3.2.79 \(\int \frac {(b x^{2/3}+a x)^{3/2}}{x} \, dx\) [179]

Optimal. Leaf size=84 \[ \frac {16 b^2 \left (b x^{2/3}+a x\right )^{5/2}}{105 a^3 x^{5/3}}-\frac {8 b \left (b x^{2/3}+a x\right )^{5/2}}{21 a^2 x^{4/3}}+\frac {2 \left (b x^{2/3}+a x\right )^{5/2}}{3 a x} \]

[Out]

16/105*b^2*(b*x^(2/3)+a*x)^(5/2)/a^3/x^(5/3)-8/21*b*(b*x^(2/3)+a*x)^(5/2)/a^2/x^(4/3)+2/3*(b*x^(2/3)+a*x)^(5/2
)/a/x

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Rubi [A]
time = 0.10, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2041, 2039} \begin {gather*} \frac {16 b^2 \left (a x+b x^{2/3}\right )^{5/2}}{105 a^3 x^{5/3}}-\frac {8 b \left (a x+b x^{2/3}\right )^{5/2}}{21 a^2 x^{4/3}}+\frac {2 \left (a x+b x^{2/3}\right )^{5/2}}{3 a x} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(b*x^(2/3) + a*x)^(3/2)/x,x]

[Out]

(16*b^2*(b*x^(2/3) + a*x)^(5/2))/(105*a^3*x^(5/3)) - (8*b*(b*x^(2/3) + a*x)^(5/2))/(21*a^2*x^(4/3)) + (2*(b*x^
(2/3) + a*x)^(5/2))/(3*a*x)

Rule 2039

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] &&
NeQ[n, j] && EqQ[m + n*p + n - j + 1, 0] && (IntegerQ[j] || GtQ[c, 0])

Rule 2041

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 \frac {\left (b x^{2/3}+a x\right )^{3/2}}{x} \, dx &=\frac {2 \left (b x^{2/3}+a x\right )^{5/2}}{3 a x}-\frac {(4 b) \int \frac {\left (b x^{2/3}+a x\right )^{3/2}}{x^{4/3}} \, dx}{9 a}\\ &=-\frac {8 b \left (b x^{2/3}+a x\right )^{5/2}}{21 a^2 x^{4/3}}+\frac {2 \left (b x^{2/3}+a x\right )^{5/2}}{3 a x}+\frac {\left (8 b^2\right ) \int \frac {\left (b x^{2/3}+a x\right )^{3/2}}{x^{5/3}} \, dx}{63 a^2}\\ &=\frac {16 b^2 \left (b x^{2/3}+a x\right )^{5/2}}{105 a^3 x^{5/3}}-\frac {8 b \left (b x^{2/3}+a x\right )^{5/2}}{21 a^2 x^{4/3}}+\frac {2 \left (b x^{2/3}+a x\right )^{5/2}}{3 a x}\\ \end {align*}

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Mathematica [A]
time = 4.75, size = 59, normalized size = 0.70 \begin {gather*} \frac {2 \left (b+a \sqrt [3]{x}\right ) \left (8 b^2-20 a b \sqrt [3]{x}+35 a^2 x^{2/3}\right ) \left (b x^{2/3}+a x\right )^{3/2}}{105 a^3 x} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(b + a*x^(1/3))*(8*b^2 - 20*a*b*x^(1/3) + 35*a^2*x^(2/3))*(b*x^(2/3) + a*x)^(3/2))/(105*a^3*x)

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Maple [A]
time = 0.46, size = 48, normalized size = 0.57

method	result	size
derivativedivides	\(\frac {2 \left (b \,x^{\frac {2}{3}}+a x \right )^{\frac {3}{2}} \left (b +a \,x^{\frac {1}{3}}\right ) \left (35 a^{2} x^{\frac {2}{3}}-20 a b \,x^{\frac {1}{3}}+8 b^{2}\right )}{105 x \,a^{3}}\)	\(48\)
default	\(\frac {2 \left (b \,x^{\frac {2}{3}}+a x \right )^{\frac {3}{2}} \left (b +a \,x^{\frac {1}{3}}\right ) \left (35 a^{2} x^{\frac {2}{3}}-20 a b \,x^{\frac {1}{3}}+8 b^{2}\right )}{105 x \,a^{3}}\)	\(48\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/105*(b*x^(2/3)+a*x)^(3/2)*(b+a*x^(1/3))*(35*a^2*x^(2/3)-20*a*b*x^(1/3)+8*b^2)/x/a^3

