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

Optimal. Leaf size=195 \[ \frac {2048 b^6 \left (a x+b x^{2/3}\right )^{3/2}}{15015 a^7 x}-\frac {1024 b^5 \left (a x+b x^{2/3}\right )^{3/2}}{5005 a^6 x^{2/3}}+\frac {256 b^4 \left (a x+b x^{2/3}\right )^{3/2}}{1001 a^5 \sqrt [3]{x}}-\frac {128 b^3 \left (a x+b x^{2/3}\right )^{3/2}}{429 a^4}+\frac {48 b^2 \sqrt [3]{x} \left (a x+b x^{2/3}\right )^{3/2}}{143 a^3}-\frac {24 b x^{2/3} \left (a x+b x^{2/3}\right )^{3/2}}{65 a^2}+\frac {2 x \left (a x+b x^{2/3}\right )^{3/2}}{5 a} \]

[Out]

-128/429*b^3*(b*x^(2/3)+a*x)^(3/2)/a^4+2048/15015*b^6*(b*x^(2/3)+a*x)^(3/2)/a^7/x-1024/5005*b^5*(b*x^(2/3)+a*x
)^(3/2)/a^6/x^(2/3)+256/1001*b^4*(b*x^(2/3)+a*x)^(3/2)/a^5/x^(1/3)+48/143*b^2*x^(1/3)*(b*x^(2/3)+a*x)^(3/2)/a^
3-24/65*b*x^(2/3)*(b*x^(2/3)+a*x)^(3/2)/a^2+2/5*x*(b*x^(2/3)+a*x)^(3/2)/a

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Rubi [A]  time = 0.27, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 3, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.176, Rules used = {2016, 2002, 2014} \[ \frac {2048 b^6 \left (a x+b x^{2/3}\right )^{3/2}}{15015 a^7 x}-\frac {1024 b^5 \left (a x+b x^{2/3}\right )^{3/2}}{5005 a^6 x^{2/3}}+\frac {256 b^4 \left (a x+b x^{2/3}\right )^{3/2}}{1001 a^5 \sqrt [3]{x}}-\frac {128 b^3 \left (a x+b x^{2/3}\right )^{3/2}}{429 a^4}+\frac {48 b^2 \sqrt [3]{x} \left (a x+b x^{2/3}\right )^{3/2}}{143 a^3}-\frac {24 b x^{2/3} \left (a x+b x^{2/3}\right )^{3/2}}{65 a^2}+\frac {2 x \left (a x+b x^{2/3}\right )^{3/2}}{5 a} \]

Antiderivative was successfully verified.

[In]

Int[x*Sqrt[b*x^(2/3) + a*x],x]

[Out]

(-128*b^3*(b*x^(2/3) + a*x)^(3/2))/(429*a^4) + (2048*b^6*(b*x^(2/3) + a*x)^(3/2))/(15015*a^7*x) - (1024*b^5*(b
*x^(2/3) + a*x)^(3/2))/(5005*a^6*x^(2/3)) + (256*b^4*(b*x^(2/3) + a*x)^(3/2))/(1001*a^5*x^(1/3)) + (48*b^2*x^(
1/3)*(b*x^(2/3) + a*x)^(3/2))/(143*a^3) - (24*b*x^(2/3)*(b*x^(2/3) + a*x)^(3/2))/(65*a^2) + (2*x*(b*x^(2/3) +
a*x)^(3/2))/(5*a)

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 x \sqrt {b x^{2/3}+a x} \, dx &=\frac {2 x \left (b x^{2/3}+a x\right )^{3/2}}{5 a}-\frac {(4 b) \int x^{2/3} \sqrt {b x^{2/3}+a x} \, dx}{5 a}\\ &=-\frac {24 b x^{2/3} \left (b x^{2/3}+a x\right )^{3/2}}{65 a^2}+\frac {2 x \left (b x^{2/3}+a x\right )^{3/2}}{5 a}+\frac {\left (8 b^2\right ) \int \sqrt [3]{x} \sqrt {b x^{2/3}+a x} \, dx}{13 a^2}\\ &=\frac {48 b^2 \sqrt [3]{x} \left (b x^{2/3}+a x\right )^{3/2}}{143 a^3}-\frac {24 b x^{2/3} \left (b x^{2/3}+a x\right )^{3/2}}{65 a^2}+\frac {2 x \left (b x^{2/3}+a x\right )^{3/2}}{5 a}-\frac {\left (64 b^3\right ) \int \sqrt {b x^{2/3}+a x} \, dx}{143 a^3}\\ &=-\frac {128 b^3 \left (b x^{2/3}+a x\right )^{3/2}}{429 a^4}+\frac {48 b^2 \sqrt [3]{x} \left (b x^{2/3}+a x\right )^{3/2}}{143 a^3}-\frac {24 b x^{2/3} \left (b x^{2/3}+a x\right )^{3/2}}{65 a^2}+\frac {2 x \left (b x^{2/3}+a x\right )^{3/2}}{5 a}+\frac {\left (128 b^4\right ) \int \frac {\sqrt {b x^{2/3}+a x}}{\sqrt [3]{x}} \, dx}{429 a^4}\\ &=-\frac {128 b^3 \left (b x^{2/3}+a x\right )^{3/2}}{429 a^4}+\frac {256 b^4 \left (b x^{2/3}+a x\right )^{3/2}}{1001 a^5 \sqrt [3]{x}}+\frac {48 b^2 \sqrt [3]{x} \left (b x^{2/3}+a x\right )^{3/2}}{143 a^3}-\frac {24 b x^{2/3} \left (b x^{2/3}+a x\right )^{3/2}}{65 a^2}+\frac {2 x \left (b x^{2/3}+a x\right )^{3/2}}{5 a}-\frac {\left (512 b^5\right ) \int \frac {\sqrt {b x^{2/3}+a x}}{x^{2/3}} \, dx}{3003 a^5}\\ &=-\frac {128 b^3 \left (b x^{2/3}+a x\right )^{3/2}}{429 a^4}-\frac {1024 b^5 \left (b x^{2/3}+a x\right )^{3/2}}{5005 a^6 x^{2/3}}+\frac {256 b^4 \left (b x^{2/3}+a x\right )^{3/2}}{1001 a^5 \sqrt [3]{x}}+\frac {48 b^2 \sqrt [3]{x} \left (b x^{2/3}+a x\right )^{3/2}}{143 a^3}-\frac {24 b x^{2/3} \left (b x^{2/3}+a x\right )^{3/2}}{65 a^2}+\frac {2 x \left (b x^{2/3}+a x\right )^{3/2}}{5 a}+\frac {\left (1024 b^6\right ) \int \frac {\sqrt {b x^{2/3}+a x}}{x} \, dx}{15015 a^6}\\ &=-\frac {128 b^3 \left (b x^{2/3}+a x\right )^{3/2}}{429 a^4}+\frac {2048 b^6 \left (b x^{2/3}+a x\right )^{3/2}}{15015 a^7 x}-\frac {1024 b^5 \left (b x^{2/3}+a x\right )^{3/2}}{5005 a^6 x^{2/3}}+\frac {256 b^4 \left (b x^{2/3}+a x\right )^{3/2}}{1001 a^5 \sqrt [3]{x}}+\frac {48 b^2 \sqrt [3]{x} \left (b x^{2/3}+a x\right )^{3/2}}{143 a^3}-\frac {24 b x^{2/3} \left (b x^{2/3}+a x\right )^{3/2}}{65 a^2}+\frac {2 x \left (b x^{2/3}+a x\right )^{3/2}}{5 a}\\ \end {align*}

