Integrand size = 20, antiderivative size = 92 \[ \int \frac {\sqrt [4]{a-b x^2}}{\sqrt {c x}} \, dx=\frac {\sqrt {c x} \sqrt [4]{a-b x^2}}{c}-\frac {\sqrt {a} \sqrt {b} \left (1-\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2} \operatorname {EllipticF}\left (\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right ),2\right )}{c^2 \left (a-b x^2\right )^{3/4}} \] Output:
(c*x)^(1/2)*(-b*x^2+a)^(1/4)/c-a^(1/2)*b^(1/2)*(1-a/b/x^2)^(3/4)*(c*x)^(3/ 2)*InverseJacobiAM(1/2*arccsc(b^(1/2)*x/a^(1/2)),2^(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 = 55, normalized size of antiderivative = 0.60 \[ \int \frac {\sqrt [4]{a-b x^2}}{\sqrt {c x}} \, dx=\frac {2 x \sqrt [4]{a-b x^2} \operatorname {Hypergeometric2F1}\left (-\frac {1}{4},\frac {1}{4},\frac {5}{4},\frac {b x^2}{a}\right )}{\sqrt {c x} \sqrt [4]{1-\frac {b x^2}{a}}} \] Input:
Integrate[(a - b*x^2)^(1/4)/Sqrt[c*x],x]
Output:
(2*x*(a - b*x^2)^(1/4)*Hypergeometric2F1[-1/4, 1/4, 5/4, (b*x^2)/a])/(Sqrt [c*x]*(1 - (b*x^2)/a)^(1/4))
Time = 0.26 (sec) , antiderivative size = 95, normalized size of antiderivative = 1.03, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.300, Rules used = {248, 266, 768, 858, 807, 230}
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.
\(\displaystyle \int \frac {\sqrt [4]{a-b x^2}}{\sqrt {c x}} \, dx\) |
\(\Big \downarrow \) 248 |
\(\displaystyle \frac {1}{2} a \int \frac {1}{\sqrt {c x} \left (a-b x^2\right )^{3/4}}dx+\frac {\sqrt {c x} \sqrt [4]{a-b x^2}}{c}\) |
\(\Big \downarrow \) 266 |
\(\displaystyle \frac {a \int \frac {1}{\left (a-b x^2\right )^{3/4}}d\sqrt {c x}}{c}+\frac {\sqrt {c x} \sqrt [4]{a-b x^2}}{c}\) |
\(\Big \downarrow \) 768 |
\(\displaystyle \frac {a (c x)^{3/2} \left (1-\frac {a}{b x^2}\right )^{3/4} \int \frac {1}{\left (1-\frac {a}{b x^2}\right )^{3/4} (c x)^{3/2}}d\sqrt {c x}}{c \left (a-b x^2\right )^{3/4}}+\frac {\sqrt {c x} \sqrt [4]{a-b x^2}}{c}\) |
\(\Big \downarrow \) 858 |
\(\displaystyle \frac {\sqrt {c x} \sqrt [4]{a-b x^2}}{c}-\frac {a (c x)^{3/2} \left (1-\frac {a}{b x^2}\right )^{3/4} \int \frac {1}{\sqrt {c x} \left (1-\frac {a c^4 x^2}{b}\right )^{3/4}}d\frac {1}{\sqrt {c x}}}{c \left (a-b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 807 |
\(\displaystyle \frac {\sqrt {c x} \sqrt [4]{a-b x^2}}{c}-\frac {a (c x)^{3/2} \left (1-\frac {a}{b x^2}\right )^{3/4} \int \frac {1}{\left (1-\frac {a c^3 x}{b}\right )^{3/4}}d(c x)}{2 c \left (a-b x^2\right )^{3/4}}\) |
\(\Big \downarrow \) 230 |
\(\displaystyle \frac {\sqrt {c x} \sqrt [4]{a-b x^2}}{c}-\frac {\sqrt {a} \sqrt {b} (c x)^{3/2} \left (1-\frac {a}{b x^2}\right )^{3/4} \operatorname {EllipticF}\left (\frac {1}{2} \arcsin \left (\frac {\sqrt {a} c^2 x}{\sqrt {b}}\right ),2\right )}{c^2 \left (a-b x^2\right )^{3/4}}\) |
