3.596 \(\int \frac {\sqrt {a+b x^2}}{(c x)^{7/2}} \, dx\)

Optimal. Leaf size=303 \[ \frac {2 b^{5/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} \operatorname {EllipticF}\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right ),\frac {1}{2}\right )}{5 a^{3/4} c^{7/2} \sqrt {a+b x^2}}-\frac {4 b^{5/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{5 a^{3/4} c^{7/2} \sqrt {a+b x^2}}+\frac {4 b^{3/2} \sqrt {c x} \sqrt {a+b x^2}}{5 a c^4 \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {4 b \sqrt {a+b x^2}}{5 a c^3 \sqrt {c x}}-\frac {2 \sqrt {a+b x^2}}{5 c (c x)^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.22, antiderivative size = 303, normalized size of antiderivative = 1.00, number of steps used = 6, 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} \[ \frac {2 b^{5/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{5 a^{3/4} c^{7/2} \sqrt {a+b x^2}}-\frac {4 b^{5/4} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {c x}}{\sqrt [4]{a} \sqrt {c}}\right )|\frac {1}{2}\right )}{5 a^{3/4} c^{7/2} \sqrt {a+b x^2}}+\frac {4 b^{3/2} \sqrt {c x} \sqrt {a+b x^2}}{5 a c^4 \left (\sqrt {a}+\sqrt {b} x\right )}-\frac {4 b \sqrt {a+b x^2}}{5 a c^3 \sqrt {c x}}-\frac {2 \sqrt {a+b x^2}}{5 c (c x)^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*x^2]/(c*x)^(7/2),x]

[Out]

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

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Mathematica [C]  time = 0.01, size = 56, normalized size = 0.18 \[ -\frac {2 x \sqrt {a+b x^2} \, _2F_1\left (-\frac {5}{4},-\frac {1}{2};-\frac {1}{4};-\frac {b x^2}{a}\right )}{5 (c x)^{7/2} \sqrt {\frac {b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.90, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{2} + a} \sqrt {c x}}{c^{4} x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.03, size = 219, normalized size = 0.72 \[ \frac {-\frac {4 b^{2} x^{4}}{5}+\frac {4 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, a b \,x^{2} \EllipticE \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{5}-\frac {2 \sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {2}\, \sqrt {\frac {-b x +\sqrt {-a b}}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, a b \,x^{2} \EllipticF \left (\sqrt {\frac {b x +\sqrt {-a b}}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{5}-\frac {6 a b \,x^{2}}{5}-\frac {2 a^{2}}{5}}{\sqrt {b \,x^{2}+a}\, \sqrt {c x}\, a \,c^{3} x^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x^2)^(1/2)/(c*x)^(7/2),x)

[Out]

int((a + b*x^2)^(1/2)/(c*x)^(7/2), x)

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sympy [C]  time = 9.75, size = 53, normalized size = 0.17 \[ \frac {\sqrt {a} \Gamma \left (- \frac {5}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} - \frac {5}{4}, - \frac {1}{2} \\ - \frac {1}{4} \end {matrix}\middle | {\frac {b x^{2} e^{i \pi }}{a}} \right )}}{2 c^{\frac {7}{2}} x^{\frac {5}{2}} \Gamma \left (- \frac {1}{4}\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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