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

Optimal. Leaf size=333 \[ \frac {7 \sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{4 a^{11/4} c^{3/2} \sqrt {a+b x^2}}-\frac {7 \sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{2 a^{11/4} c^{3/2} \sqrt {a+b x^2}}+\frac {7 \sqrt {b} \sqrt {c x} \sqrt {a+b x^2}}{2 a^3 c^2 \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {7 \sqrt {a+b x^2}}{2 a^3 c \sqrt {c x}}+\frac {7}{6 a^2 c \sqrt {c x} \sqrt {a+b x^2}}+\frac {1}{3 a c \sqrt {c x} \left (a+b x^2\right )^{3/2}} \]

[Out]

1/3/a/c/(b*x^2+a)^(3/2)/(c*x)^(1/2)+7/6/a^2/c/(c*x)^(1/2)/(b*x^2+a)^(1/2)-7/2*(b*x^2+a)^(1/2)/a^3/c/(c*x)^(1/2
)+7/2*b^(1/2)*(c*x)^(1/2)*(b*x^2+a)^(1/2)/a^3/c^2/(a^(1/2)+x*b^(1/2))-7/2*b^(1/4)*(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)))*EllipticE(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^
(11/4)/c^(3/2)/(b*x^2+a)^(1/2)+7/4*b^(1/4)*(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^(11/4)/c^(3/2)/(b*x^2+a)^(1/2)

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Rubi [A]  time = 0.26, antiderivative size = 333, normalized size of antiderivative = 1.00, 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 = {290, 325, 329, 305, 220, 1196} \[ \frac {7 \sqrt {b} \sqrt {c x} \sqrt {a+b x^2}}{2 a^3 c^2 \left (\sqrt {a}+\sqrt {b} x\right )}+\frac {7 \sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{4 a^{11/4} c^{3/2} \sqrt {a+b x^2}}-\frac {7 \sqrt [4]{b} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{2 a^{11/4} c^{3/2} \sqrt {a+b x^2}}-\frac {7 \sqrt {a+b x^2}}{2 a^3 c \sqrt {c x}}+\frac {7}{6 a^2 c \sqrt {c x} \sqrt {a+b x^2}}+\frac {1}{3 a c \sqrt {c x} \left (a+b x^2\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

1/(3*a*c*Sqrt[c*x]*(a + b*x^2)^(3/2)) + 7/(6*a^2*c*Sqrt[c*x]*Sqrt[a + b*x^2]) - (7*Sqrt[a + b*x^2])/(2*a^3*c*S
qrt[c*x]) + (7*Sqrt[b]*Sqrt[c*x]*Sqrt[a + b*x^2])/(2*a^3*c^2*(Sqrt[a] + Sqrt[b]*x)) - (7*b^(1/4)*(Sqrt[a] + Sq
rt[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])/(2*a^(11/4)*c^(3/2)*Sqrt[a + b*x^2]) + (7*b^(1/4)*(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])/(4*a^(11/4)*c^(3/2)*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 290

