3.3003 \(\int \sqrt [3]{a+b x} (c+d x)^{2/3} \, dx\)

Optimal. Leaf size=219 \[ \frac{(b c-a d)^2 \log (a+b x)}{18 b^{5/3} d^{4/3}}+\frac{(b c-a d)^2 \log \left (\frac{\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{6 b^{5/3} d^{4/3}}+\frac{(b c-a d)^2 \tan ^{-1}\left (\frac{2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt{3} \sqrt [3]{d} \sqrt [3]{a+b x}}+\frac{1}{\sqrt{3}}\right )}{3 \sqrt{3} b^{5/3} d^{4/3}}+\frac{\sqrt [3]{a+b x} (c+d x)^{2/3} (b c-a d)}{3 b d}+\frac{(a+b x)^{4/3} (c+d x)^{2/3}}{2 b} \]

[Out]

((b*c - a*d)*(a + b*x)^(1/3)*(c + d*x)^(2/3))/(3*b*d) + ((a + b*x)^(4/3)*(c + d*x)^(2/3))/(2*b) + ((b*c - a*d)
^2*ArcTan[1/Sqrt[3] + (2*b^(1/3)*(c + d*x)^(1/3))/(Sqrt[3]*d^(1/3)*(a + b*x)^(1/3))])/(3*Sqrt[3]*b^(5/3)*d^(4/
3)) + ((b*c - a*d)^2*Log[a + b*x])/(18*b^(5/3)*d^(4/3)) + ((b*c - a*d)^2*Log[-1 + (b^(1/3)*(c + d*x)^(1/3))/(d
^(1/3)*(a + b*x)^(1/3))])/(6*b^(5/3)*d^(4/3))

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Rubi [A]  time = 0.0732004, antiderivative size = 219, normalized size of antiderivative = 1., 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 = {50, 59} \[ \frac{(b c-a d)^2 \log (a+b x)}{18 b^{5/3} d^{4/3}}+\frac{(b c-a d)^2 \log \left (\frac{\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{6 b^{5/3} d^{4/3}}+\frac{(b c-a d)^2 \tan ^{-1}\left (\frac{2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt{3} \sqrt [3]{d} \sqrt [3]{a+b x}}+\frac{1}{\sqrt{3}}\right )}{3 \sqrt{3} b^{5/3} d^{4/3}}+\frac{\sqrt [3]{a+b x} (c+d x)^{2/3} (b c-a d)}{3 b d}+\frac{(a+b x)^{4/3} (c+d x)^{2/3}}{2 b} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(1/3)*(c + d*x)^(2/3),x]

[Out]

((b*c - a*d)*(a + b*x)^(1/3)*(c + d*x)^(2/3))/(3*b*d) + ((a + b*x)^(4/3)*(c + d*x)^(2/3))/(2*b) + ((b*c - a*d)
^2*ArcTan[1/Sqrt[3] + (2*b^(1/3)*(c + d*x)^(1/3))/(Sqrt[3]*d^(1/3)*(a + b*x)^(1/3))])/(3*Sqrt[3]*b^(5/3)*d^(4/
3)) + ((b*c - a*d)^2*Log[a + b*x])/(18*b^(5/3)*d^(4/3)) + ((b*c - a*d)^2*Log[-1 + (b^(1/3)*(c + d*x)^(1/3))/(d
^(1/3)*(a + b*x)^(1/3))])/(6*b^(5/3)*d^(4/3))

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 59

Int[1/(((a_.) + (b_.)*(x_))^(1/3)*((c_.) + (d_.)*(x_))^(2/3)), x_Symbol] :> With[{q = Rt[d/b, 3]}, -Simp[(Sqrt
[3]*q*ArcTan[(2*q*(a + b*x)^(1/3))/(Sqrt[3]*(c + d*x)^(1/3)) + 1/Sqrt[3]])/d, x] + (-Simp[(3*q*Log[(q*(a + b*x
)^(1/3))/(c + d*x)^(1/3) - 1])/(2*d), x] - Simp[(q*Log[c + d*x])/(2*d), x])] /; FreeQ[{a, b, c, d}, x] && NeQ[
b*c - a*d, 0] && PosQ[d/b]

