Integrand size = 19, antiderivative size = 66 \[ \int \frac {1}{\sqrt {c x} \left (a+b x^2\right )^{3/4}} \, dx=-\frac {2 \sqrt {b} \left (1+\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2} \operatorname {EllipticF}\left (\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{\sqrt {a} c^2 \left (a+b x^2\right )^{3/4}} \] Output:
-2*b^(1/2)*(1+a/b/x^2)^(3/4)*(c*x)^(3/2)*InverseJacobiAM(1/2*arccot(b^(1/2 )*x/a^(1/2)),2^(1/2))/a^(1/2)/c^2/(b*x^2+a)^(3/4)
Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
Time = 10.01 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.82 \[ \int \frac {1}{\sqrt {c x} \left (a+b x^2\right )^{3/4}} \, dx=\frac {2 x \left (1+\frac {b x^2}{a}\right )^{3/4} \operatorname {Hypergeometric2F1}\left (\frac {1}{4},\frac {3}{4},\frac {5}{4},-\frac {b x^2}{a}\right )}{\sqrt {c x} \left (a+b x^2\right )^{3/4}} \] Input:
Integrate[1/(Sqrt[c*x]*(a + b*x^2)^(3/4)),x]
Output:
(2*x*(1 + (b*x^2)/a)^(3/4)*Hypergeometric2F1[1/4, 3/4, 5/4, -((b*x^2)/a)]) /(Sqrt[c*x]*(a + b*x^2)^(3/4))
Time = 0.23 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.05, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.263, Rules used = {266, 768, 858, 807, 229}
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.
\(\displaystyle \int \frac {1}{\sqrt {c x} \left (a+b x^2\right )^{3/4}} \, dx\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {2 \int \frac {1}{\left (b x^2+a\right )^{3/4}}d\sqrt {c x}}{c}\) |
\(\Big \downarrow \) 768 |
\(\displaystyle \frac {2 (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\left (\frac {a}{b x^2}+1\right )^{3/4} (c x)^{3/2}}d\sqrt {c x}}{c \left (a+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle -\frac {2 (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\sqrt {c x} \left (\frac {a x^2 c^4}{b}+1\right )^{3/4}}d\frac {1}{\sqrt {c x}}}{c \left (a+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle -\frac {(c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \int \frac {1}{\left (\frac {a x c^3}{b}+1\right )^{3/4}}d(c x)}{c \left (a+b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 229 |
\(\displaystyle -\frac {2 \sqrt {b} (c x)^{3/2} \left (\frac {a}{b x^2}+1\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arctan \left (\frac {\sqrt {a} c^2 x}{\sqrt {b}}\right ),2\right )}{\sqrt {a} c^2 \left (a+b x^2\right )^{3/4}}\) |
Input:
Int[1/(Sqrt[c*x]*(a + b*x^2)^(3/4)),x]
Output:
(-2*Sqrt[b]*(1 + a/(b*x^2))^(3/4)*(c*x)^(3/2)*EllipticF[ArcTan[(Sqrt[a]*c^ 2*x)/Sqrt[b]]/2, 2])/(Sqrt[a]*c^2*(a + b*x^2)^(3/4))
Int[((a_) + (b_.)*(x_)^2)^(-3/4), x_Symbol] :> Simp[(2/(a^(3/4)*Rt[b/a, 2]) )*EllipticF[(1/2)*ArcTan[Rt[b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a , 0] && PosQ[b/a]
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]
Int[((a_) + (b_.)*(x_)^4)^(-3/4), x_Symbol] :> Simp[x^3*((1 + a/(b*x^4))^(3 /4)/(a + b*x^4)^(3/4)) Int[1/(x^3*(1 + a/(b*x^4))^(3/4)), x], x] /; FreeQ [{a, b}, x]
Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Simp[1/k Subst[Int[x^((m + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]
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]
\[\int \frac {1}{\sqrt {c x}\, \left (b \,x^{2}+a \right )^{\frac {3}{4}}}d x\]
Input:
int(1/(c*x)^(1/2)/(b*x^2+a)^(3/4),x)
Output:
int(1/(c*x)^(1/2)/(b*x^2+a)^(3/4),x)
\[ \int \frac {1}{\sqrt {c x} \left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x}} \,d x } \] Input:
integrate(1/(c*x)^(1/2)/(b*x^2+a)^(3/4),x, algorithm="fricas")
Output:
integral((b*x^2 + a)^(1/4)*sqrt(c*x)/(b*c*x^3 + a*c*x), x)
Result contains complex when optimal does not.
Time = 0.85 (sec) , antiderivative size = 31, normalized size of antiderivative = 0.47 \[ \int \frac {1}{\sqrt {c x} \left (a+b x^2\right )^{3/4}} \, dx=- \frac {{{}_{2}F_{1}\left (\begin {matrix} \frac {1}{2}, \frac {3}{4} \\ \frac {3}{2} \end {matrix}\middle | {\frac {a e^{i \pi }}{b x^{2}}} \right )}}{b^{\frac {3}{4}} \sqrt {c} x} \] Input:
integrate(1/(c*x)**(1/2)/(b*x**2+a)**(3/4),x)
Output:
-hyper((1/2, 3/4), (3/2,), a*exp_polar(I*pi)/(b*x**2))/(b**(3/4)*sqrt(c)*x )
\[ \int \frac {1}{\sqrt {c x} \left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x}} \,d x } \] Input:
integrate(1/(c*x)^(1/2)/(b*x^2+a)^(3/4),x, algorithm="maxima")
Output:
integrate(1/((b*x^2 + a)^(3/4)*sqrt(c*x)), x)
\[ \int \frac {1}{\sqrt {c x} \left (a+b x^2\right )^{3/4}} \, dx=\int { \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x}} \,d x } \] Input:
integrate(1/(c*x)^(1/2)/(b*x^2+a)^(3/4),x, algorithm="giac")
Output:
integrate(1/((b*x^2 + a)^(3/4)*sqrt(c*x)), x)
Timed out. \[ \int \frac {1}{\sqrt {c x} \left (a+b x^2\right )^{3/4}} \, dx=\int \frac {1}{\sqrt {c\,x}\,{\left (b\,x^2+a\right )}^{3/4}} \,d x \] Input:
int(1/((c*x)^(1/2)*(a + b*x^2)^(3/4)),x)
Output:
int(1/((c*x)^(1/2)*(a + b*x^2)^(3/4)), x)
\[ \int \frac {1}{\sqrt {c x} \left (a+b x^2\right )^{3/4}} \, dx=\frac {\sqrt {c}\, \left (\int \frac {\sqrt {x}\, \left (b \,x^{2}+a \right )^{\frac {5}{4}}}{b^{2} x^{5}+2 a b \,x^{3}+a^{2} x}d x \right )}{c} \] Input:
int(1/(c*x)^(1/2)/(b*x^2+a)^(3/4),x)
Output:
(sqrt(c)*int((sqrt(x)*(a + b*x**2)**(5/4))/(a**2*x + 2*a*b*x**3 + b**2*x** 5),x))/c