3.258 \(\int \frac {(c \sqrt {a+b x^2})^{3/2}}{x^4} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 193, normalized size of antiderivative = 1.25, number of steps used = 6, number of rules used = 6, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.286, Rules used = {6720, 277, 325, 229, 227, 196} \[ \frac {b^2 c x \sqrt {c \sqrt {a+b x^2}}}{2 a \sqrt {a+b x^2}}-\frac {b^{3/2} c \sqrt [4]{\frac {b x^2}{a}+1} \sqrt {c \sqrt {a+b x^2}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 \sqrt {a} \sqrt {a+b x^2}}-\frac {b c \sqrt {a+b x^2} \sqrt {c \sqrt {a+b x^2}}}{2 a x}-\frac {c \sqrt {a+b x^2} \sqrt {c \sqrt {a+b x^2}}}{3 x^3} \]

Antiderivative was successfully verified.

[In]

Int[(c*Sqrt[a + b*x^2])^(3/2)/x^4,x]

[Out]

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

Rule 196

Int[((a_) + (b_.)*(x_)^2)^(-5/4), x_Symbol] :> Simp[(2*EllipticE[(1*ArcTan[Rt[b/a, 2]*x])/2, 2])/(a^(5/4)*Rt[b
/a, 2]), x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 227

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2*x)/(a + b*x^2)^(1/4), x] - Dist[a, Int[1/(a + b*x^2)^(5
/4), x], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && PosQ[b/a]

Rule 229

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Dist[(1 + (b*x^2)/a)^(1/4)/(a + b*x^2)^(1/4), Int[1/(1 + (b*x^2
)/a)^(1/4), x], x] /; FreeQ[{a, b}, x] && PosQ[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 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 6720

Int[(u_.)*((a_.)*(v_)^(m_.))^(p_), x_Symbol] :> Dist[(a^IntPart[p]*(a*v^m)^FracPart[p])/v^(m*FracPart[p]), Int
[u*v^(m*p), x], x] /; FreeQ[{a, m, p}, x] &&  !IntegerQ[p] &&  !FreeQ[v, x] &&  !(EqQ[a, 1] && EqQ[m, 1]) &&
!(EqQ[v, x] && EqQ[m, 1])

Rubi steps

\begin {align*} \int \frac {\left (c \sqrt {a+b x^2}\right )^{3/2}}{x^4} \, dx &=\frac {\left (c \sqrt {c \sqrt {a+b x^2}}\right ) \int \frac {\left (a+b x^2\right )^{3/4}}{x^4} \, dx}{\sqrt [4]{a+b x^2}}\\ &=-\frac {c \sqrt {c \sqrt {a+b x^2}} \sqrt {a+b x^2}}{3 x^3}+\frac {\left (b c \sqrt {c \sqrt {a+b x^2}}\right ) \int \frac {1}{x^2 \sqrt [4]{a+b x^2}} \, dx}{2 \sqrt [4]{a+b x^2}}\\ &=-\frac {c \sqrt {c \sqrt {a+b x^2}} \sqrt {a+b x^2}}{3 x^3}-\frac {b c \sqrt {c \sqrt {a+b x^2}} \sqrt {a+b x^2}}{2 a x}+\frac {\left (b^2 c \sqrt {c \sqrt {a+b x^2}}\right ) \int \frac {1}{\sqrt [4]{a+b x^2}} \, dx}{4 a \sqrt [4]{a+b x^2}}\\ &=-\frac {c \sqrt {c \sqrt {a+b x^2}} \sqrt {a+b x^2}}{3 x^3}-\frac {b c \sqrt {c \sqrt {a+b x^2}} \sqrt {a+b x^2}}{2 a x}+\frac {\left (b^2 c \sqrt {c \sqrt {a+b x^2}} \sqrt [4]{1+\frac {b x^2}{a}}\right ) \int \frac {1}{\sqrt [4]{1+\frac {b x^2}{a}}} \, dx}{4 a \sqrt {a+b x^2}}\\ &=\frac {b^2 c x \sqrt {c \sqrt {a+b x^2}}}{2 a \sqrt {a+b x^2}}-\frac {c \sqrt {c \sqrt {a+b x^2}} \sqrt {a+b x^2}}{3 x^3}-\frac {b c \sqrt {c \sqrt {a+b x^2}} \sqrt {a+b x^2}}{2 a x}-\frac {\left (b^2 c \sqrt {c \sqrt {a+b x^2}} \sqrt [4]{1+\frac {b x^2}{a}}\right ) \int \frac {1}{\left (1+\frac {b x^2}{a}\right )^{5/4}} \, dx}{4 a \sqrt {a+b x^2}}\\ &=\frac {b^2 c x \sqrt {c \sqrt {a+b x^2}}}{2 a \sqrt {a+b x^2}}-\frac {c \sqrt {c \sqrt {a+b x^2}} \sqrt {a+b x^2}}{3 x^3}-\frac {b c \sqrt {c \sqrt {a+b x^2}} \sqrt {a+b x^2}}{2 a x}-\frac {b^{3/2} c \sqrt {c \sqrt {a+b x^2}} \sqrt [4]{1+\frac {b x^2}{a}} E\left (\left .\frac {1}{2} \tan ^{-1}\left (\frac {\sqrt {b} x}{\sqrt {a}}\right )\right |2\right )}{2 \sqrt {a} \sqrt {a+b x^2}}\\ \end {align*}

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Mathematica [C]  time = 0.01, size = 57, normalized size = 0.37 \[ -\frac {\left (c \sqrt {a+b x^2}\right )^{3/2} \, _2F_1\left (-\frac {3}{2},-\frac {3}{4};-\frac {1}{2};-\frac {b x^2}{a}\right )}{3 x^3 \left (\frac {b x^2}{a}+1\right )^{3/4}} \]

Antiderivative was successfully verified.

[In]

Integrate[(c*Sqrt[a + b*x^2])^(3/2)/x^4,x]

[Out]

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

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fricas [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(b*x^2+a)^(1/2))^(3/2)/x^4,x, algorithm="fricas")

[Out]

Exception raised: TypeError >>  Error detected within library code:   Shouldn't happen

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\sqrt {b x^{2} + a} c\right )^{\frac {3}{2}}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(b*x^2+a)^(1/2))^(3/2)/x^4,x, algorithm="giac")

[Out]

integrate((sqrt(b*x^2 + a)*c)^(3/2)/x^4, x)

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maple [F]  time = 0.01, size = 0, normalized size = 0.00 \[ \int \frac {\left (\sqrt {b \,x^{2}+a}\, c \right )^{\frac {3}{2}}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((b*x^2+a)^(1/2)*c)^(3/2)/x^4,x)

[Out]

int(((b*x^2+a)^(1/2)*c)^(3/2)/x^4,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (\sqrt {b x^{2} + a} c\right )^{\frac {3}{2}}}{x^{4}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(b*x^2+a)^(1/2))^(3/2)/x^4,x, algorithm="maxima")

[Out]

integrate((sqrt(b*x^2 + a)*c)^(3/2)/x^4, x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (c\,\sqrt {b\,x^2+a}\right )}^{3/2}}{x^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c*(a + b*x^2)^(1/2))^(3/2)/x^4,x)

[Out]

int((c*(a + b*x^2)^(1/2))^(3/2)/x^4, x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\left (c \sqrt {a + b x^{2}}\right )^{\frac {3}{2}}}{x^{4}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((c*(b*x**2+a)**(1/2))**(3/2)/x**4,x)

[Out]

Integral((c*sqrt(a + b*x**2))**(3/2)/x**4, x)

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