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 501 vs. \(2 (62) = 124\).
time = 234.82, size = 501, normalized size = 5.96 \begin {gather*} -\frac {{\left (201326592 \, b^{10} + 41943040 \, b^{9} + 196608 \, {\left (6784 \, a^{3} - 3\right )} b^{7} - 3932160 \, b^{8} + 1024 \, {\left (257536 \, a^{3} + 53\right )} b^{6} - 407680 \, a^{6} - 384 \, {\left (72704 \, a^{3} + 1\right )} b^{5} + 12 \, {\left (94371840 \, a^{6} - 437248 \, a^{3} - 3\right )} b^{4} + 896 \, {\left (442368 \, a^{6} + 449 \, a^{3}\right )} b^{3} + 24 \, {\left (1105920 \, a^{6} - 151 \, a^{3}\right )} b^{2} - 15 \, {\left (253952 \, a^{6} + 15 \, a^{3}\right )} b\right )} x - 2 \, {\left (35 \, {\left (16777216 \, a^{4} b^{6} + 6291456 \, a^{4} b^{5} + 196608 \, a^{4} b^{4} - 262144 \, a^{7} - 114688 \, a^{4} b^{3} - 2304 \, a^{4} b^{2} + 864 \, a^{4} b - 27 \, a^{4}\right )} x^{2} + 3 \, {\left (16777216 \, a^{2} b^{8} + 6291456 \, a^{2} b^{7} + 196608 \, a^{2} b^{6} - 114688 \, a^{2} b^{5} - 2304 \, a^{2} b^{4} + 864 \, a^{2} b^{3} - {\left (262144 \, a^{5} + 27 \, a^{2}\right )} b^{2}\right )} x^{\frac {4}{3}} - 4 \, {\left (16777216 \, a b^{9} + 6291456 \, a b^{8} + 196608 \, a b^{7} - 114688 \, a b^{6} - 2304 \, a b^{5} + 864 \, a b^{4} - {\left (262144 \, a^{4} + 27 \, a\right )} b^{3}\right )} x + 2 \, {\left (67108864 \, b^{10} + 25165824 \, b^{9} + 786432 \, b^{8} - 458752 \, b^{7} - 9216 \, b^{6} - 4 \, {\left (262144 \, a^{3} + 27\right )} b^{4} + 3456 \, b^{5} + 25 \, {\left (16777216 \, a^{3} b^{7} + 6291456 \, a^{3} b^{6} + 196608 \, a^{3} b^{5} - 114688 \, a^{3} b^{4} - 2304 \, a^{3} b^{3} + 864 \, a^{3} b^{2} - {\left (262144 \, a^{6} + 27 \, a^{3}\right )} b\right )} x\right )} x^{\frac {2}{3}}\right )} \sqrt {a x + b x^{\frac {2}{3}}}}{105 \, {\left (16777216 \, a^{3} b^{6} + 6291456 \, a^{3} b^{5} + 196608 \, a^{3} b^{4} - 262144 \, a^{6} - 114688 \, a^{3} b^{3} - 2304 \, a^{3} b^{2} + 864 \, a^{3} b - 27 \, a^{3}\right )} x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/105*((201326592*b^10 + 41943040*b^9 + 196608*(6784*a^3 - 3)*b^7 - 3932160*b^8 + 1024*(257536*a^3 + 53)*b^6
- 407680*a^6 - 384*(72704*a^3 + 1)*b^5 + 12*(94371840*a^6 - 437248*a^3 - 3)*b^4 + 896*(442368*a^6 + 449*a^3)*b
^3 + 24*(1105920*a^6 - 151*a^3)*b^2 - 15*(253952*a^6 + 15*a^3)*b)*x - 2*(35*(16777216*a^4*b^6 + 6291456*a^4*b^
5 + 196608*a^4*b^4 - 262144*a^7 - 114688*a^4*b^3 - 2304*a^4*b^2 + 864*a^4*b - 27*a^4)*x^2 + 3*(16777216*a^2*b^
8 + 6291456*a^2*b^7 + 196608*a^2*b^6 - 114688*a^2*b^5 - 2304*a^2*b^4 + 864*a^2*b^3 - (262144*a^5 + 27*a^2)*b^2
)*x^(4/3) - 4*(16777216*a*b^9 + 6291456*a*b^8 + 196608*a*b^7 - 114688*a*b^6 - 2304*a*b^5 + 864*a*b^4 - (262144
*a^4 + 27*a)*b^3)*x + 2*(67108864*b^10 + 25165824*b^9 + 786432*b^8 - 458752*b^7 - 9216*b^6 - 4*(262144*a^3 + 2
7)*b^4 + 3456*b^5 + 25*(16777216*a^3*b^7 + 6291456*a^3*b^6 + 196608*a^3*b^5 - 114688*a^3*b^4 - 2304*a^3*b^3 +
864*a^3*b^2 - (262144*a^6 + 27*a^3)*b)*x)*x^(2/3))*sqrt(a*x + b*x^(2/3)))/((16777216*a^3*b^6 + 6291456*a^3*b^5
+ 196608*a^3*b^4 - 262144*a^6 - 114688*a^3*b^3 - 2304*a^3*b^2 + 864*a^3*b - 27*a^3)*x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 265 vs. \(2 (62) = 124\).
time = 1.84, size = 265, normalized size = 3.15 \begin {gather*} -\frac {2}{35} \, b {\left (\frac {8 \, b^{\frac {7}{2}}}{a^{3}} - \frac {\frac {7 \, {\left (3 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {5}{2}} - 10 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {3}{2}} b + 15 \, \sqrt {a x^{\frac {1}{3}} + b} b^{2}\right )} b}{a^{2}} + \frac {3 \, {\left (5 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {7}{2}} - 21 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {5}{2}} b + 35 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {3}{2}} b^{2} - 35 \, \sqrt {a x^{\frac {1}{3}} + b} b^{3}\right )}}{a^{2}}}{a}\right )} + \frac {2}{105} \, a {\left (\frac {16 \, b^{\frac {9}{2}}}{a^{4}} + \frac {\frac {9 \, {\left (5 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {7}{2}} - 21 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {5}{2}} b + 35 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {3}{2}} b^{2} - 35 \, \sqrt {a x^{\frac {1}{3}} + b} b^{3}\right )} b}{a^{3}} + \frac {35 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {9}{2}} - 180 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {7}{2}} b + 378 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {5}{2}} b^{2} - 420 \, {\left (a x^{\frac {1}{3}} + b\right )}^{\frac {3}{2}} b^{3} + 315 \, \sqrt {a x^{\frac {1}{3}} + b} b^{4}}{a^{3}}}{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,x, algorithm="giac")