Input:
Int[(a - b*x^2)^(1/4)/Sqrt[c*x],x]
Output:
(Sqrt[c*x]*(a - b*x^2)^(1/4))/c - (Sqrt[a]*Sqrt[b]*(1 - a/(b*x^2))^(3/4)*( c*x)^(3/2)*EllipticF[ArcSin[(Sqrt[a]*c^2*x)/Sqrt[b]]/2, 2])/(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)*ArcSin[Rt[-b/a, 2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ [a, 0] && NegQ[b/a]
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ (m + 1)*((a + b*x^2)^p/(c*(m + 2*p + 1))), x] + Simp[2*a*(p/(m + 2*p + 1)) Int[(c*x)^m*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x] && GtQ[ p, 0] && NeQ[m + 2*p + 1, 0] && IntBinomialQ[a, b, c, 2, m, p, x]
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]
\[\int \frac {\left (-b \,x^{2}+a \right )^{\frac {1}{4}}}{\sqrt {c x}}d x\]
Input:
int((-b*x^2+a)^(1/4)/(c*x)^(1/2),x)
Output:
int((-b*x^2+a)^(1/4)/(c*x)^(1/2),x)
\[ \int \frac {\sqrt [4]{a-b x^2}}{\sqrt {c x}} \, dx=\int { \frac {{\left (-b x^{2} + a\right )}^{\frac {1}{4}}}{\sqrt {c x}} \,d x } \] Input:
integrate((-b*x^2+a)^(1/4)/(c*x)^(1/2),x, algorithm="fricas")
Output:
integral((-b*x^2 + a)^(1/4)*sqrt(c*x)/(c*x), x)
Result contains complex when optimal does not.
Time = 0.68 (sec) , antiderivative size = 39, normalized size of antiderivative = 0.42 \[ \int \frac {\sqrt [4]{a-b x^2}}{\sqrt {c x}} \, dx=- \frac {i \sqrt [4]{b} x e^{\frac {3 i \pi }{4}} {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{2}, - \frac {1}{4} \\ \frac {1}{2} \end {matrix}\middle | {\frac {a}{b x^{2}}} \right )}}{\sqrt {c}} \] Input:
integrate((-b*x**2+a)**(1/4)/(c*x)**(1/2),x)
Output:
-I*b**(1/4)*x*exp(3*I*pi/4)*hyper((-1/2, -1/4), (1/2,), a/(b*x**2))/sqrt(c )
\[ \int \frac {\sqrt [4]{a-b x^2}}{\sqrt {c x}} \, dx=\int { \frac {{\left (-b x^{2} + a\right )}^{\frac {1}{4}}}{\sqrt {c x}} \,d x } \] Input:
integrate((-b*x^2+a)^(1/4)/(c*x)^(1/2),x, algorithm="maxima")
Output:
integrate((-b*x^2 + a)^(1/4)/sqrt(c*x), x)
\[ \int \frac {\sqrt [4]{a-b x^2}}{\sqrt {c x}} \, dx=\int { \frac {{\left (-b x^{2} + a\right )}^{\frac {1}{4}}}{\sqrt {c x}} \,d x } \] Input:
integrate((-b*x^2+a)^(1/4)/(c*x)^(1/2),x, algorithm="giac")
Output:
integrate((-b*x^2 + a)^(1/4)/sqrt(c*x), x)
Timed out. \[ \int \frac {\sqrt [4]{a-b x^2}}{\sqrt {c x}} \, dx=\int \frac {{\left (a-b\,x^2\right )}^{1/4}}{\sqrt {c\,x}} \,d x \] Input:
int((a - b*x^2)^(1/4)/(c*x)^(1/2),x)
Output:
int((a - b*x^2)^(1/4)/(c*x)^(1/2), x)
\[ \int \frac {\sqrt [4]{a-b x^2}}{\sqrt {c x}} \, dx=\frac {\sqrt {c}\, \left (\int \frac {\sqrt {x}\, \left (-b \,x^{2}+a \right )^{\frac {1}{4}}}{x}d x \right )}{c} \] Input:
int((-b*x^2+a)^(1/4)/(c*x)^(1/2),x)
Output:
(sqrt(c)*int((sqrt(x)*(a - b*x**2)**(1/4))/x,x))/c