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

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

________________________________________________________________________________________

Rubi [A]  time = 0.25, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 3, integrand size = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2002, 2016, 2014} \begin {gather*} -\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} \end {gather*}

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 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])

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])

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 = 0.07, size = 98, normalized size = 0.58 \begin {gather*} \frac {2 \left (a \sqrt [3]{x}+b\right )^2 \sqrt {a x+b x^{2/3}} \left (3003 a^5 x^{5/3}-2310 a^4 b x^{4/3}+1680 a^3 b^2 x-1120 a^2 b^3 x^{2/3}+640 a b^4 \sqrt [3]{x}-256 b^5\right )}{15015 a^6 \sqrt [3]{x}} \end {gather*}

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))

________________________________________________________________________________________

IntegrateAlgebraic [A]  time = 4.47, size = 98, normalized size = 0.58 \begin {gather*} \frac {2 \left (a \sqrt [3]{x}+b\right ) \left (x^{2/3} \left (a \sqrt [3]{x}+b\right )\right )^{3/2} \left (3003 a^5 x^{5/3}-2310 a^4 b x^{4/3}+1680 a^3 b^2 x-1120 a^2 b^3 x^{2/3}+640 a b^4 \sqrt [3]{x}-256 b^5\right )}{15015 a^6 x} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(b*x^(2/3) + a*x)^(3/2),x]

[Out]

(2*(b + a*x^(1/3))*((b + a*x^(1/3))*x^(2/3))^(3/2)*(-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)

________________________________________________________________________________________

fricas [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

________________________________________________________________________________________

giac [B]  time = 0.23, size = 434, normalized size = 2.57 \begin {gather*} \frac {2}{3003} \, b {\left (\frac {256 \, b^{\frac {13}{2}}}{a^{6}} + \frac {\frac {13 \, {\left (63 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {11}{2}} - 385 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {9}{2}} b + 990 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {7}{2}} b^{2} - 1386 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {5}{2}} b^{3} + 1155 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {3}{2}} b^{4} - 693 \, \sqrt {a x^{\frac {1}{3}} + b} b^{5}\right )} b}{a^{5}} + \frac {3 \, {\left (231 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {13}{2}} - 1638 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {11}{2}} b + 5005 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {9}{2}} b^{2} - 8580 \, {\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} - 6006 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {3}{2}} b^{5} + 3003 \, \sqrt {a x^{\frac {1}{3}} + b} b^{6}\right )}}{a^{5}}}{a}\right )} - \frac {2}{15015} \, a {\left (\frac {1024 \, b^{\frac {15}{2}}}{a^{7}} - \frac {\frac {15 \, {\left (231 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {13}{2}} - 1638 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {11}{2}} b + 5005 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {9}{2}} b^{2} - 8580 \, {\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} - 6006 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {3}{2}} b^{5} + 3003 \, \sqrt {a x^{\frac {1}{3}} + b} b^{6}\right )} b}{a^{6}} + \frac {7 \, {\left (429 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {15}{2}} - 3465 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {13}{2}} b + 12285 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {11}{2}} b^{2} - 25025 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {9}{2}} b^{3} + 32175 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {7}{2}} b^{4} - 27027 \, {\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} - 6435 \, \sqrt {a x^{\frac {1}{3}} + b} b^{7}\right )}}{a^{6}}}{a}\right )} \end {gather*}

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 + (13*(63*(a*x^(1/3) + b)^(11/2) - 385*(a*x^(1/3) + b)^(9/2)*b + 990*(a*x^(1/3) + b
)^(7/2)*b^2 - 1386*(a*x^(1/3) + b)^(5/2)*b^3 + 1155*(a*x^(1/3) + b)^(3/2)*b^4 - 693*sqrt(a*x^(1/3) + b)*b^5)*b
/a^5 + 3*(231*(a*x^(1/3) + b)^(13/2) - 1638*(a*x^(1/3) + b)^(11/2)*b + 5005*(a*x^(1/3) + b)^(9/2)*b^2 - 8580*(
a*x^(1/3) + b)^(7/2)*b^3 + 9009*(a*x^(1/3) + b)^(5/2)*b^4 - 6006*(a*x^(1/3) + b)^(3/2)*b^5 + 3003*sqrt(a*x^(1/
3) + b)*b^6)/a^5)/a) - 2/15015*a*(1024*b^(15/2)/a^7 - (15*(231*(a*x^(1/3) + b)^(13/2) - 1638*(a*x^(1/3) + b)^(
11/2)*b + 5005*(a*x^(1/3) + b)^(9/2)*b^2 - 8580*(a*x^(1/3) + b)^(7/2)*b^3 + 9009*(a*x^(1/3) + b)^(5/2)*b^4 - 6
006*(a*x^(1/3) + b)^(3/2)*b^5 + 3003*sqrt(a*x^(1/3) + b)*b^6)*b/a^6 + 7*(429*(a*x^(1/3) + b)^(15/2) - 3465*(a*
x^(1/3) + b)^(13/2)*b + 12285*(a*x^(1/3) + b)^(11/2)*b^2 - 25025*(a*x^(1/3) + b)^(9/2)*b^3 + 32175*(a*x^(1/3)
+ b)^(7/2)*b^4 - 27027*(a*x^(1/3) + b)^(5/2)*b^5 + 15015*(a*x^(1/3) + b)^(3/2)*b^6 - 6435*sqrt(a*x^(1/3) + b)*
b^7)/a^6)/a)

________________________________________________________________________________________

maple [A]  time = 0.05, size = 79, normalized size = 0.47 \begin {gather*} \frac {2 \left (a x +b \,x^{\frac {2}{3}}\right )^{\frac {3}{2}} \left (a \,x^{\frac {1}{3}}+b \right ) \left (3003 a^{5} x^{\frac {5}{3}}-2310 a^{4} b \,x^{\frac {4}{3}}+1680 a^{3} b^{2} x -1120 a^{2} b^{3} x^{\frac {2}{3}}+640 a \,b^{4} x^{\frac {1}{3}}-256 b^{5}\right )}{15015 a^{6} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

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)

________________________________________________________________________________________

mupad [B]  time = 5.14, size = 40, normalized size = 0.24 \begin {gather*} \frac {x\,{\left (a\,x+b\,x^{2/3}\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},6;\ 7;\ -\frac {a\,x^{1/3}}{b}\right )}{2\,{\left (\frac {a\,x^{1/3}}{b}+1\right )}^{3/2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + b*x^(2/3))^(3/2),x)

[Out]

(x*(a*x + b*x^(2/3))^(3/2)*hypergeom([-3/2, 6], 7, -(a*x^(1/3))/b))/(2*((a*x^(1/3))/b + 1)^(3/2))

________________________________________________________________________________________

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

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)

________________________________________________________________________________________