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

Optimal. Leaf size=134 \[ \frac {4 a^{7/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{7 \sqrt [4]{b} \sqrt {a x+b x^3}}+\frac {4}{7} a \sqrt {a x+b x^3}+\frac {2 \left (a x+b x^3\right )^{3/2}}{7 x} \]

[Out]

2/7*(b*x^3+a*x)^(3/2)/x+4/7*a*(b*x^3+a*x)^(1/2)+4/7*a^(7/4)*(cos(2*arctan(b^(1/4)*x^(1/2)/a^(1/4)))^2)^(1/2)/c
os(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^(1/4)/(b*x^3+a*x)^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 134, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {2021, 2011, 329, 220} \[ \frac {4 a^{7/4} \sqrt {x} \left (\sqrt {a}+\sqrt {b} x\right ) \sqrt {\frac {a+b x^2}{\left (\sqrt {a}+\sqrt {b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac {\sqrt [4]{b} \sqrt {x}}{\sqrt [4]{a}}\right )|\frac {1}{2}\right )}{7 \sqrt [4]{b} \sqrt {a x+b x^3}}+\frac {4}{7} a \sqrt {a x+b x^3}+\frac {2 \left (a x+b x^3\right )^{3/2}}{7 x} \]

Antiderivative was successfully verified.

[In]

Int[(a*x + b*x^3)^(3/2)/x^2,x]

[Out]

(4*a*Sqrt[a*x + b*x^3])/7 + (2*(a*x + b*x^3)^(3/2))/(7*x) + (4*a^(7/4)*Sqrt[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])/(7*b^(1/4)*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 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 2011

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Dist[(a*x^j + b*x^n)^FracPart[p]/(x^(j*FracPart[p
])*(a + b*x^(n - j))^FracPart[p]), Int[x^(j*p)*(a + b*x^(n - j))^p, x], x] /; FreeQ[{a, b, j, n, p}, x] &&  !I
ntegerQ[p] && NeQ[n, j] && PosQ[n - j]

Rule 2021

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 + n*p + 1)), x] + Dist[(a*(n - j)*p)/(c^j*(m + n*p + 1)), Int[(c*x)^(m + j)*(a*x^j + b*x^n)^(p
- 1), x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[p] && LtQ[0, j, n] && (IntegersQ[j, n] || GtQ[c, 0]) && G
tQ[p, 0] && NeQ[m + n*p + 1, 0]

Rubi steps

\begin {align*} \int \frac {\left (a x+b x^3\right )^{3/2}}{x^2} \, dx &=\frac {2 \left (a x+b x^3\right )^{3/2}}{7 x}+\frac {1}{7} (6 a) \int \frac {\sqrt {a x+b x^3}}{x} \, dx\\ &=\frac {4}{7} a \sqrt {a x+b x^3}+\frac {2 \left (a x+b x^3\right )^{3/2}}{7 x}+\frac {1}{7} \left (4 a^2\right ) \int \frac {1}{\sqrt {a x+b x^3}} \, dx\\ &=\frac {4}{7} a \sqrt {a x+b x^3}+\frac {2 \left (a x+b x^3\right )^{3/2}}{7 x}+\frac {\left (4 a^2 \sqrt {x} \sqrt {a+b x^2}\right ) \int \frac {1}{\sqrt {x} \sqrt {a+b x^2}} \, dx}{7 \sqrt {a x+b x^3}}\\ &=\frac {4}{7} a \sqrt {a x+b x^3}+\frac {2 \left (a x+b x^3\right )^{3/2}}{7 x}+\frac {\left (8 a^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 )}{7 \sqrt {a x+b x^3}}\\ &=\frac {4}{7} a \sqrt {a x+b x^3}+\frac {2 \left (a x+b x^3\right )^{3/2}}{7 x}+\frac {4 a^{7/4} \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 )}{7 \sqrt [4]{b} \sqrt {a x+b x^3}}\\ \end {align*}

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Mathematica [C]  time = 0.02, size = 49, normalized size = 0.37 \[ \frac {2 a \sqrt {x \left (a+b x^2\right )} \, _2F_1\left (-\frac {3}{2},\frac {1}{4};\frac {5}{4};-\frac {b x^2}{a}\right )}{\sqrt {\frac {b x^2}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a*x + b*x^3)^(3/2)/x^2,x]

[Out]

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

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fricas [F]  time = 0.58, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b x^{3} + a x} {\left (b x^{2} + a\right )}}{x}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a*x)^(3/2)/x^2, x)

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maple [A]  time = 0.08, size = 144, normalized size = 1.07 \[ \frac {2 \sqrt {b \,x^{3}+a x}\, b \,x^{2}}{7}+\frac {4 \sqrt {-a b}\, \sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {2 \left (x -\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}\, \sqrt {-\frac {b x}{\sqrt {-a b}}}\, a^{2} \EllipticF \left (\sqrt {\frac {\left (x +\frac {\sqrt {-a b}}{b}\right ) b}{\sqrt {-a b}}}, \frac {\sqrt {2}}{2}\right )}{7 \sqrt {b \,x^{3}+a x}\, b}+\frac {6 \sqrt {b \,x^{3}+a x}\, a}{7} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/7*b*x^2*(b*x^3+a*x)^(1/2)+6/7*a*(b*x^3+a*x)^(1/2)+4/7*a^2*(-a*b)^(1/2)/b*((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)*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 \[ \int \frac {{\left (b x^{3} + a x\right )}^{\frac {3}{2}}}{x^{2}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*x^3 + a*x)^(3/2)/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + b*x^3)^(3/2)/x^2,x)

[Out]

int((a*x + b*x^3)^(3/2)/x^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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