3.628 \(\int \frac {1}{(c x)^{7/2} (a+b x^2)^{3/2}} \, dx\)

Optimal. Leaf size=331 \[ -\frac {21 b^{5/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{10 a^{11/4} c^{7/2} \sqrt {a+b x^2}}+\frac {21 b^{5/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{5 a^{11/4} c^{7/2} \sqrt {a+b x^2}}-\frac {21 b^{3/2} \sqrt {c x} \sqrt {a+b x^2}}{5 a^3 c^4 \left (\sqrt {a}+\sqrt {b} x\right )}+\frac {21 b \sqrt {a+b x^2}}{5 a^3 c^3 \sqrt {c x}}-\frac {7 \sqrt {a+b x^2}}{5 a^2 c (c x)^{5/2}}+\frac {1}{a c (c x)^{5/2} \sqrt {a+b x^2}} \]

[Out]

1/a/c/(c*x)^(5/2)/(b*x^2+a)^(1/2)-7/5*(b*x^2+a)^(1/2)/a^2/c/(c*x)^(5/2)+21/5*b*(b*x^2+a)^(1/2)/a^3/c^3/(c*x)^(
1/2)-21/5*b^(3/2)*(c*x)^(1/2)*(b*x^2+a)^(1/2)/a^3/c^4/(a^(1/2)+x*b^(1/2))+21/5*b^(5/4)*(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*arcta
n(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^(7/2)/(b*x^2+a)^(1/2)-21/10*b^(5/4)*(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^(7/2)/(b*x^2+a)^(1/2)

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Rubi [A]  time = 0.25, antiderivative size = 331, 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} \[ -\frac {21 b^{3/2} \sqrt {c x} \sqrt {a+b x^2}}{5 a^3 c^4 \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {21 b^{5/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{10 a^{11/4} c^{7/2} \sqrt {a+b x^2}}+\frac {21 b^{5/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{5 a^{11/4} c^{7/2} \sqrt {a+b x^2}}+\frac {21 b \sqrt {a+b x^2}}{5 a^3 c^3 \sqrt {c x}}-\frac {7 \sqrt {a+b x^2}}{5 a^2 c (c x)^{5/2}}+\frac {1}{a c (c x)^{5/2} \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((c*x)^(7/2)*(a + b*x^2)^(3/2)),x]

[Out]

1/(a*c*(c*x)^(5/2)*Sqrt[a + b*x^2]) - (7*Sqrt[a + b*x^2])/(5*a^2*c*(c*x)^(5/2)) + (21*b*Sqrt[a + b*x^2])/(5*a^
3*c^3*Sqrt[c*x]) - (21*b^(3/2)*Sqrt[c*x]*Sqrt[a + b*x^2])/(5*a^3*c^4*(Sqrt[a] + Sqrt[b]*x)) + (21*b^(5/4)*(Sqr
t[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)*Sq
rt[c])], 1/2])/(5*a^(11/4)*c^(7/2)*Sqrt[a + b*x^2]) - (21*b^(5/4)*(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])/(10*a^(11/4)*c^(7/2)*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)^{7/2} \left (a+b x^2\right )^{3/2}} \, dx &=\frac {1}{a c (c x)^{5/2} \sqrt {a+b x^2}}+\frac {7 \int \frac {1}{(c x)^{7/2} \sqrt {a+b x^2}} \, dx}{2 a}\\ &=\frac {1}{a c (c x)^{5/2} \sqrt {a+b x^2}}-\frac {7 \sqrt {a+b x^2}}{5 a^2 c (c x)^{5/2}}-\frac {(21 b) \int \frac {1}{(c x)^{3/2} \sqrt {a+b x^2}} \, dx}{10 a^2 c^2}\\ &=\frac {1}{a c (c x)^{5/2} \sqrt {a+b x^2}}-\frac {7 \sqrt {a+b x^2}}{5 a^2 c (c x)^{5/2}}+\frac {21 b \sqrt {a+b x^2}}{5 a^3 c^3 \sqrt {c x}}-\frac {\left (21 b^2\right ) \int \frac {\sqrt {c x}}{\sqrt {a+b x^2}} \, dx}{10 a^3 c^4}\\ &=\frac {1}{a c (c x)^{5/2} \sqrt {a+b x^2}}-\frac {7 \sqrt {a+b x^2}}{5 a^2 c (c x)^{5/2}}+\frac {21 b \sqrt {a+b x^2}}{5 a^3 c^3 \sqrt {c x}}-\frac {\left (21 b^2\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+\frac {b x^4}{c^2}}} \, dx,x,\sqrt {c x}\right )}{5 a^3 c^5}\\ &=\frac {1}{a c (c x)^{5/2} \sqrt {a+b x^2}}-\frac {7 \sqrt {a+b x^2}}{5 a^2 c (c x)^{5/2}}+\frac {21 b \sqrt {a+b x^2}}{5 a^3 c^3 \sqrt {c x}}-\frac {\left (21 b^{3/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^4}{c^2}}} \, dx,x,\sqrt {c x}\right )}{5 a^{5/2} c^4}+\frac {\left (21 b^{3/2}\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 )}{5 a^{5/2} c^4}\\ &=\frac {1}{a c (c x)^{5/2} \sqrt {a+b x^2}}-\frac {7 \sqrt {a+b x^2}}{5 a^2 c (c x)^{5/2}}+\frac {21 b \sqrt {a+b x^2}}{5 a^3 c^3 \sqrt {c x}}-\frac {21 b^{3/2} \sqrt {c x} \sqrt {a+b x^2}}{5 a^3 c^4 \left (\sqrt {a}+\sqrt {b} x\right )}+\frac {21 b^{5/4} \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 )}{5 a^{11/4} c^{7/2} \sqrt {a+b x^2}}-\frac {21 b^{5/4} \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 )}{10 a^{11/4} c^{7/2} \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 59, normalized size = 0.18 \[ -\frac {2 x \sqrt {\frac {b x^2}{a}+1} \, _2F_1\left (-\frac {5}{4},\frac {3}{2};-\frac {1}{4};-\frac {b x^2}{a}\right )}{5 a (c x)^{7/2} \sqrt {a+b x^2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((c*x)^(7/2)*(a + b*x^2)^(3/2)),x]

