3.606 \(\int \frac{(a+b x^2)^{3/2}}{(c x)^{11/2}} \, dx\)

Optimal. Leaf size=331 \[ \frac{4 b^{9/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right ),\frac{1}{2}\right )}{15 a^{3/4} c^{11/2} \sqrt{a+b x^2}}-\frac{8 b^{9/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{15 a^{3/4} c^{11/2} \sqrt{a+b x^2}}+\frac{8 b^{5/2} \sqrt{c x} \sqrt{a+b x^2}}{15 a c^6 \left (\sqrt{a}+\sqrt{b} x\right )}-\frac{8 b^2 \sqrt{a+b x^2}}{15 a c^5 \sqrt{c x}}-\frac{4 b \sqrt{a+b x^2}}{15 c^3 (c x)^{5/2}}-\frac{2 \left (a+b x^2\right )^{3/2}}{9 c (c x)^{9/2}} \]

[Out]

(-4*b*Sqrt[a + b*x^2])/(15*c^3*(c*x)^(5/2)) - (8*b^2*Sqrt[a + b*x^2])/(15*a*c^5*Sqrt[c*x]) + (8*b^(5/2)*Sqrt[c
*x]*Sqrt[a + b*x^2])/(15*a*c^6*(Sqrt[a] + Sqrt[b]*x)) - (2*(a + b*x^2)^(3/2))/(9*c*(c*x)^(9/2)) - (8*b^(9/4)*(
Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[c*x])/(a^(1/4)
*Sqrt[c])], 1/2])/(15*a^(3/4)*c^(11/2)*Sqrt[a + b*x^2]) + (4*b^(9/4)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(S
qrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/(15*a^(3/4)*c^(11/2)*S
qrt[a + b*x^2])

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Rubi [A]  time = 0.270284, antiderivative size = 331, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.316, Rules used = {277, 325, 329, 305, 220, 1196} \[ \frac{4 b^{9/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{15 a^{3/4} c^{11/2} \sqrt{a+b x^2}}-\frac{8 b^{9/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{15 a^{3/4} c^{11/2} \sqrt{a+b x^2}}+\frac{8 b^{5/2} \sqrt{c x} \sqrt{a+b x^2}}{15 a c^6 \left (\sqrt{a}+\sqrt{b} x\right )}-\frac{8 b^2 \sqrt{a+b x^2}}{15 a c^5 \sqrt{c x}}-\frac{4 b \sqrt{a+b x^2}}{15 c^3 (c x)^{5/2}}-\frac{2 \left (a+b x^2\right )^{3/2}}{9 c (c x)^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/2)/(c*x)^(11/2),x]

[Out]

(-4*b*Sqrt[a + b*x^2])/(15*c^3*(c*x)^(5/2)) - (8*b^2*Sqrt[a + b*x^2])/(15*a*c^5*Sqrt[c*x]) + (8*b^(5/2)*Sqrt[c
*x]*Sqrt[a + b*x^2])/(15*a*c^6*(Sqrt[a] + Sqrt[b]*x)) - (2*(a + b*x^2)^(3/2))/(9*c*(c*x)^(9/2)) - (8*b^(9/4)*(
Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[c*x])/(a^(1/4)
*Sqrt[c])], 1/2])/(15*a^(3/4)*c^(11/2)*Sqrt[a + b*x^2]) + (4*b^(9/4)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(S
qrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/(15*a^(3/4)*c^(11/2)*S
qrt[a + b*x^2])

Rule 277

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

Rule 325

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^(p + 1))/(a*
c*(m + 1)), x] - Dist[(b*(m + n*(p + 1) + 1))/(a*c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^p, x], x] /; Free
Q[{a, b, c, p}, x] && IGtQ[n, 0] && LtQ[m, -1] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 305

Int[(x_)^2/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 2]}, Dist[1/q, Int[1/Sqrt[a + b*x^4], x],
x] - Dist[1/q, Int[(1 - q*x^2)/Sqrt[a + b*x^4], x], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 220

Int[1/Sqrt[(a_) + (b_.)*(x_)^4], x_Symbol] :> With[{q = Rt[b/a, 4]}, Simp[((1 + q^2*x^2)*Sqrt[(a + b*x^4)/(a*(
1 + q^2*x^2)^2)]*EllipticF[2*ArcTan[q*x], 1/2])/(2*q*Sqrt[a + b*x^4]), x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 1196

Int[((d_) + (e_.)*(x_)^2)/Sqrt[(a_) + (c_.)*(x_)^4], x_Symbol] :> With[{q = Rt[c/a, 4]}, -Simp[(d*x*Sqrt[a + c
*x^4])/(a*(1 + q^2*x^2)), x] + Simp[(d*(1 + q^2*x^2)*Sqrt[(a + c*x^4)/(a*(1 + q^2*x^2)^2)]*EllipticE[2*ArcTan[
q*x], 1/2])/(q*Sqrt[a + c*x^4]), x] /; EqQ[e + d*q^2, 0]] /; FreeQ[{a, c, d, e}, x] && PosQ[c/a]

