3.65 \(\int \frac{x^7}{(a x+b x^3)^{3/2}} \, dx\)

Optimal. Leaf size=161 \[ \frac{15 a^{7/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} \text{EllipticF}\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right ),\frac{1}{2}\right )}{14 b^{13/4} \sqrt{a x+b x^3}}+\frac{9 x^2 \sqrt{a x+b x^3}}{7 b^2}-\frac{15 a \sqrt{a x+b x^3}}{7 b^3}-\frac{x^5}{b \sqrt{a x+b x^3}} \]

[Out]

-(x^5/(b*Sqrt[a*x + b*x^3])) - (15*a*Sqrt[a*x + b*x^3])/(7*b^3) + (9*x^2*Sqrt[a*x + b*x^3])/(7*b^2) + (15*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])/(14*b^(13/4)*Sqrt[a*x + b*x^3])

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Rubi [A]  time = 0.193408, antiderivative size = 161, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 17, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.294, Rules used = {2022, 2024, 2011, 329, 220} \[ \frac{15 a^{7/4} \sqrt{x} \left (\sqrt{a}+\sqrt{b} x\right ) \sqrt{\frac{a+b x^2}{\left (\sqrt{a}+\sqrt{b} x\right )^2}} F\left (2 \tan ^{-1}\left (\frac{\sqrt [4]{b} \sqrt{x}}{\sqrt [4]{a}}\right )|\frac{1}{2}\right )}{14 b^{13/4} \sqrt{a x+b x^3}}+\frac{9 x^2 \sqrt{a x+b x^3}}{7 b^2}-\frac{15 a \sqrt{a x+b x^3}}{7 b^3}-\frac{x^5}{b \sqrt{a x+b x^3}} \]

Antiderivative was successfully verified.

[In]

Int[x^7/(a*x + b*x^3)^(3/2),x]

[Out]

-(x^5/(b*Sqrt[a*x + b*x^3])) - (15*a*Sqrt[a*x + b*x^3])/(7*b^3) + (9*x^2*Sqrt[a*x + b*x^3])/(7*b^2) + (15*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])/(14*b^(13/4)*Sqrt[a*x + b*x^3])

Rule 2022

Int[((c_.)*(x_))^(m_.)*((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n +
1)*(a*x^j + b*x^n)^(p + 1))/(b*(n - j)*(p + 1)), x] - Dist[(c^n*(m + j*p - n + j + 1))/(b*(n - j)*(p + 1)), I
nt[(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] && (I
ntegersQ[j, n] || GtQ[c, 0]) && LtQ[p, -1] && GtQ[m + j*p + 1, n - j]

Rule 2024

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

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 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 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]

Rubi steps

\begin{align*} \int \frac{x^7}{\left (a x+b x^3\right )^{3/2}} \, dx &=-\frac{x^5}{b \sqrt{a x+b x^3}}+\frac{9 \int \frac{x^4}{\sqrt{a x+b x^3}} \, dx}{2 b}\\ &=-\frac{x^5}{b \sqrt{a x+b x^3}}+\frac{9 x^2 \sqrt{a x+b x^3}}{7 b^2}-\frac{(45 a) \int \frac{x^2}{\sqrt{a x+b x^3}} \, dx}{14 b^2}\\ &=-\frac{x^5}{b \sqrt{a x+b x^3}}-\frac{15 a \sqrt{a x+b x^3}}{7 b^3}+\frac{9 x^2 \sqrt{a x+b x^3}}{7 b^2}+\frac{\left (15 a^2\right ) \int \frac{1}{\sqrt{a x+b x^3}} \, dx}{14 b^3}\\ &=-\frac{x^5}{b \sqrt{a x+b x^3}}-\frac{15 a \sqrt{a x+b x^3}}{7 b^3}+\frac{9 x^2 \sqrt{a x+b x^3}}{7 b^2}+\frac{\left (15 a^2 \sqrt{x} \sqrt{a+b x^2}\right ) \int \frac{1}{\sqrt{x} \sqrt{a+b x^2}} \, dx}{14 b^3 \sqrt{a x+b x^3}}\\ &=-\frac{x^5}{b \sqrt{a x+b x^3}}-\frac{15 a \sqrt{a x+b x^3}}{7 b^3}+\frac{9 x^2 \sqrt{a x+b x^3}}{7 b^2}+\frac{\left (15 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 b^3 \sqrt{a x+b x^3}}\\ &=-\frac{x^5}{b \sqrt{a x+b x^3}}-\frac{15 a \sqrt{a x+b x^3}}{7 b^3}+\frac{9 x^2 \sqrt{a x+b x^3}}{7 b^2}+\frac{15 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 )}{14 b^{13/4} \sqrt{a x+b x^3}}\\ \end{align*}

