(7/2)), x)
\[ \int \frac {1}{(c x)^{7/2} \left (a+b x^2\right )^{5/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{4}} \left (c x\right )^{\frac {7}{2}}} \,d x } \] Input:
integrate(1/(c*x)^(7/2)/(b*x^2+a)^(5/4),x, algorithm="giac")
Output:
integrate(1/((b*x^2 + a)^(5/4)*(c*x)^(7/2)), x)
Timed out. \[ \int \frac {1}{(c x)^{7/2} \left (a+b x^2\right )^{5/4}} \, dx=\int \frac {1}{{\left (c\,x\right )}^{7/2}\,{\left (b\,x^2+a\right )}^{5/4}} \,d x \] Input:
int(1/((c*x)^(7/2)*(a + b*x^2)^(5/4)),x)
Output:
int(1/((c*x)^(7/2)*(a + b*x^2)^(5/4)), x)
Time = 0.20 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.37 \[ \int \frac {1}{(c x)^{7/2} \left (a+b x^2\right )^{5/4}} \, dx=\frac {2 \sqrt {c}\, \left (b \,x^{2}+a \right )^{\frac {1}{4}} \left (4 b \,x^{2}-a \right )}{5 \sqrt {x}\, \sqrt {b \,x^{2}+a}\, a^{2} c^{4} x^{2}} \] Input:
int(1/(c*x)^(7/2)/(b*x^2+a)^(5/4),x)
Output:
(2*sqrt(c)*(a + b*x**2)**(1/4)*( - a + 4*b*x**2))/(5*sqrt(x)*sqrt(a + b*x* *2)*a**2*c**4*x**2)