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

Optimal. Leaf size=169 \[ -\frac{512 b^5 \left (a x+b x^{2/3}\right )^{5/2}}{15015 a^6 x^{5/3}}+\frac{256 b^4 \left (a x+b x^{2/3}\right )^{5/2}}{3003 a^5 x^{4/3}}-\frac{64 b^3 \left (a x+b x^{2/3}\right )^{5/2}}{429 a^4 x}+\frac{32 b^2 \left (a x+b x^{2/3}\right )^{5/2}}{143 a^3 x^{2/3}}-\frac{4 b \left (a x+b x^{2/3}\right )^{5/2}}{13 a^2 \sqrt [3]{x}}+\frac{2 \left (a x+b x^{2/3}\right )^{5/2}}{5 a} \]

[Out]

(2*(b*x^(2/3) + a*x)^(5/2))/(5*a) - (512*b^5*(b*x^(2/3) + a*x)^(5/2))/(15015*a^6*x^(5/3)) + (256*b^4*(b*x^(2/3
) + a*x)^(5/2))/(3003*a^5*x^(4/3)) - (64*b^3*(b*x^(2/3) + a*x)^(5/2))/(429*a^4*x) + (32*b^2*(b*x^(2/3) + a*x)^
(5/2))/(143*a^3*x^(2/3)) - (4*b*(b*x^(2/3) + a*x)^(5/2))/(13*a^2*x^(1/3))

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Rubi [A]  time = 0.249476, antiderivative size = 169, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2002, 2016, 2014} \[ -\frac{512 b^5 \left (a x+b x^{2/3}\right )^{5/2}}{15015 a^6 x^{5/3}}+\frac{256 b^4 \left (a x+b x^{2/3}\right )^{5/2}}{3003 a^5 x^{4/3}}-\frac{64 b^3 \left (a x+b x^{2/3}\right )^{5/2}}{429 a^4 x}+\frac{32 b^2 \left (a x+b x^{2/3}\right )^{5/2}}{143 a^3 x^{2/3}}-\frac{4 b \left (a x+b x^{2/3}\right )^{5/2}}{13 a^2 \sqrt [3]{x}}+\frac{2 \left (a x+b x^{2/3}\right )^{5/2}}{5 a} \]

Antiderivative was successfully verified.

[In]

Int[(b*x^(2/3) + a*x)^(3/2),x]

[Out]

(2*(b*x^(2/3) + a*x)^(5/2))/(5*a) - (512*b^5*(b*x^(2/3) + a*x)^(5/2))/(15015*a^6*x^(5/3)) + (256*b^4*(b*x^(2/3
) + a*x)^(5/2))/(3003*a^5*x^(4/3)) - (64*b^3*(b*x^(2/3) + a*x)^(5/2))/(429*a^4*x) + (32*b^2*(b*x^(2/3) + a*x)^
(5/2))/(143*a^3*x^(2/3)) - (4*b*(b*x^(2/3) + a*x)^(5/2))/(13*a^2*x^(1/3))

Rule 2002

Int[((a_.)*(x_)^(j_.) + (b_.)*(x_)^(n_.))^(p_), x_Symbol] :> Simp[(a*x^j + b*x^n)^(p + 1)/(a*(j*p + 1)*x^(j -
1)), x] - Dist[(b*(n*p + n - j + 1))/(a*(j*p + 1)), Int[x^(n - j)*(a*x^j + b*x^n)^p, x], x] /; FreeQ[{a, b, j,
n, p}, x] &&  !IntegerQ[p] && NeQ[n, j] && ILtQ[Simplify[(n*p + n - j + 1)/(n - j)], 0] && NeQ[j*p + 1, 0]

Rule 2016

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

Rule 2014

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

Rubi steps

\begin{align*} \int \left (b x^{2/3}+a x\right )^{3/2} \, dx &=\frac{2 \left (b x^{2/3}+a x\right )^{5/2}}{5 a}-\frac{(2 b) \int \frac{\left (b x^{2/3}+a x\right )^{3/2}}{\sqrt [3]{x}} \, dx}{3 a}\\ &=\frac{2 \left (b x^{2/3}+a x\right )^{5/2}}{5 a}-\frac{4 b \left (b x^{2/3}+a x\right )^{5/2}}{13 a^2 \sqrt [3]{x}}+\frac{\left (16 b^2\right ) \int \frac{\left (b x^{2/3}+a x\right )^{3/2}}{x^{2/3}} \, dx}{39 a^2}\\ &=\frac{2 \left (b x^{2/3}+a x\right )^{5/2}}{5 a}+\frac{32 b^2 \left (b x^{2/3}+a x\right )^{5/2}}{143 a^3 x^{2/3}}-\frac{4 b \left (b x^{2/3}+a x\right )^{5/2}}{13 a^2 \sqrt [3]{x}}-\frac{\left (32 b^3\right ) \int \frac{\left (b x^{2/3}+a x\right )^{3/2}}{x} \, dx}{143 a^3}\\ &=\frac{2 \left (b x^{2/3}+a x\right )^{5/2}}{5 a}-\frac{64 b^3 \left (b x^{2/3}+a x\right )^{5/2}}{429 a^4 x}+\frac{32 b^2 \left (b x^{2/3}+a x\right )^{5/2}}{143 a^3 x^{2/3}}-\frac{4 b \left (b x^{2/3}+a x\right )^{5/2}}{13 a^2 \sqrt [3]{x}}+\frac{\left (128 b^4\right ) \int \frac{\left (b x^{2/3}+a x\right )^{3/2}}{x^{4/3}} \, dx}{1287 a^4}\\ &=\frac{2 \left (b x^{2/3}+a x\right )^{5/2}}{5 a}+\frac{256 b^4 \left (b x^{2/3}+a x\right )^{5/2}}{3003 a^5 x^{4/3}}-\frac{64 b^3 \left (b x^{2/3}+a x\right )^{5/2}}{429 a^4 x}+\frac{32 b^2 \left (b x^{2/3}+a x\right )^{5/2}}{143 a^3 x^{2/3}}-\frac{4 b \left (b x^{2/3}+a x\right )^{5/2}}{13 a^2 \sqrt [3]{x}}-\frac{\left (256 b^5\right ) \int \frac{\left (b x^{2/3}+a x\right )^{3/2}}{x^{5/3}} \, dx}{9009 a^5}\\ &=\frac{2 \left (b x^{2/3}+a x\right )^{5/2}}{5 a}-\frac{512 b^5 \left (b x^{2/3}+a x\right )^{5/2}}{15015 a^6 x^{5/3}}+\frac{256 b^4 \left (b x^{2/3}+a x\right )^{5/2}}{3003 a^5 x^{4/3}}-\frac{64 b^3 \left (b x^{2/3}+a x\right )^{5/2}}{429 a^4 x}+\frac{32 b^2 \left (b x^{2/3}+a x\right )^{5/2}}{143 a^3 x^{2/3}}-\frac{4 b \left (b x^{2/3}+a x\right )^{5/2}}{13 a^2 \sqrt [3]{x}}\\ \end{align*}

