3.1575 \(\int (a+b x)^{2/3} \sqrt [3]{c+d x} \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.0875222, 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} \[ \frac{(b c-a d)^2 \log (c+d x)}{18 b^{4/3} d^{5/3}}+\frac{(b c-a d)^2 \log \left (\frac{\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}-1\right )}{6 b^{4/3} d^{5/3}}+\frac{(b c-a d)^2 \tan ^{-1}\left (\frac{2 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt{3} \sqrt [3]{b} \sqrt [3]{c+d x}}+\frac{1}{\sqrt{3}}\right )}{3 \sqrt{3} b^{4/3} d^{5/3}}+\frac{(a+b x)^{2/3} \sqrt [3]{c+d x} (b c-a d)}{6 b d}+\frac{(a+b x)^{5/3} \sqrt [3]{c+d x}}{2 b} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [C]  time = 0.0314631, size = 73, normalized size = 0.33 \[ \frac{3 (a+b x)^{5/3} \sqrt [3]{c+d x} \, _2F_1\left (-\frac{1}{3},\frac{5}{3};\frac{8}{3};\frac{d (a+b x)}{a d-b c}\right )}{5 b \sqrt [3]{\frac{b (c+d x)}{b c-a d}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(a + b*x)^(5/3)*(c + d*x)^(1/3)*Hypergeometric2F1[-1/3, 5/3, 8/3, (d*(a + b*x))/(-(b*c) + a*d)])/(5*b*((b*(
c + d*x))/(b*c - a*d))^(1/3))

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Maple [F]  time = 0.024, size = 0, normalized size = 0. \begin{align*} \int \left ( bx+a \right ) ^{{\frac{2}{3}}}\sqrt [3]{dx+c}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.06942, size = 1800, normalized size = 8.22 \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)^(2/3)*(d*x+c)^(1/3),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*d^2)^(1/3)/b)*log(-3*b*d^2*x - 2*b*c*d - a
*d^2 + 3*(b*d^2)^(1/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3)*d + 3*sqrt(1/3)*(2*(b*x + a)^(1/3)*(d*x + c)^(2/3)*b*d
- (b*d^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) - (b*d^2)^(1/3)*(b*d*x + a*d))*sqrt(-(b*d^2)^(1/3)/b)) + 2*(b^
2*c^2 - 2*a*b*c*d + a^2*d^2)*(b*d^2)^(2/3)*log(((b*x + a)^(2/3)*(d*x + c)^(1/3)*b*d - (b*d^2)^(2/3)*(b*x + a))
/(b*x + a)) - (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(b*d^2)^(2/3)*log(((b*x + a)^(1/3)*(d*x + c)^(2/3)*b*d + (b*d^2)
^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) + (b*d^2)^(1/3)*(b*d*x + a*d))/(b*x + a)) + 3*(3*b^2*d^3*x + b^2*c*d^2
+ 2*a*b*d^3)*(b*x + a)^(2/3)*(d*x + c)^(1/3))/(b^2*d^3), -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*d^2)^(1/3)/b)*arctan(sqrt(1/3)*(2*(b*d^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) + (b*d^2)^(1/3)*
(b*d*x + a*d))*sqrt((b*d^2)^(1/3)/b)/(b*d^2*x + a*d^2)) - 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(b*d^2)^(2/3)*log(
((b*x + a)^(2/3)*(d*x + c)^(1/3)*b*d - (b*d^2)^(2/3)*(b*x + a))/(b*x + a)) + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*(
b*d^2)^(2/3)*log(((b*x + a)^(1/3)*(d*x + c)^(2/3)*b*d + (b*d^2)^(2/3)*(b*x + a)^(2/3)*(d*x + c)^(1/3) + (b*d^2
)^(1/3)*(b*d*x + a*d))/(b*x + a)) - 3*(3*b^2*d^3*x + b^2*c*d^2 + 2*a*b*d^3)*(b*x + a)^(2/3)*(d*x + c)^(1/3))/(
b^2*d^3)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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