3.992 \(\int \frac {(c x)^{9/2}}{(a+b x^2)^{5/4}} \, dx\)

Optimal. Leaf size=124 \[ -\frac {7 a^{3/2} c^4 \sqrt {c x} \sqrt [4]{\frac {a}{b x^2}+1} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 b^{5/2} \sqrt [4]{a+b x^2}}-\frac {7 a c^3 (c x)^{3/2}}{6 b^2 \sqrt [4]{a+b x^2}}+\frac {c (c x)^{7/2}}{3 b \sqrt [4]{a+b x^2}} \]

[Out]

-7/6*a*c^3*(c*x)^(3/2)/b^2/(b*x^2+a)^(1/4)+1/3*c*(c*x)^(7/2)/b/(b*x^2+a)^(1/4)-7/2*a^(3/2)*c^4*(1+a/b/x^2)^(1/
4)*(cos(1/2*arccot(x*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccot(x*b^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arccot(x*
b^(1/2)/a^(1/2))),2^(1/2))*(c*x)^(1/2)/b^(5/2)/(b*x^2+a)^(1/4)

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Rubi [A]  time = 0.05, antiderivative size = 124, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {285, 284, 335, 196} \[ -\frac {7 a^{3/2} c^4 \sqrt {c x} \sqrt [4]{\frac {a}{b x^2}+1} E\left (\left .\frac {1}{2} \cot ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 b^{5/2} \sqrt [4]{a+b x^2}}-\frac {7 a c^3 (c x)^{3/2}}{6 b^2 \sqrt [4]{a+b x^2}}+\frac {c (c x)^{7/2}}{3 b \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[(c*x)^(9/2)/(a + b*x^2)^(5/4),x]

[Out]

(-7*a*c^3*(c*x)^(3/2))/(6*b^2*(a + b*x^2)^(1/4)) + (c*(c*x)^(7/2))/(3*b*(a + b*x^2)^(1/4)) - (7*a^(3/2)*c^4*(1
+ a/(b*x^2))^(1/4)*Sqrt[c*x]*EllipticE[ArcCot[(Sqrt[b]*x)/Sqrt[a]]/2, 2])/(2*b^(5/2)*(a + b*x^2)^(1/4))

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 284

Int[Sqrt[(c_.)*(x_)]/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Dist[(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]

Rule 285

Int[((c_.)*(x_))^(m_)/((a_) + (b_.)*(x_)^2)^(5/4), x_Symbol] :> Simp[(2*c*(c*x)^(m - 1))/(b*(2*m - 3)*(a + b*x
^2)^(1/4)), x] - Dist[(2*a*c^2*(m - 1))/(b*(2*m - 3)), Int[(c*x)^(m - 2)/(a + b*x^2)^(5/4), x], x] /; FreeQ[{a
, b, c}, x] && PosQ[b/a] && IntegerQ[2*m] && GtQ[m, 3/2]

Rule 335

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] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {(c x)^{9/2}}{\left (a+b x^2\right )^{5/4}} \, dx &=\frac {c (c x)^{7/2}}{3 b \sqrt [4]{a+b x^2}}-\frac {\left (7 a c^2\right ) \int \frac {(c x)^{5/2}}{\left (a+b x^2\right )^{5/4}} \, dx}{6 b}\\ &=-\frac {7 a c^3 (c x)^{3/2}}{6 b^2 \sqrt [4]{a+b x^2}}+\frac {c (c x)^{7/2}}{3 b \sqrt [4]{a+b x^2}}+\frac {\left (7 a^2 c^4\right ) \int \frac {\sqrt {c x}}{\left (a+b x^2\right )^{5/4}} \, dx}{4 b^2}\\ &=-\frac {7 a c^3 (c x)^{3/2}}{6 b^2 \sqrt [4]{a+b x^2}}+\frac {c (c x)^{7/2}}{3 b \sqrt [4]{a+b x^2}}+\frac {\left (7 a^2 c^4 \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {c x}\right ) \int \frac {1}{\left (1+\frac {a}{b x^2}\right )^{5/4} x^2} \, dx}{4 b^3 \sqrt [4]{a+b x^2}}\\ &=-\frac {7 a c^3 (c x)^{3/2}}{6 b^2 \sqrt [4]{a+b x^2}}+\frac {c (c x)^{7/2}}{3 b \sqrt [4]{a+b x^2}}-\frac {\left (7 a^2 c^4 \sqrt [4]{1+\frac {a}{b x^2}} \sqrt {c x}\right ) \operatorname {Subst}\left (\int \frac {1}{\left (1+\frac {a x^2}{b}\right )^{5/4}} \, dx,x,\frac {1}{x}\right )}{4 b^3 \sqrt [4]{a+b x^2}}\\ &=-\frac {7 a c^3 (c x)^{3/2}}{6 b^2 \sqrt [4]{a+b x^2}}+\frac {c (c x)^{7/2}}{3 b \sqrt [4]{a+b x^2}}-\frac {7 a^{3/2} c^4 \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 )}{2 b^{5/2} \sqrt [4]{a+b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.03, size = 74, normalized size = 0.60 \[ \frac {c^3 (c x)^{3/2} \left (7 a \sqrt [4]{\frac {b x^2}{a}+1} \, _2F_1\left (\frac {3}{4},\frac {5}{4};\frac {7}{4};-\frac {b x^2}{a}\right )-7 a+2 b x^2\right )}{6 b^2 \sqrt [4]{a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*x)^(9/2)/(a + b*x^2)^(5/4),x]

[Out]

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

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{4}} \sqrt {c x} c^{4} x^{4}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(9/2)/(b*x^2+a)^(5/4),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x\right )^{\frac {9}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(9/2)/(b*x^2+a)^(5/4),x, algorithm="giac")

[Out]

integrate((c*x)^(9/2)/(b*x^2 + a)^(5/4), x)

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maple [F]  time = 0.33, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x \right )^{\frac {9}{2}}}{\left (b \,x^{2}+a \right )^{\frac {5}{4}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(9/2)/(b*x^2+a)^(5/4),x)

[Out]

int((c*x)^(9/2)/(b*x^2+a)^(5/4),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c x\right )^{\frac {9}{2}}}{{\left (b x^{2} + a\right )}^{\frac {5}{4}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)^(9/2)/(b*x^2+a)^(5/4),x, algorithm="maxima")

[Out]

integrate((c*x)^(9/2)/(b*x^2 + a)^(5/4), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,x\right )}^{9/2}}{{\left (b\,x^2+a\right )}^{5/4}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*x)^(9/2)/(a + b*x^2)^(5/4),x)

[Out]

int((c*x)^(9/2)/(a + b*x^2)^(5/4), x)

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sympy [C]  time = 64.02, size = 44, normalized size = 0.35 \[ \frac {c^{\frac {9}{2}} x^{\frac {11}{2}} \Gamma \left (\frac {11}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {5}{4}, \frac {11}{4} \\ \frac {15}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{4}} \Gamma \left (\frac {15}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*x)**(9/2)/(b*x**2+a)**(5/4),x)

[Out]

c**(9/2)*x**(11/2)*gamma(11/4)*hyper((5/4, 11/4), (15/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(5/4)*gamma(15/4))

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