3.10.99 \(\int \frac {(c x)^{3/4}}{\sqrt [4]{a+b x^2}} \, dx\) [999]

Optimal. Leaf size=58 \[ \frac {4 (c x)^{7/4} \sqrt [4]{1+\frac {b x^2}{a}} \, _2F_1\left (\frac {1}{4},\frac {7}{8};\frac {15}{8};-\frac {b x^2}{a}\right )}{7 c \sqrt [4]{a+b x^2}} \]

[Out]

4/7*(c*x)^(7/4)*(1+b*x^2/a)^(1/4)*hypergeom([1/4, 7/8],[15/8],-b*x^2/a)/c/(b*x^2+a)^(1/4)

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Rubi [A]
time = 0.01, antiderivative size = 58, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {372, 371} \begin {gather*} \frac {4 (c x)^{7/4} \sqrt [4]{\frac {b x^2}{a}+1} \, _2F_1\left (\frac {1}{4},\frac {7}{8};\frac {15}{8};-\frac {b x^2}{a}\right )}{7 c \sqrt [4]{a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*x)^(3/4)/(a + b*x^2)^(1/4),x]

[Out]

(4*(c*x)^(7/4)*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[1/4, 7/8, 15/8, -((b*x^2)/a)])/(7*c*(a + b*x^2)^(1/4))

Rule 371

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[a^p*((c*x)^(m + 1)/(c*(m + 1)))*Hyperg
eometric2F1[-p, (m + 1)/n, (m + 1)/n + 1, (-b)*(x^n/a)], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rule 372

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Dist[a^IntPart[p]*((a + b*x^n)^FracPart[p]/
(1 + b*(x^n/a))^FracPart[p]), Int[(c*x)^m*(1 + b*(x^n/a))^p, x], x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[
p, 0] &&  !(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

\begin {align*} \int \frac {(c x)^{3/4}}{\sqrt [4]{a+b x^2}} \, dx &=\frac {\sqrt [4]{1+\frac {b x^2}{a}} \int \frac {(c x)^{3/4}}{\sqrt [4]{1+\frac {b x^2}{a}}} \, dx}{\sqrt [4]{a+b x^2}}\\ &=\frac {4 (c x)^{7/4} \sqrt [4]{1+\frac {b x^2}{a}} \, _2F_1\left (\frac {1}{4},\frac {7}{8};\frac {15}{8};-\frac {b x^2}{a}\right )}{7 c \sqrt [4]{a+b x^2}}\\ \end {align*}

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Mathematica [A]
time = 10.02, size = 56, normalized size = 0.97 \begin {gather*} \frac {4 x (c x)^{3/4} \sqrt [4]{1+\frac {b x^2}{a}} \, _2F_1\left (\frac {1}{4},\frac {7}{8};\frac {15}{8};-\frac {b x^2}{a}\right )}{7 \sqrt [4]{a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*x*(c*x)^(3/4)*(1 + (b*x^2)/a)^(1/4)*Hypergeometric2F1[1/4, 7/8, 15/8, -((b*x^2)/a)])/(7*(a + b*x^2)^(1/4))

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Maple [F]
time = 0.01, size = 0, normalized size = 0.00 \[\int \frac {\left (c x \right )^{\frac {3}{4}}}{\left (b \,x^{2}+a \right )^{\frac {1}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 1.85, size = 44, normalized size = 0.76 \begin {gather*} \frac {c^{\frac {3}{4}} x^{\frac {7}{4}} \Gamma \left (\frac {7}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {1}{4}, \frac {7}{8} \\ \frac {15}{8} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 \sqrt [4]{a} \Gamma \left (\frac {15}{8}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c**(3/4)*x**(7/4)*gamma(7/8)*hyper((1/4, 7/8), (15/8,), b*x**2*exp_polar(I*pi)/a)/(2*a**(1/4)*gamma(15/8))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (c\,x\right )}^{3/4}}{{\left (b\,x^2+a\right )}^{1/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(3/4)/(a + b*x^2)^(1/4),x)

[Out]

int((c*x)^(3/4)/(a + b*x^2)^(1/4), x)

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