[Out]

-2/35*b*(8*b^(7/2)/a^3 - (7*(3*(a*x^(1/3) + b)^(5/2) - 10*(a*x^(1/3) + b)^(3/2)*b + 15*sqrt(a*x^(1/3) + b)*b^2
)*b/a^2 + 3*(5*(a*x^(1/3) + b)^(7/2) - 21*(a*x^(1/3) + b)^(5/2)*b + 35*(a*x^(1/3) + b)^(3/2)*b^2 - 35*sqrt(a*x
^(1/3) + b)*b^3)/a^2)/a) + 2/105*a*(16*b^(9/2)/a^4 + (9*(5*(a*x^(1/3) + b)^(7/2) - 21*(a*x^(1/3) + b)^(5/2)*b
+ 35*(a*x^(1/3) + b)^(3/2)*b^2 - 35*sqrt(a*x^(1/3) + b)*b^3)*b/a^3 + (35*(a*x^(1/3) + b)^(9/2) - 180*(a*x^(1/3
) + b)^(7/2)*b + 378*(a*x^(1/3) + b)^(5/2)*b^2 - 420*(a*x^(1/3) + b)^(3/2)*b^3 + 315*sqrt(a*x^(1/3) + b)*b^4)/
a^3)/a)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a*x + b*x^(2/3))^(3/2)/x,x)

[Out]

int((a*x + b*x^(2/3))^(3/2)/x, x)

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