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Mathematica [A]  time = 0.09, size = 107, normalized size = 0.55 \[ \frac {2 \left (a \sqrt [3]{x}+b\right ) \sqrt {a x+b x^{2/3}} \left (3003 a^6 x^2-2772 a^5 b x^{5/3}+2520 a^4 b^2 x^{4/3}-2240 a^3 b^3 x+1920 a^2 b^4 x^{2/3}-1536 a b^5 \sqrt [3]{x}+1024 b^6\right )}{15015 a^7 \sqrt [3]{x}} \]

Antiderivative was successfully verified.

[In]

Integrate[x*Sqrt[b*x^(2/3) + a*x],x]

[Out]

(2*(b + a*x^(1/3))*Sqrt[b*x^(2/3) + a*x]*(1024*b^6 - 1536*a*b^5*x^(1/3) + 1920*a^2*b^4*x^(2/3) - 2240*a^3*b^3*
x + 2520*a^4*b^2*x^(4/3) - 2772*a^5*b*x^(5/3) + 3003*a^6*x^2))/(15015*a^7*x^(1/3))

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x^(2/3)+a*x)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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giac [A]  time = 0.19, size = 228, normalized size = 1.17 \[ -\frac {2048 \, b^{\frac {15}{2}}}{15015 \, a^{7}} + \frac {2 \, {\left (\frac {15 \, {\left (231 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {13}{2}} - 1638 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {11}{2}} b + 5005 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {9}{2}} b^{2} - 8580 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {7}{2}} b^{3} + 9009 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {5}{2}} b^{4} - 6006 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {3}{2}} b^{5} + 3003 \, \sqrt {a x^{\frac {1}{3}} + b} b^{6}\right )} b}{a^{6}} + \frac {7 \, {\left (429 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {15}{2}} - 3465 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {13}{2}} b + 12285 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {11}{2}} b^{2} - 25025 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {9}{2}} b^{3} + 32175 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {7}{2}} b^{4} - 27027 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {5}{2}} b^{5} + 15015 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {3}{2}} b^{6} - 6435 \, \sqrt {a x^{\frac {1}{3}} + b} b^{7}\right )}}{a^{6}}\right )}}{15015 \, a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x^(2/3)+a*x)^(1/2),x, algorithm="giac")

[Out]

-2048/15015*b^(15/2)/a^7 + 2/15015*(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 - 6006*(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 - 27
027*(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

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maple [A]  time = 0.05, size = 90, normalized size = 0.46 \[ -\frac {2 \sqrt {a x +b \,x^{\frac {2}{3}}}\, \left (a \,x^{\frac {1}{3}}+b \right ) \left (-3003 a^{6} x^{2}+2772 a^{5} b \,x^{\frac {5}{3}}-2520 a^{4} b^{2} x^{\frac {4}{3}}+2240 a^{3} b^{3} x -1920 a^{2} b^{4} x^{\frac {2}{3}}+1536 a \,b^{5} x^{\frac {1}{3}}-1024 b^{6}\right )}{15015 a^{7} x^{\frac {1}{3}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/15015*(a*x+b*x^(2/3))^(1/2)*(a*x^(1/3)+b)*(2772*x^(5/3)*a^5*b-2520*x^(4/3)*a^4*b^2-1920*x^(2/3)*a^2*b^4-300
3*x^2*a^6+1536*x^(1/3)*a*b^5+2240*x*a^3*b^3-1024*b^6)/x^(1/3)/a^7

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x^(2/3)+a*x)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(a*x + b*x^(2/3))*x, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x*(a*x + b*x^(2/3))^(1/2),x)

[Out]

int(x*(a*x + b*x^(2/3))^(1/2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x*(b*x**(2/3)+a*x)**(1/2),x)

[Out]

Integral(x*sqrt(a*x + b*x**(2/3)), x)

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