3.606 \(\int \frac {(a+b x^2)^{3/2}}{(c x)^{11/2}} \, dx\)

Optimal. Leaf size=331 \[ \frac {4 b^{9/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{15 a^{3/4} c^{11/2} \sqrt {a+b x^2}}-\frac {8 b^{9/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{15 a^{3/4} c^{11/2} \sqrt {a+b x^2}}+\frac {8 b^{5/2} \sqrt {c x} \sqrt {a+b x^2}}{15 a c^6 \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {8 b^2 \sqrt {a+b x^2}}{15 a c^5 \sqrt {c x}}-\frac {4 b \sqrt {a+b x^2}}{15 c^3 (c x)^{5/2}}-\frac {2 \left (a+b x^2\right )^{3/2}}{9 c (c x)^{9/2}} \]

[Out]

-2/9*(b*x^2+a)^(3/2)/c/(c*x)^(9/2)-4/15*b*(b*x^2+a)^(1/2)/c^3/(c*x)^(5/2)-8/15*b^2*(b*x^2+a)^(1/2)/a/c^5/(c*x)
^(1/2)+8/15*b^(5/2)*(c*x)^(1/2)*(b*x^2+a)^(1/2)/a/c^6/(a^(1/2)+x*b^(1/2))-8/15*b^(9/4)*(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^(3/4)/c^(11/2)/(b*x^2+a)^(1/2)+4/15*b^(9/4)*(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^(3/4)/c^(11/2)/(b*x^2+a)^(1/2)

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Rubi [A]  time = 0.27, 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 = {277, 325, 329, 305, 220, 1196} \[ \frac {4 b^{9/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{15 a^{3/4} c^{11/2} \sqrt {a+b x^2}}-\frac {8 b^{9/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{15 a^{3/4} c^{11/2} \sqrt {a+b x^2}}+\frac {8 b^{5/2} \sqrt {c x} \sqrt {a+b x^2}}{15 a c^6 \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {8 b^2 \sqrt {a+b x^2}}{15 a c^5 \sqrt {c x}}-\frac {4 b \sqrt {a+b x^2}}{15 c^3 (c x)^{5/2}}-\frac {2 \left (a+b x^2\right )^{3/2}}{9 c (c x)^{9/2}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x^2)^(3/2)/(c*x)^(11/2),x]

[Out]

(-4*b*Sqrt[a + b*x^2])/(15*c^3*(c*x)^(5/2)) - (8*b^2*Sqrt[a + b*x^2])/(15*a*c^5*Sqrt[c*x]) + (8*b^(5/2)*Sqrt[c
*x]*Sqrt[a + b*x^2])/(15*a*c^6*(Sqrt[a] + Sqrt[b]*x)) - (2*(a + b*x^2)^(3/2))/(9*c*(c*x)^(9/2)) - (8*b^(9/4)*(
Sqrt[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)
*Sqrt[c])], 1/2])/(15*a^(3/4)*c^(11/2)*Sqrt[a + b*x^2]) + (4*b^(9/4)*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(S
qrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[c*x])/(a^(1/4)*Sqrt[c])], 1/2])/(15*a^(3/4)*c^(11/2)*S
qrt[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 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 {\left (a+b x^2\right )^{3/2}}{(c x)^{11/2}} \, dx &=-\frac {2 \left (a+b x^2\right )^{3/2}}{9 c (c x)^{9/2}}+\frac {(2 b) \int \frac {\sqrt {a+b x^2}}{(c x)^{7/2}} \, dx}{3 c^2}\\ &=-\frac {4 b \sqrt {a+b x^2}}{15 c^3 (c x)^{5/2}}-\frac {2 \left (a+b x^2\right )^{3/2}}{9 c (c x)^{9/2}}+\frac {\left (4 b^2\right ) \int \frac {1}{(c x)^{3/2} \sqrt {a+b x^2}} \, dx}{15 c^4}\\ &=-\frac {4 b \sqrt {a+b x^2}}{15 c^3 (c x)^{5/2}}-\frac {8 b^2 \sqrt {a+b x^2}}{15 a c^5 \sqrt {c x}}-\frac {2 \left (a+b x^2\right )^{3/2}}{9 c (c x)^{9/2}}+\frac {\left (4 b^3\right ) \int \frac {\sqrt {c x}}{\sqrt {a+b x^2}} \, dx}{15 a c^6}\\ &=-\frac {4 b \sqrt {a+b x^2}}{15 c^3 (c x)^{5/2}}-\frac {8 b^2 \sqrt {a+b x^2}}{15 a c^5 \sqrt {c x}}-\frac {2 \left (a+b x^2\right )^{3/2}}{9 c (c x)^{9/2}}+\frac {\left (8 b^3\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+\frac {b x^4}{c^2}}} \, dx,x,\sqrt {c x}\right )}{15 a c^7}\\ &=-\frac {4 b \sqrt {a+b x^2}}{15 c^3 (c x)^{5/2}}-\frac {8 b^2 \sqrt {a+b x^2}}{15 a c^5 \sqrt {c x}}-\frac {2 \left (a+b x^2\right )^{3/2}}{9 c (c x)^{9/2}}+\frac {\left (8 b^{5/2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+\frac {b x^4}{c^2}}} \, dx,x,\sqrt {c x}\right )}{15 \sqrt {a} c^6}-\frac {\left (8 b^{5/2}\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 )}{15 \sqrt {a} c^6}\\ &=-\frac {4 b \sqrt {a+b x^2}}{15 c^3 (c x)^{5/2}}-\frac {8 b^2 \sqrt {a+b x^2}}{15 a c^5 \sqrt {c x}}+\frac {8 b^{5/2} \sqrt {c x} \sqrt {a+b x^2}}{15 a c^6 \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {2 \left (a+b x^2\right )^{3/2}}{9 c (c x)^{9/2}}-\frac {8 b^{9/4} \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 )}{15 a^{3/4} c^{11/2} \sqrt {a+b x^2}}+\frac {4 b^{9/4} \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 )}{15 a^{3/4} c^{11/2} \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 57, normalized size = 0.17 \[ -\frac {2 a x \sqrt {a+b x^2} \, _2F_1\left (-\frac {9}{4},-\frac {3}{2};-\frac {5}{4};-\frac {b x^2}{a}\right )}{9 (c x)^{11/2} \sqrt {\frac {b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x^2)^(3/2)/(c*x)^(11/2),x]