Mathematica [A]  time = 0.0578055, size = 98, normalized size = 0.58 \[ \frac{2 \left (a \sqrt [3]{x}+b\right )^2 \sqrt{a x+b x^{2/3}} \left (-1120 a^2 b^3 x^{2/3}+1680 a^3 b^2 x-2310 a^4 b x^{4/3}+3003 a^5 x^{5/3}+640 a b^4 \sqrt [3]{x}-256 b^5\right )}{15015 a^6 \sqrt [3]{x}} \]

Antiderivative was successfully verified.

[In]

Integrate[(b*x^(2/3) + a*x)^(3/2),x]

[Out]

(2*(b + a*x^(1/3))^2*Sqrt[b*x^(2/3) + a*x]*(-256*b^5 + 640*a*b^4*x^(1/3) - 1120*a^2*b^3*x^(2/3) + 1680*a^3*b^2
*x - 2310*a^4*b*x^(4/3) + 3003*a^5*x^(5/3)))/(15015*a^6*x^(1/3))

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Maple [A]  time = 0.003, size = 79, normalized size = 0.5 \begin{align*}{\frac{2}{15015\,x{a}^{6}} \left ( b{x}^{{\frac{2}{3}}}+ax \right ) ^{{\frac{3}{2}}} \left ( b+a\sqrt [3]{x} \right ) \left ( 3003\,{x}^{5/3}{a}^{5}-2310\,{x}^{4/3}{a}^{4}b+1680\,x{a}^{3}{b}^{2}-1120\,{x}^{2/3}{a}^{2}{b}^{3}+640\,\sqrt [3]{x}a{b}^{4}-256\,{b}^{5} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/15015*(b*x^(2/3)+a*x)^(3/2)*(b+a*x^(1/3))*(3003*x^(5/3)*a^5-2310*x^(4/3)*a^4*b+1680*x*a^3*b^2-1120*x^(2/3)*a
^2*b^3+640*x^(1/3)*a*b^4-256*b^5)/x/a^6

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((a*x + b*x^(2/3))^(3/2), x)

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Fricas [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.22769, size = 281, normalized size = 1.66 \begin{align*} \frac{2}{3003} \, b{\left (\frac{256 \, b^{\frac{13}{2}}}{a^{6}} + \frac{693 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{13}{2}} - 4095 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{11}{2}} b + 10010 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{9}{2}} b^{2} - 12870 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{7}{2}} b^{3} + 9009 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{5}{2}} b^{4} - 3003 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{3}{2}} b^{5}}{a^{6}}\right )} - \frac{2}{15015} \, a{\left (\frac{1024 \, b^{\frac{15}{2}}}{a^{7}} - \frac{3003 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{15}{2}} - 20790 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{13}{2}} b + 61425 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{11}{2}} b^{2} - 100100 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{9}{2}} b^{3} + 96525 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{7}{2}} b^{4} - 54054 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{5}{2}} b^{5} + 15015 \,{\left (a x^{\frac{1}{3}} + b\right )}^{\frac{3}{2}} b^{6}}{a^{7}}\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3003*b*(256*b^(13/2)/a^6 + (693*(a*x^(1/3) + b)^(13/2) - 4095*(a*x^(1/3) + b)^(11/2)*b + 10010*(a*x^(1/3) +
b)^(9/2)*b^2 - 12870*(a*x^(1/3) + b)^(7/2)*b^3 + 9009*(a*x^(1/3) + b)^(5/2)*b^4 - 3003*(a*x^(1/3) + b)^(3/2)*b
^5)/a^6) - 2/15015*a*(1024*b^(15/2)/a^7 - (3003*(a*x^(1/3) + b)^(15/2) - 20790*(a*x^(1/3) + b)^(13/2)*b + 6142
5*(a*x^(1/3) + b)^(11/2)*b^2 - 100100*(a*x^(1/3) + b)^(9/2)*b^3 + 96525*(a*x^(1/3) + b)^(7/2)*b^4 - 54054*(a*x
^(1/3) + b)^(5/2)*b^5 + 15015*(a*x^(1/3) + b)^(3/2)*b^6)/a^7)