3.603 \(\int \frac {(a+b x^2)^{3/2}}{(c x)^{5/2}} \, dx\)

Optimal. Leaf size=152 \[ \frac {4 a^{3/4} b^{3/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{3 c^{5/2} \sqrt {a+b x^2}}+\frac {4 b \sqrt {c x} \sqrt {a+b x^2}}{3 c^3}-\frac {2 \left (a+b x^2\right )^{3/2}}{3 c (c x)^{3/2}} \]

[Out]

-2/3*(b*x^2+a)^(3/2)/c/(c*x)^(3/2)+4/3*b*(c*x)^(1/2)*(b*x^2+a)^(1/2)/c^3+4/3*a^(3/4)*b^(3/4)*(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)/c^(5/2)/(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 = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {277, 279, 329, 220} \[ \frac {4 a^{3/4} b^{3/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{3 c^{5/2} \sqrt {a+b x^2}}+\frac {4 b \sqrt {c x} \sqrt {a+b x^2}}{3 c^3}-\frac {2 \left (a+b x^2\right )^{3/2}}{3 c (c x)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/2)/(c*x)^(5/2),x]

[Out]

(4*b*Sqrt[c*x]*Sqrt[a + b*x^2])/(3*c^3) - (2*(a + b*x^2)^(3/2))/(3*c*(c*x)^(3/2)) + (4*a^(3/4)*b^(3/4)*(Sqrt[a
] + Sqrt[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])/(3*c^(5/2)*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 279

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
n*p + 1)), x] + Dist[(a*n*p)/(m + n*p + 1), Int[(c*x)^m*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c, m}, x]
&& IGtQ[n, 0] && GtQ[p, 0] && NeQ[m + n*p + 1, 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)^{5/2}} \, dx &=-\frac {2 \left (a+b x^2\right )^{3/2}}{3 c (c x)^{3/2}}+\frac {(2 b) \int \frac {\sqrt {a+b x^2}}{\sqrt {c x}} \, dx}{c^2}\\ &=\frac {4 b \sqrt {c x} \sqrt {a+b x^2}}{3 c^3}-\frac {2 \left (a+b x^2\right )^{3/2}}{3 c (c x)^{3/2}}+\frac {(4 a b) \int \frac {1}{\sqrt {c x} \sqrt {a+b x^2}} \, dx}{3 c^2}\\ &=\frac {4 b \sqrt {c x} \sqrt {a+b x^2}}{3 c^3}-\frac {2 \left (a+b x^2\right )^{3/2}}{3 c (c x)^{3/2}}+\frac {(8 a b) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^4}{c^2}}} \, dx,x,\sqrt {c x}\right )}{3 c^3}\\ &=\frac {4 b \sqrt {c x} \sqrt {a+b x^2}}{3 c^3}-\frac {2 \left (a+b x^2\right )^{3/2}}{3 c (c x)^{3/2}}+\frac {4 a^{3/4} b^{3/4} \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 )}{3 c^{5/2} \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 57, normalized size = 0.38 \[ -\frac {2 a x \sqrt {a+b x^2} \, _2F_1\left (-\frac {3}{2},-\frac {3}{4};\frac {1}{4};-\frac {b x^2}{a}\right )}{3 (c x)^{5/2} \sqrt {\frac {b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(3/2)/(c*x)^(5/2),x]

[Out]

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

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fricas [F]  time = 0.76, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} \sqrt {c x}}{c^{3} x^{3}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/(c*x)^(5/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}}}{\left (c x\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/(c*x)^(5/2),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.01, size = 125, normalized size = 0.82 \[ \frac {\frac {2 b^{2} x^{4}}{3}+\frac {4 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \sqrt {-a b}\, a x \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{3}-\frac {2 a^{2}}{3}}{\sqrt {b \,x^{2}+a}\, \sqrt {c x}\, c^{2} x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}}}{\left (c x\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/(c*x)^(5/2),x, algorithm="maxima")

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (b\,x^2+a\right )}^{3/2}}{{\left (c\,x\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(3/2)/(c*x)^(5/2),x)

[Out]

int((a + b*x^2)^(3/2)/(c*x)^(5/2), x)

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sympy [C]  time = 4.50, size = 49, normalized size = 0.32 \[ \frac {a^{\frac {3}{2}} \Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{2}, - \frac {3}{4} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 c^{\frac {5}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(3/2)/(c*x)**(5/2),x)

[Out]

a**(3/2)*gamma(-3/4)*hyper((-3/2, -3/4), (1/4,), b*x**2*exp_polar(I*pi)/a)/(2*c**(5/2)*x**(3/2)*gamma(1/4))

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