Rubi steps

\begin{align*} \int \sqrt [3]{a+b x} (c+d x)^{2/3} \, dx &=\frac{(a+b x)^{4/3} (c+d x)^{2/3}}{2 b}+\frac{(b c-a d) \int \frac{\sqrt [3]{a+b x}}{\sqrt [3]{c+d x}} \, dx}{3 b}\\ &=\frac{(b c-a d) \sqrt [3]{a+b x} (c+d x)^{2/3}}{3 b d}+\frac{(a+b x)^{4/3} (c+d x)^{2/3}}{2 b}-\frac{(b c-a d)^2 \int \frac{1}{(a+b x)^{2/3} \sqrt [3]{c+d x}} \, dx}{9 b d}\\ &=\frac{(b c-a d) \sqrt [3]{a+b x} (c+d x)^{2/3}}{3 b d}+\frac{(a+b x)^{4/3} (c+d x)^{2/3}}{2 b}+\frac{(b c-a d)^2 \tan ^{-1}\left (\frac{1}{\sqrt{3}}+\frac{2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt{3} \sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{3 \sqrt{3} b^{5/3} d^{4/3}}+\frac{(b c-a d)^2 \log (a+b x)}{18 b^{5/3} d^{4/3}}+\frac{(b c-a d)^2 \log \left (-1+\frac{\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{6 b^{5/3} d^{4/3}}\\ \end{align*}

Mathematica [C]  time = 0.0244762, size = 73, normalized size = 0.33 \[ \frac{3 (a+b x)^{4/3} (c+d x)^{2/3} \, _2F_1\left (-\frac{2}{3},\frac{4}{3};\frac{7}{3};\frac{d (a+b x)}{a d-b c}\right )}{4 b \left (\frac{b (c+d x)}{b c-a d}\right )^{2/3}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(1/3)*(c + d*x)^(2/3),x]

[Out]

(3*(a + b*x)^(4/3)*(c + d*x)^(2/3)*Hypergeometric2F1[-2/3, 4/3, 7/3, (d*(a + b*x))/(-(b*c) + a*d)])/(4*b*((b*(
c + d*x))/(b*c - a*d))^(2/3))

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Maple [F]  time = 0.016, size = 0, normalized size = 0. \begin{align*} \int \sqrt [3]{bx+a} \left ( dx+c \right ) ^{{\frac{2}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x + a\right )}^{\frac{1}{3}}{\left (d x + c\right )}^{\frac{2}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.24569, size = 1798, normalized size = 8.21 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/18*(3*sqrt(1/3)*(b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*sqrt(-(b^2*d)^(1/3)/d)*log(3*b^2*d*x + b^2*c + 2*a*
b*d - 3*(b^2*d)^(1/3)*(b*x + a)^(1/3)*(d*x + c)^(2/3)*b - 3*sqrt(1/3)*(2*(b*x + a)^(2/3)*(d*x + c)^(1/3)*b*d -
(b^2*d)^(2/3)*(b*x + a)^(1/3)*(d*x + c)^(2/3) - (b^2*d)^(1/3)*(b*d*x + b*c))*sqrt(-(b^2*d)^(1/3)/d)) - (b^2*c
^2 - 2*a*b*c*d + a^2*d^2)*(b^2*d)^(2/3)*log(((b*x + a)^(2/3)*(d*x + c)^(1/3)*b*d + (b^2*d)^(2/3)*(b*x + a)^(1/
3)*(d*x + c)^(2/3) + (b^2*d)^(1/3)*(b*d*x + b*c))/(d*x + c)) + 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(b^2*d)^(2/3)
*log(((b*x + a)^(1/3)*(d*x + c)^(2/3)*b*d - (b^2*d)^(2/3)*(d*x + c))/(d*x + c)) + 3*(3*b^3*d^2*x + 2*b^3*c*d +
a*b^2*d^2)*(b*x + a)^(1/3)*(d*x + c)^(2/3))/(b^3*d^2), -1/18*(6*sqrt(1/3)*(b^3*c^2*d - 2*a*b^2*c*d^2 + a^2*b*
d^3)*sqrt((b^2*d)^(1/3)/d)*arctan(sqrt(1/3)*(2*(b^2*d)^(2/3)*(b*x + a)^(1/3)*(d*x + c)^(2/3) + (b^2*d)^(1/3)*(
b*d*x + b*c))*sqrt((b^2*d)^(1/3)/d)/(b^2*d*x + b^2*c)) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(b^2*d)^(2/3)*log(((b
*x + a)^(2/3)*(d*x + c)^(1/3)*b*d + (b^2*d)^(2/3)*(b*x + a)^(1/3)*(d*x + c)^(2/3) + (b^2*d)^(1/3)*(b*d*x + b*c
))/(d*x + c)) - 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(b^2*d)^(2/3)*log(((b*x + a)^(1/3)*(d*x + c)^(2/3)*b*d - (b^
2*d)^(2/3)*(d*x + c))/(d*x + c)) - 3*(3*b^3*d^2*x + 2*b^3*c*d + a*b^2*d^2)*(b*x + a)^(1/3)*(d*x + c)^(2/3))/(b
^3*d^2)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((a + b*x)**(1/3)*(c + d*x)**(2/3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b x + a\right )}^{\frac{1}{3}}{\left (d x + c\right )}^{\frac{2}{3}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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