Mathematica [C]  time = 0.0332699, size = 80, normalized size = 0.5 \[ \frac{x \left (15 a^2 \sqrt{\frac{b x^2}{a}+1} \, _2F_1\left (\frac{1}{4},\frac{1}{2};\frac{5}{4};-\frac{b x^2}{a}\right )-15 a^2-6 a b x^2+2 b^2 x^4\right )}{7 b^3 \sqrt{x \left (a+b x^2\right )}} \]

Antiderivative was successfully verified.

[In]

Integrate[x^7/(a*x + b*x^3)^(3/2),x]

[Out]

(x*(-15*a^2 - 6*a*b*x^2 + 2*b^2*x^4 + 15*a^2*Sqrt[1 + (b*x^2)/a]*Hypergeometric2F1[1/4, 1/2, 5/4, -((b*x^2)/a)
]))/(7*b^3*Sqrt[x*(a + b*x^2)])

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Maple [A]  time = 0.018, size = 172, normalized size = 1.1 \begin{align*} -{\frac{{a}^{2}x}{{b}^{3}}{\frac{1}{\sqrt{ \left ({\frac{a}{b}}+{x}^{2} \right ) bx}}}}+{\frac{2\,{x}^{2}}{7\,{b}^{2}}\sqrt{b{x}^{3}+ax}}-{\frac{8\,a}{7\,{b}^{3}}\sqrt{b{x}^{3}+ax}}+{\frac{15\,{a}^{2}}{14\,{b}^{4}}\sqrt{-ab}\sqrt{{b \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ){\frac{1}{\sqrt{-ab}}}}}\sqrt{-2\,{\frac{b}{\sqrt{-ab}} \left ( x-{\frac{\sqrt{-ab}}{b}} \right ) }}\sqrt{-{bx{\frac{1}{\sqrt{-ab}}}}}{\it EllipticF} \left ( \sqrt{{b \left ( x+{\frac{1}{b}\sqrt{-ab}} \right ){\frac{1}{\sqrt{-ab}}}}},{\frac{\sqrt{2}}{2}} \right ){\frac{1}{\sqrt{b{x}^{3}+ax}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^7/(b*x^3+a*x)^(3/2),x)

[Out]

-1/b^3*x*a^2/((1/b*a+x^2)*b*x)^(1/2)+2/7*x^2*(b*x^3+a*x)^(1/2)/b^2-8/7*a*(b*x^3+a*x)^(1/2)/b^3+15/14*a^2/b^4*(
-a*b)^(1/2)*((x+1/b*(-a*b)^(1/2))*b/(-a*b)^(1/2))^(1/2)*(-2*(x-1/b*(-a*b)^(1/2))*b/(-a*b)^(1/2))^(1/2)*(-x*b/(
-a*b)^(1/2))^(1/2)/(b*x^3+a*x)^(1/2)*EllipticF(((x+1/b*(-a*b)^(1/2))*b/(-a*b)^(1/2))^(1/2),1/2*2^(1/2))

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{7}}{{\left (b x^{3} + a x\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(b*x^3+a*x)^(3/2),x, algorithm="maxima")

[Out]

integrate(x^7/(b*x^3 + a*x)^(3/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{b x^{3} + a x} x^{5}}{b^{2} x^{4} + 2 \, a b x^{2} + a^{2}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(b*x^3+a*x)^(3/2),x, algorithm="fricas")

[Out]

integral(sqrt(b*x^3 + a*x)*x^5/(b^2*x^4 + 2*a*b*x^2 + a^2), x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{7}}{\left (x \left (a + b x^{2}\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**7/(b*x**3+a*x)**(3/2),x)

[Out]

Integral(x**7/(x*(a + b*x**2))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{x^{7}}{{\left (b x^{3} + a x\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^7/(b*x^3+a*x)^(3/2),x, algorithm="giac")

[Out]

integrate(x^7/(b*x^3 + a*x)^(3/2), x)