3.11.4 \(\int \frac {(c x)^{5/4}}{(a+b x^2)^{7/4}} \, dx\) [1004]

Optimal. Leaf size=61 \[ \frac {4 (c x)^{9/4} \left (1+\frac {b x^2}{a}\right )^{3/4} \, _2F_1\left (\frac {9}{8},\frac {7}{4};\frac {17}{8};-\frac {b x^2}{a}\right )}{9 a c \left (a+b x^2\right )^{3/4}} \]

[Out]

4/9*(c*x)^(9/4)*(1+b*x^2/a)^(3/4)*hypergeom([9/8, 7/4],[17/8],-b*x^2/a)/a/c/(b*x^2+a)^(3/4)

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Rubi [A]
time = 0.01, antiderivative size = 61, 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)^{9/4} \left (\frac {b x^2}{a}+1\right )^{3/4} \, _2F_1\left (\frac {9}{8},\frac {7}{4};\frac {17}{8};-\frac {b x^2}{a}\right )}{9 a c \left (a+b x^2\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c*x)^(5/4)/(a + b*x^2)^(7/4),x]

[Out]

(4*(c*x)^(9/4)*(1 + (b*x^2)/a)^(3/4)*Hypergeometric2F1[9/8, 7/4, 17/8, -((b*x^2)/a)])/(9*a*c*(a + b*x^2)^(3/4)
)

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)^{5/4}}{\left (a+b x^2\right )^{7/4}} \, dx &=\frac {\left (1+\frac {b x^2}{a}\right )^{3/4} \int \frac {(c x)^{5/4}}{\left (1+\frac {b x^2}{a}\right )^{7/4}} \, dx}{a \left (a+b x^2\right )^{3/4}}\\ &=\frac {4 (c x)^{9/4} \left (1+\frac {b x^2}{a}\right )^{3/4} \, _2F_1\left (\frac {9}{8},\frac {7}{4};\frac {17}{8};-\frac {b x^2}{a}\right )}{9 a c \left (a+b x^2\right )^{3/4}}\\ \end {align*}

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Mathematica [A]
time = 10.03, size = 59, normalized size = 0.97 \begin {gather*} \frac {4 x (c x)^{5/4} \left (1+\frac {b x^2}{a}\right )^{3/4} \, _2F_1\left (\frac {9}{8},\frac {7}{4};\frac {17}{8};-\frac {b x^2}{a}\right )}{9 a \left (a+b x^2\right )^{3/4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^(5/4)/(a + b*x^2)^(7/4),x]

[Out]

(4*x*(c*x)^(5/4)*(1 + (b*x^2)/a)^(3/4)*Hypergeometric2F1[9/8, 7/4, 17/8, -((b*x^2)/a)])/(9*a*(a + b*x^2)^(3/4)
)

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {\left (c x \right )^{\frac {5}{4}}}{\left (b \,x^{2}+a \right )^{\frac {7}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(5/4)/(b*x^2+a)^(7/4),x)

[Out]

int((c*x)^(5/4)/(b*x^2+a)^(7/4),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)^(5/4)/(b*x^2+a)^(7/4),x, algorithm="maxima")

[Out]

integrate((c*x)^(5/4)/(b*x^2 + a)^(7/4), 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)^(5/4)/(b*x^2+a)^(7/4),x, algorithm="fricas")

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 14.30, size = 44, normalized size = 0.72 \begin {gather*} \frac {c^{\frac {5}{4}} x^{\frac {9}{4}} \Gamma \left (\frac {9}{8}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {9}{8}, \frac {7}{4} \\ \frac {17}{8} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {7}{4}} \Gamma \left (\frac {17}{8}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**(5/4)/(b*x**2+a)**(7/4),x)

[Out]

c**(5/4)*x**(9/4)*gamma(9/8)*hyper((9/8, 7/4), (17/8,), b*x**2*exp_polar(I*pi)/a)/(2*a**(7/4)*gamma(17/8))

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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)^(5/4)/(b*x^2+a)^(7/4),x, algorithm="giac")

[Out]

integrate((c*x)^(5/4)/(b*x^2 + a)^(7/4), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \frac {{\left (c\,x\right )}^{5/4}}{{\left (b\,x^2+a\right )}^{7/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(5/4)/(a + b*x^2)^(7/4),x)

[Out]

int((c*x)^(5/4)/(a + b*x^2)^(7/4), x)

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