[Out]

(-2*x*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[-5/4, 3/2, -1/4, -((b*x^2)/a)])/(5*a*(c*x)^(7/2)*Sqrt[a + b*x^2])

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fricas [F]  time = 1.04, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{2} + a} \sqrt {c x}}{b^{2} c^{4} x^{8} + 2 \, a b c^{4} x^{6} + a^{2} c^{4} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(7/2)/(b*x^2+a)^(3/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} \left (c x\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(7/2)/(b*x^2+a)^(3/2),x, algorithm="giac")

[Out]

integrate(1/((b*x^2 + a)^(3/2)*(c*x)^(7/2)), x)

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maple [A]  time = 0.02, size = 219, normalized size = 0.66 \[ -\frac {-42 b^{2} x^{4}+42 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, a b \,x^{2} \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-21 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, a b \,x^{2} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )-28 a b \,x^{2}+4 a^{2}}{10 \sqrt {b \,x^{2}+a}\, \sqrt {c x}\, a^{3} c^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(c*x)^(7/2)/(b*x^2+a)^(3/2),x)

[Out]

-1/10/x^2*(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)*Elliptic
F(((b*x+(-a*b)^(1/2))/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))*x^2*a*b-42*b^2*x^4-28*a*b*x^2+4*a^2)/(b*x^2+a)^(1/2)/c^
3/(c*x)^(1/2)/a^3

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b x^{2} + a\right )}^{\frac {3}{2}} \left (c x\right )^{\frac {7}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)^(7/2)/(b*x^2+a)^(3/2),x, algorithm="maxima")

[Out]

integrate(1/((b*x^2 + a)^(3/2)*(c*x)^(7/2)), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\left (c\,x\right )}^{7/2}\,{\left (b\,x^2+a\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/((c*x)^(7/2)*(a + b*x^2)^(3/2)),x)

[Out]

int(1/((c*x)^(7/2)*(a + b*x^2)^(3/2)), x)

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sympy [C]  time = 24.98, size = 51, normalized size = 0.15 \[ \frac {\Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, \frac {3}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 a^{\frac {3}{2}} c^{\frac {7}{2}} x^{\frac {5}{2}} \Gamma \left (- \frac {1}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(c*x)**(7/2)/(b*x**2+a)**(3/2),x)

[Out]

gamma(-5/4)*hyper((-5/4, 3/2), (-1/4,), b*x**2*exp_polar(I*pi)/a)/(2*a**(3/2)*c**(7/2)*x**(5/2)*gamma(-1/4))

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