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

Optimal. Leaf size=277 \[ \frac {24 b^{3/2} x \left (a+b x^2\right )}{5 \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {a x+b x^3}}+\frac {12 \sqrt [4]{a} b^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 \sqrt {a x+b x^3}}-\frac {24 \sqrt [4]{a} b^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 \sqrt {a x+b x^3}}-\frac {12 b \sqrt {a x+b x^3}}{5 x}-\frac {2 \left (a x+b x^3\right )^{3/2}}{5 x^4} \]

[Out]

-2/5*(b*x^3+a*x)^(3/2)/x^4+24/5*b^(3/2)*x*(b*x^2+a)/(a^(1/2)+x*b^(1/2))/(b*x^3+a*x)^(1/2)-12/5*b*(b*x^3+a*x)^(
1/2)/x-24/5*a^(1/4)*b^(5/4)*(cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)))^2)^(1/2)/cos(2*arctan(b^(1/4)*x^(1/2)/a^(1
/4)))*EllipticE(sin(2*arctan(b^(1/4)*x^(1/2)/a^(1/4))),1/2*2^(1/2))*(a^(1/2)+x*b^(1/2))*x^(1/2)*((b*x^2+a)/(a^
(1/2)+x*b^(1/2))^2)^(1/2)/(b*x^3+a*x)^(1/2)+12/5*a^(1/4)*b^(5/4)*(cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)))^2)^(1
/2)/cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)))*EllipticF(sin(2*arctan(b^(1/4)*x^(1/2)/a^(1/4))),1/2*2^(1/2))*(a^(1
/2)+x*b^(1/2))*x^(1/2)*((b*x^2+a)/(a^(1/2)+x*b^(1/2))^2)^(1/2)/(b*x^3+a*x)^(1/2)

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Rubi [A]  time = 0.26, antiderivative size = 277, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.353, Rules used = {2020, 2032, 329, 305, 220, 1196} \[ \frac {24 b^{3/2} x \left (a+b x^2\right )}{5 \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {a x+b x^3}}+\frac {12 \sqrt [4]{a} b^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 \sqrt {a x+b x^3}}-\frac {24 \sqrt [4]{a} b^{5/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} E\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 \sqrt {a x+b x^3}}-\frac {12 b \sqrt {a x+b x^3}}{5 x}-\frac {2 \left (a x+b x^3\right )^{3/2}}{5 x^4} \]

Antiderivative was successfully verified.

[In]

Int[(a*x + b*x^3)^(3/2)/x^5,x]

[Out]

(24*b^(3/2)*x*(a + b*x^2))/(5*(Sqrt[a] + Sqrt[b]*x)*Sqrt[a*x + b*x^3]) - (12*b*Sqrt[a*x + b*x^3])/(5*x) - (2*(
a*x + b*x^3)^(3/2))/(5*x^4) - (24*a^(1/4)*b^(5/4)*Sqrt[x]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sq
rt[b]*x)^2]*EllipticE[2*ArcTan[(b^(1/4)*Sqrt[x])/a^(1/4)], 1/2])/(5*Sqrt[a*x + b*x^3]) + (12*a^(1/4)*b^(5/4)*S
qrt[x]*(Sqrt[a] + Sqrt[b]*x)*Sqrt[(a + b*x^2)/(Sqrt[a] + Sqrt[b]*x)^2]*EllipticF[2*ArcTan[(b^(1/4)*Sqrt[x])/a^
(1/4)], 1/2])/(5*Sqrt[a*x + b*x^3])

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 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 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]

Rule 2020

Int[((c_.)*(x_))^(m_)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a*x^j + b*
x^n)^p)/(c*(m + j*p + 1)), x] - Dist[(b*p*(n - j))/(c^n*(m + j*p + 1)), Int[(c*x)^(m + n)*(a*x^j + b*x^n)^(p -
1), x], x] /; FreeQ[{a, b, c}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && GtQ[p
, 0] && LtQ[m + j*p + 1, 0]

Rule 2032

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(c^IntPart[m]*(c*x)^FracP
art[m]*(a*x^j + b*x^n)^FracPart[p])/(x^(FracPart[m] + j*FracPart[p])*(a + b*x^(n - j))^FracPart[p]), Int[x^(m
+ j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, c, j, m, n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && PosQ[n
- j]

