3.7.27 \(\int \frac {1}{(c x)^{5/2} (a+b x^2)^{3/2}} \, dx\) [627]

Optimal. Leaf size=154 \[ \frac {1}{a c (c x)^{3/2} \sqrt {a+b x^2}}-\frac {5 \sqrt {a+b x^2}}{3 a^2 c (c x)^{3/2}}-\frac {5 b^{3/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{6 a^{9/4} c^{5/2} \sqrt {a+b x^2}} \]

[Out]

1/a/c/(c*x)^(3/2)/(b*x^2+a)^(1/2)-5/3*(b*x^2+a)^(1/2)/a^2/c/(c*x)^(3/2)-5/6*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)/
a^(9/4)/c^(5/2)/(b*x^2+a)^(1/2)

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Rubi [A]
time = 0.06, antiderivative size = 154, 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 = {296, 331, 335, 226} \begin {gather*} -\frac {5 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 \text {ArcTan}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{6 a^{9/4} c^{5/2} \sqrt {a+b x^2}}-\frac {5 \sqrt {a+b x^2}}{3 a^2 c (c x)^{3/2}}+\frac {1}{a c (c x)^{3/2} \sqrt {a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[1/((c*x)^(5/2)*(a + b*x^2)^(3/2)),x]

[Out]

1/(a*c*(c*x)^(3/2)*Sqrt[a + b*x^2]) - (5*Sqrt[a + b*x^2])/(3*a^2*c*(c*x)^(3/2)) - (5*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]
)/(6*a^(9/4)*c^(5/2)*Sqrt[a + b*x^2])

Rule 226

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)]/(2*q*Sqrt[a + b*x^4]))*EllipticF[2*ArcTan[q*x], 1/2], x]] /; FreeQ[{a, b}, x] && PosQ[b/a]

Rule 296

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

Rule 331

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 335

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.02, size = 59, normalized size = 0.38 \begin {gather*} -\frac {2 x \sqrt {1+\frac {b x^2}{a}} \, _2F_1\left (-\frac {3}{4},\frac {3}{2};\frac {1}{4};-\frac {b x^2}{a}\right )}{3 a (c x)^{5/2} \sqrt {a+b x^2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[1/((c*x)^(5/2)*(a + b*x^2)^(3/2)),x]

[Out]

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

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Maple [A]
time = 0.08, size = 124, normalized size = 0.81

method	result	size
default	\(-\frac {5 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right ) \sqrt {-a b}\, x +10 b \,x^{2}+4 a}{6 x \sqrt {b \,x^{2}+a}\, a^{2} c^{2} \sqrt {c x}}\)	\(124\)
elliptic	\(\frac {\sqrt {c x \left (b \,x^{2}+a \right )}\, \left (-\frac {b x}{c^{2} a^{2} \sqrt {\left (x^{2}+\frac {a}{b}\right ) b c x}}-\frac {2 \sqrt {b c \,x^{3}+a c x}}{3 a^{2} c^{3} x^{2}}-\frac {5 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{6 a^{2} c^{2} \sqrt {b c \,x^{3}+a c x}}\right )}{\sqrt {c x}\, \sqrt {b \,x^{2}+a}}\)	\(191\)
risch	\(-\frac {2 \sqrt {b \,x^{2}+a}}{3 a^{2} x \,c^{2} \sqrt {c x}}-\frac {b \left (\frac {\sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b \sqrt {b c \,x^{3}+a c x}}+3 a \left (\frac {x}{a \sqrt {\left (x^{2}+\frac {a}{b}\right ) b c x}}+\frac {\sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {x b}{\sqrt {-a b}}}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{2 a b \sqrt {b c \,x^{3}+a c x}}\right )\right ) \sqrt {c x \left (b \,x^{2}+a \right )}}{3 a^{2} c^{2} \sqrt {c x}\, \sqrt {b \,x^{2}+a}}\)	\(309\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x)^(5/2)/(b*x^2+a)^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/6/x*(5*((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))*(-a*b)^(1/2)*x+10*b*x^2+4*a)/(b*
x^2+a)^(1/2)/a^2/c^2/(c*x)^(1/2)

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

[Out]

integrate(1/((b*x^2 + a)^(3/2)*(c*x)^(5/2)), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 0.17, size = 79, normalized size = 0.51 \begin {gather*} -\frac {5 \, {\left (b x^{4} + a x^{2}\right )} \sqrt {b c} {\rm weierstrassPInverse}\left (-\frac {4 \, a}{b}, 0, x\right ) + {\left (5 \, b x^{2} + 2 \, a\right )} \sqrt {b x^{2} + a} \sqrt {c x}}{3 \, {\left (a^{2} b c^{3} x^{4} + a^{3} c^{3} x^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(5/2)/(b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

-1/3*(5*(b*x^4 + a*x^2)*sqrt(b*c)*weierstrassPInverse(-4*a/b, 0, x) + (5*b*x^2 + 2*a)*sqrt(b*x^2 + a)*sqrt(c*x
))/(a^2*b*c^3*x^4 + a^3*c^3*x^2)

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Sympy [C] Result contains complex when optimal does not.
time = 4.47, size = 48, normalized size = 0.31 \begin {gather*} \frac {\Gamma \left (- \frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {3}{4}, \frac {3}{2} \\ \frac {1}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} c^{\frac {5}{2}} x^{\frac {3}{2}} \Gamma \left (\frac {1}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)**(5/2)/(b*x**2+a)**(3/2),x)

[Out]

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

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

[Out]

integrate(1/((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 \begin {gather*} \int \frac {1}{{\left (c\,x\right )}^{5/2}\,{\left (b\,x^2+a\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((c*x)^(5/2)*(a + b*x^2)^(3/2)),x)

[Out]

int(1/((c*x)^(5/2)*(a + b*x^2)^(3/2)), x)

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