[Out]

(-2*a*x*Sqrt[a + b*x^2]*Hypergeometric2F1[-9/4, -3/2, -5/4, -((b*x^2)/a)])/(9*(c*x)^(11/2)*Sqrt[1 + (b*x^2)/a]
)

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fricas [F]  time = 0.89, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}} \sqrt {c x}}{c^{6} x^{6}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/(c*x)^(11/2),x, algorithm="fricas")

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}}}{\left (c x\right )^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/(c*x)^(11/2),x, algorithm="giac")

[Out]

integrate((b*x^2 + a)^(3/2)/(c*x)^(11/2), x)

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maple [A]  time = 0.04, size = 234, normalized size = 0.71 \[ \frac {-\frac {8 b^{3} x^{6}}{15}+\frac {8 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, a \,b^{2} x^{4} \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{15}-\frac {4 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, a \,b^{2} x^{4} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{15}-\frac {46 a \,b^{2} x^{4}}{45}-\frac {32 a^{2} b \,x^{2}}{45}-\frac {2 a^{3}}{9}}{\sqrt {b \,x^{2}+a}\, \sqrt {c x}\, a \,c^{5} x^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x^2+a)^(3/2)/(c*x)^(11/2),x)

[Out]

2/45/x^4*(12*((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^4*a*b^2-6*((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^4*a*b^2-12*b^3*x^6-23*a*b^2*x^4-16*a^2*b*x^2-5*a^3)/(
b*x^2+a)^(1/2)/a/c^5/(c*x)^(1/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x^{2} + a\right )}^{\frac {3}{2}}}{\left (c x\right )^{\frac {11}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x^2+a)^(3/2)/(c*x)^(11/2),x, algorithm="maxima")

[Out]

integrate((b*x^2 + a)^(3/2)/(c*x)^(11/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (b\,x^2+a\right )}^{3/2}}{{\left (c\,x\right )}^{11/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(3/2)/(c*x)^(11/2),x)

[Out]

int((a + b*x^2)^(3/2)/(c*x)^(11/2), x)

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sympy [C]  time = 83.79, size = 53, normalized size = 0.16 \[ \frac {a^{\frac {3}{2}} \Gamma \left (- \frac {9}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {9}{4}, - \frac {3}{2} \\ - \frac {5}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 c^{\frac {11}{2}} x^{\frac {9}{2}} \Gamma \left (- \frac {5}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x**2+a)**(3/2)/(c*x)**(11/2),x)

[Out]

a**(3/2)*gamma(-9/4)*hyper((-9/4, -3/2), (-5/4,), b*x**2*exp_polar(I*pi)/a)/(2*c**(11/2)*x**(9/2)*gamma(-5/4))

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