3.605 \(\int \frac {(a+b x^2)^{3/2}}{(c x)^{9/2}} \, dx\)

Optimal. Leaf size=152 \[ \frac {4 b^{7/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{7 \sqrt [4]{a} c^{9/2} \sqrt {a+b x^2}}-\frac {4 b \sqrt {a+b x^2}}{7 c^3 (c x)^{3/2}}-\frac {2 \left (a+b x^2\right )^{3/2}}{7 c (c x)^{7/2}} \]

[Out]

-2/7*(b*x^2+a)^(3/2)/c/(c*x)^(7/2)-4/7*b*(b*x^2+a)^(1/2)/c^3/(c*x)^(3/2)+4/7*b^(7/4)*(cos(2*arctan(b^(1/4)*(c*
x)^(1/2)/a^(1/4)/c^(1/2)))^2)^(1/2)/cos(2*arctan(b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2)))*EllipticF(sin(2*arctan(
b^(1/4)*(c*x)^(1/2)/a^(1/4)/c^(1/2))),1/2*2^(1/2))*(a^(1/2)+x*b^(1/2))*((b*x^2+a)/(a^(1/2)+x*b^(1/2))^2)^(1/2)
/a^(1/4)/c^(9/2)/(b*x^2+a)^(1/2)

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Rubi [A]  time = 0.09, antiderivative size = 152, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {277, 329, 220} \[ \frac {4 b^{7/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{7 \sqrt [4]{a} c^{9/2} \sqrt {a+b x^2}}-\frac {4 b \sqrt {a+b x^2}}{7 c^3 (c x)^{3/2}}-\frac {2 \left (a+b x^2\right )^{3/2}}{7 c (c x)^{7/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/2)/(c*x)^(9/2),x]

[Out]

(-4*b*Sqrt[a + b*x^2])/(7*c^3*(c*x)^(3/2)) - (2*(a + b*x^2)^(3/2))/(7*c*(c*x)^(7/2)) + (4*b^(7/4)*(Sqrt[a] + S
qrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])],
1/2])/(7*a^(1/4)*c^(9/2)*Sqrt[a + b*x^2])

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 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 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]

Rubi steps

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

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Mathematica [C]  time = 0.01, size = 57, normalized size = 0.38 \[ -\frac {2 a x \sqrt {a+b x^2} \, _2F_1\left (-\frac {7}{4},-\frac {3}{2};-\frac {3}{4};-\frac {b x^2}{a}\right )}{7 (c x)^{9/2} \sqrt {\frac {b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(3/2)/(c*x)^(9/2),x]

[Out]

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

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fricas [F]  time = 0.64, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} \sqrt {c x}}{c^{5} x^{5}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/(c*x)^(9/2),x, algorithm="fricas")

[Out]

integral((b*x^2 + a)^(3/2)*sqrt(c*x)/(c^5*x^5), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}}}{\left (c x\right )^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/(c*x)^(9/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(3/2)/(c*x)^(9/2), x)

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maple [A]  time = 0.04, size = 135, normalized size = 0.89 \[ \frac {-\frac {6 b^{2} x^{4}}{7}+\frac {4 \sqrt {-a b}\, \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, b \,x^{3} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{7}-\frac {8 a b \,x^{2}}{7}-\frac {2 a^{2}}{7}}{\sqrt {b \,x^{2}+a}\, \sqrt {c x}\, c^{4} x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(3/2)/(c*x)^(9/2),x)

[Out]

2/7/(b*x^2+a)^(1/2)/x^3*(2*(-a*b)^(1/2)*((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)*(-1/(-a*b)^(1/2)*b*x)^(1/2)*EllipticF(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*
x^3*b-3*b^2*x^4-4*a*b*x^2-a^2)/c^4/(c*x)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}}}{\left (c x\right )^{\frac {9}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/(c*x)^(9/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(3/2)/(c*x)^(9/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,x^2+a\right )}^{3/2}}{{\left (c\,x\right )}^{9/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(3/2)/(c*x)^(9/2),x)

[Out]

int((a + b*x^2)^(3/2)/(c*x)^(9/2), x)

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sympy [C]  time = 33.12, size = 53, normalized size = 0.35 \[ \frac {a^{\frac {3}{2}} \Gamma \left (- \frac {7}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {7}{4}, - \frac {3}{2} \\ - \frac {3}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 c^{\frac {9}{2}} x^{\frac {7}{2}} \Gamma \left (- \frac {3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(3/2)/(c*x)**(9/2),x)

[Out]

a**(3/2)*gamma(-7/4)*hyper((-7/4, -3/2), (-3/4,), b*x**2*exp_polar(I*pi)/a)/(2*c**(9/2)*x**(7/2)*gamma(-3/4))

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