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] /; FreeQ[
{a, b, c, m}, x] && IGtQ[n, 0] && LtQ[p, -1] && 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 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 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 {1}{(c x)^{3/2} \left (a+b x^2\right )^{5/2}} \, dx &=\frac {1}{3 a c \sqrt {c x} \left (a+b x^2\right )^{3/2}}+\frac {7 \int \frac {1}{(c x)^{3/2} \left (a+b x^2\right )^{3/2}} \, dx}{6 a}\\ &=\frac {1}{3 a c \sqrt {c x} \left (a+b x^2\right )^{3/2}}+\frac {7}{6 a^2 c \sqrt {c x} \sqrt {a+b x^2}}+\frac {7 \int \frac {1}{(c x)^{3/2} \sqrt {a+b x^2}} \, dx}{4 a^2}\\ &=\frac {1}{3 a c \sqrt {c x} \left (a+b x^2\right )^{3/2}}+\frac {7}{6 a^2 c \sqrt {c x} \sqrt {a+b x^2}}-\frac {7 \sqrt {a+b x^2}}{2 a^3 c \sqrt {c x}}+\frac {(7 b) \int \frac {\sqrt {c x}}{\sqrt {a+b x^2}} \, dx}{4 a^3 c^2}\\ &=\frac {1}{3 a c \sqrt {c x} \left (a+b x^2\right )^{3/2}}+\frac {7}{6 a^2 c \sqrt {c x} \sqrt {a+b x^2}}-\frac {7 \sqrt {a+b x^2}}{2 a^3 c \sqrt {c x}}+\frac {(7 b) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+\frac {b x^4}{c^2}}} \, dx,x,\sqrt {c x}\right )}{2 a^3 c^3}\\ &=\frac {1}{3 a c \sqrt {c x} \left (a+b x^2\right )^{3/2}}+\frac {7}{6 a^2 c \sqrt {c x} \sqrt {a+b x^2}}-\frac {7 \sqrt {a+b x^2}}{2 a^3 c \sqrt {c x}}+\frac {\left (7 \sqrt {b}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^4}{c^2}}} \, dx,x,\sqrt {c x}\right )}{2 a^{5/2} c^2}-\frac {\left (7 \sqrt {b}\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 )}{2 a^{5/2} c^2}\\ &=\frac {1}{3 a c \sqrt {c x} \left (a+b x^2\right )^{3/2}}+\frac {7}{6 a^2 c \sqrt {c x} \sqrt {a+b x^2}}-\frac {7 \sqrt {a+b x^2}}{2 a^3 c \sqrt {c x}}+\frac {7 \sqrt {b} \sqrt {c x} \sqrt {a+b x^2}}{2 a^3 c^2 \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {7 \sqrt [4]{b} \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 )}{2 a^{11/4} c^{3/2} \sqrt {a+b x^2}}+\frac {7 \sqrt [4]{b} \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 )}{4 a^{11/4} c^{3/2} \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 57, normalized size = 0.17 \[ -\frac {2 x \sqrt {\frac {b x^2}{a}+1} \, _2F_1\left (-\frac {1}{4},\frac {5}{2};\frac {3}{4};-\frac {b x^2}{a}\right )}{a^2 (c x)^{3/2} \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 1.01, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{2} + a} \sqrt {c x}}{b^{3} c^{2} x^{8} + 3 \, a b^{2} c^{2} x^{6} + 3 \, a^{2} b c^{2} x^{4} + a^{3} c^{2} x^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(b*x^2 + a)*sqrt(c*x)/(b^3*c^2*x^8 + 3*a*b^2*c^2*x^6 + 3*a^2*b*c^2*x^4 + a^3*c^2*x^2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{2}} \left (c x\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.02, size = 384, normalized size = 1.15 \[ \frac {-42 b^{2} x^{4}+42 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, a b \,x^{2} \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-21 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, a b \,x^{2} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-70 a b \,x^{2}+42 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, a^{2} \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-21 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, a^{2} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-24 a^{2}}{12 \sqrt {c x}\, \left (b \,x^{2}+a \right )^{\frac {3}{2}} a^{3} c} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/12*(42*((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)*EllipticE(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*x^2*a*b-21*((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^2*a*b+42*((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)*EllipticE(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2)
)^(1/2),1/2*2^(1/2))*a^2-21*((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^2-42*b^2*x
^4-70*a*b*x^2-24*a^2)/a^3/c/(c*x)^(1/2)/(b*x^2+a)^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{\frac {5}{2}} \left (c x\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 7.66, size = 48, normalized size = 0.14 \[ \frac {\Gamma \left (- \frac {1}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {1}{4}, \frac {5}{2} \\ \frac {3}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {5}{2}} c^{\frac {3}{2}} \sqrt {x} \Gamma \left (\frac {3}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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