Rubi steps

\begin {align*} \int \frac {\left (a x+b x^3\right )^{3/2}}{x^5} \, dx &=-\frac {2 \left (a x+b x^3\right )^{3/2}}{5 x^4}+\frac {1}{5} (6 b) \int \frac {\sqrt {a x+b x^3}}{x^2} \, dx\\ &=-\frac {12 b \sqrt {a x+b x^3}}{5 x}-\frac {2 \left (a x+b x^3\right )^{3/2}}{5 x^4}+\frac {1}{5} \left (12 b^2\right ) \int \frac {x}{\sqrt {a x+b x^3}} \, dx\\ &=-\frac {12 b \sqrt {a x+b x^3}}{5 x}-\frac {2 \left (a x+b x^3\right )^{3/2}}{5 x^4}+\frac {\left (12 b^2 \sqrt {x} \sqrt {a+b x^2}\right ) \int \frac {\sqrt {x}}{\sqrt {a+b x^2}} \, dx}{5 \sqrt {a x+b x^3}}\\ &=-\frac {12 b \sqrt {a x+b x^3}}{5 x}-\frac {2 \left (a x+b x^3\right )^{3/2}}{5 x^4}+\frac {\left (24 b^2 \sqrt {x} \sqrt {a+b x^2}\right ) \operatorname {Subst}\left (\int \frac {x^2}{\sqrt {a+b x^4}} \, dx,x,\sqrt {x}\right )}{5 \sqrt {a x+b x^3}}\\ &=-\frac {12 b \sqrt {a x+b x^3}}{5 x}-\frac {2 \left (a x+b x^3\right )^{3/2}}{5 x^4}+\frac {\left (24 \sqrt {a} b^{3/2} \sqrt {x} \sqrt {a+b x^2}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^4}} \, dx,x,\sqrt {x}\right )}{5 \sqrt {a x+b x^3}}-\frac {\left (24 \sqrt {a} b^{3/2} \sqrt {x} \sqrt {a+b x^2}\right ) \operatorname {Subst}\left (\int \frac {1-\frac {\sqrt {b} x^2}{\sqrt {a}}}{\sqrt {a+b x^4}} \, dx,x,\sqrt {x}\right )}{5 \sqrt {a x+b x^3}}\\ &=\frac {24 b^{3/2} x \left (a+b x^2\right )}{5 \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {a x+b x^3}}-\frac {12 b \sqrt {a x+b x^3}}{5 x}-\frac {2 \left (a x+b x^3\right )^{3/2}}{5 x^4}-\frac {24 \sqrt [4]{a} b^{5/4} \sqrt {x} \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 {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 \sqrt {a x+b x^3}}+\frac {12 \sqrt [4]{a} b^{5/4} \sqrt {x} \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 {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{5 \sqrt {a x+b x^3}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 54, normalized size = 0.19 \[ -\frac {2 a \sqrt {x \left (a+b x^2\right )} \, _2F_1\left (-\frac {3}{2},-\frac {5}{4};-\frac {1}{4};-\frac {b x^2}{a}\right )}{5 x^3 \sqrt {\frac {b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x + b*x^3)^(3/2)/x^5,x]

[Out]

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

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fricas [F]  time = 0.61, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{3} + a x} {\left (b x^{2} + a\right )}}{x^{4}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a*x)^(3/2)/x^5, x)

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maple [A]  time = 0.08, size = 196, normalized size = 0.71 \[ -\frac {14 \left (b \,x^{2}+a \right ) b}{5 \sqrt {\left (b \,x^{2}+a \right ) x}}+\frac {12 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, \left (-\frac {2 \sqrt {-a b}\, \EllipticE \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}+\frac {\sqrt {-a b}\, \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{b}\right ) b}{5 \sqrt {b \,x^{3}+a x}}-\frac {2 \sqrt {b \,x^{3}+a x}\, a}{5 x^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*a*(b*x^3+a*x)^(1/2)/x^3-14/5*(b*x^2+a)*b/((b*x^2+a)*x)^(1/2)+12/5*b*(-a*b)^(1/2)*((x+(-a*b)^(1/2)/b)/(-a*
b)^(1/2)*b)^(1/2)*(-2*(x-(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2)*(-1/(-a*b)^(1/2)*b*x)^(1/2)/(b*x^3+a*x)^(1/2)*(
-2*(-a*b)^(1/2)/b*EllipticE(((x+(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2),1/2*2^(1/2))+(-a*b)^(1/2)/b*EllipticF(((
x+(-a*b)^(1/2)/b)/(-a*b)^(1/2)*b)^(1/2),1/2*2^(1/2)))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a*x)^(3/2)/x^5, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + b*x^3)^(3/2)/x^5,x)

[Out]

int((a*x + b*x^3)^(3/2)/x^5, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((x*(a + b*x**2))**(3/2)/x**5, x)

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