Rubi steps

\begin{align*} \int \frac{\left (a+b x^2\right )^{3/2}}{(c x)^{11/2}} \, dx &=-\frac{2 \left (a+b x^2\right )^{3/2}}{9 c (c x)^{9/2}}+\frac{(2 b) \int \frac{\sqrt{a+b x^2}}{(c x)^{7/2}} \, dx}{3 c^2}\\ &=-\frac{4 b \sqrt{a+b x^2}}{15 c^3 (c x)^{5/2}}-\frac{2 \left (a+b x^2\right )^{3/2}}{9 c (c x)^{9/2}}+\frac{\left (4 b^2\right ) \int \frac{1}{(c x)^{3/2} \sqrt{a+b x^2}} \, dx}{15 c^4}\\ &=-\frac{4 b \sqrt{a+b x^2}}{15 c^3 (c x)^{5/2}}-\frac{8 b^2 \sqrt{a+b x^2}}{15 a c^5 \sqrt{c x}}-\frac{2 \left (a+b x^2\right )^{3/2}}{9 c (c x)^{9/2}}+\frac{\left (4 b^3\right ) \int \frac{\sqrt{c x}}{\sqrt{a+b x^2}} \, dx}{15 a c^6}\\ &=-\frac{4 b \sqrt{a+b x^2}}{15 c^3 (c x)^{5/2}}-\frac{8 b^2 \sqrt{a+b x^2}}{15 a c^5 \sqrt{c x}}-\frac{2 \left (a+b x^2\right )^{3/2}}{9 c (c x)^{9/2}}+\frac{\left (8 b^3\right ) \operatorname{Subst}\left (\int \frac{x^2}{\sqrt{a+\frac{b x^4}{c^2}}} \, dx,x,\sqrt{c x}\right )}{15 a c^7}\\ &=-\frac{4 b \sqrt{a+b x^2}}{15 c^3 (c x)^{5/2}}-\frac{8 b^2 \sqrt{a+b x^2}}{15 a c^5 \sqrt{c x}}-\frac{2 \left (a+b x^2\right )^{3/2}}{9 c (c x)^{9/2}}+\frac{\left (8 b^{5/2}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+\frac{b x^4}{c^2}}} \, dx,x,\sqrt{c x}\right )}{15 \sqrt{a} c^6}-\frac{\left (8 b^{5/2}\right ) \operatorname{Subst}\left (\int \frac{1-\frac{\sqrt{b} x^2}{\sqrt{a} c}}{\sqrt{a+\frac{b x^4}{c^2}}} \, dx,x,\sqrt{c x}\right )}{15 \sqrt{a} c^6}\\ &=-\frac{4 b \sqrt{a+b x^2}}{15 c^3 (c x)^{5/2}}-\frac{8 b^2 \sqrt{a+b x^2}}{15 a c^5 \sqrt{c x}}+\frac{8 b^{5/2} \sqrt{c x} \sqrt{a+b x^2}}{15 a c^6 \left (\sqrt{a}+\sqrt{b} x\right )}-\frac{2 \left (a+b x^2\right )^{3/2}}{9 c (c x)^{9/2}}-\frac{8 b^{9/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{15 a^{3/4} c^{11/2} \sqrt{a+b x^2}}+\frac{4 b^{9/4} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{c x}}{\sqrt [4]{a} \sqrt{c}}\right )|\frac{1}{2}\right )}{15 a^{3/4} c^{11/2} \sqrt{a+b x^2}}\\ \end{align*}

Mathematica [C]  time = 0.0131863, size = 57, normalized size = 0.17 \[ -\frac{2 a x \sqrt{a+b x^2} \, _2F_1\left (-\frac{9}{4},-\frac{3}{2};-\frac{5}{4};-\frac{b x^2}{a}\right )}{9 (c x)^{11/2} \sqrt{\frac{b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(3/2)/(c*x)^(11/2),x]

[Out]

(-2*a*x*Sqrt[a + b*x^2]*Hypergeometric2F1[-9/4, -3/2, -5/4, -((b*x^2)/a)])/(9*(c*x)^(11/2)*Sqrt[1 + (b*x^2)/a]
)

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Maple [A]  time = 0.029, size = 234, normalized size = 0.7 \begin{align*}{\frac{2}{45\,a{x}^{4}{c}^{5}} \left ( 12\,\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticE} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ){x}^{4}a{b}^{2}-6\,\sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{2}\sqrt{{\frac{-bx+\sqrt{-ab}}{\sqrt{-ab}}}}\sqrt{-{\frac{bx}{\sqrt{-ab}}}}{\it EllipticF} \left ( \sqrt{{\frac{bx+\sqrt{-ab}}{\sqrt{-ab}}}},1/2\,\sqrt{2} \right ){x}^{4}a{b}^{2}-12\,{b}^{3}{x}^{6}-23\,a{b}^{2}{x}^{4}-16\,{a}^{2}b{x}^{2}-5\,{a}^{3} \right ){\frac{1}{\sqrt{cx}}}{\frac{1}{\sqrt{b{x}^{2}+a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(3/2)/(c*x)^(11/2),x)

[Out]

2/45/x^4*(12*((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*2^(1/2)*((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-x*b/(-
a*b)^(1/2))^(1/2)*EllipticE(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*x^4*a*b^2-6*((b*x+(-a*b)^(1/2
))/(-a*b)^(1/2))^(1/2)*2^(1/2)*((-b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2)*(-x*b/(-a*b)^(1/2))^(1/2)*EllipticF(((
b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*x^4*a*b^2-12*b^3*x^6-23*a*b^2*x^4-16*a^2*b*x^2-5*a^3)/(b*x^
2+a)^(1/2)/a/c^5/(c*x)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{\frac{3}{2}}}{\left (c x\right )^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/(c*x)^(11/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(3/2)/(c*x)^(11/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (b x^{2} + a\right )}^{\frac{3}{2}} \sqrt{c x}}{c^{6} x^{6}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/(c*x)^(11/2),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(3/2)*sqrt(c*x)/(c^6*x^6), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(3/2)/(c*x)**(11/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x^{2} + a\right )}^{\frac{3}{2}}}{\left (c x\right )^{\frac{11}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/(c*x)^(11/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(3/2)/(c*x)^(11/2), x)