3.15.68 \(\int \frac {(a+b x)^{2/3}}{(c+d x)^{2/3}} \, dx\)

Optimal. Leaf size=169 \[ \frac {(b c-a d) \log (c+d x)}{3 \sqrt [3]{b} d^{5/3}}+\frac {(b c-a d) \log \left (\frac {\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}-1\right )}{\sqrt [3]{b} d^{5/3}}+\frac {2 (b c-a d) \tan ^{-1}\left (\frac {2 \sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt {3} \sqrt [3]{b} \sqrt [3]{c+d x}}+\frac {1}{\sqrt {3}}\right )}{\sqrt {3} \sqrt [3]{b} d^{5/3}}+\frac {(a+b x)^{2/3} \sqrt [3]{c+d x}}{d} \]

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Rubi [A]  time = 0.04, antiderivative size = 169, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {50, 59} \begin {gather*} \frac {(b c-a d) \log (c+d x)}{3 \sqrt [3]{b} d^{5/3}}+\frac {(b c-a d) \log \left (\frac {\sqrt [3]{d} \sqrt [3]{a+b x}}{\sqrt [3]{b} \sqrt [3]{c+d x}}-1\right )}{\sqrt [3]{b} d^{5/3}}+\frac {2 (b c-a d) \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 )}{\sqrt {3} \sqrt [3]{b} d^{5/3}}+\frac {(a+b x)^{2/3} \sqrt [3]{c+d x}}{d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Mathematica [C]  time = 0.03, size = 73, normalized size = 0.43 \begin {gather*} \frac {3 (a+b x)^{5/3} \left (\frac {b (c+d x)}{b c-a d}\right )^{2/3} \, _2F_1\left (\frac {2}{3},\frac {5}{3};\frac {8}{3};\frac {d (a+b x)}{a d-b c}\right )}{5 b (c+d x)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(a + b*x)^(5/3)*((b*(c + d*x))/(b*c - a*d))^(2/3)*Hypergeometric2F1[2/3, 5/3, 8/3, (d*(a + b*x))/(-(b*c) +
a*d)])/(5*b*(c + d*x)^(2/3))

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IntegrateAlgebraic [A]  time = 7.90, size = 298, normalized size = 1.76 \begin {gather*} \frac {d^{2/3} (a+b x)^{2/3} \left (\frac {(a d-b c) \log \left (\sqrt [3]{b} \sqrt [3]{c+d x} \sqrt [3]{a d+b (c+d x)-b c}+(a d+b (c+d x)-b c)^{2/3}+b^{2/3} (c+d x)^{2/3}\right )}{3 \sqrt [3]{b} d^{5/3}}+\frac {\sqrt [3]{c+d x} (a d+b (c+d x)-b c)^{2/3}}{d^{5/3}}+\frac {2 (b c-a d) \log \left (\sqrt [3]{a d+b (c+d x)-b c}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{3 \sqrt [3]{b} d^{5/3}}-\frac {2 (b c-a d) \tan ^{-1}\left (\frac {\sqrt {3} \sqrt [3]{b} \sqrt [3]{c+d x}}{2 \sqrt [3]{a d+b (c+d x)-b c}+\sqrt [3]{b} \sqrt [3]{c+d x}}\right )}{\sqrt {3} \sqrt [3]{b} d^{5/3}}\right )}{(a d+b d x)^{2/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 1.53, size = 619, normalized size = 3.66 \begin {gather*} \left [\frac {3 \, {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} b d^{2} - 3 \, \sqrt {\frac {1}{3}} {\left (b^{2} c d - a b d^{2}\right )} \sqrt {\frac {\left (-b d^{2}\right )^{\frac {1}{3}}}{b}} \log \left (-3 \, b d^{2} x - 2 \, b c d - a d^{2} - 3 \, \left (-b d^{2}\right )^{\frac {1}{3}} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} d - 3 \, \sqrt {\frac {1}{3}} {\left (2 \, {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} b d - \left (-b d^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} + \left (-b d^{2}\right )^{\frac {1}{3}} {\left (b d x + a d\right )}\right )} \sqrt {\frac {\left (-b d^{2}\right )^{\frac {1}{3}}}{b}}\right ) + 2 \, \left (-b d^{2}\right )^{\frac {2}{3}} {\left (b c - a d\right )} \log \left (\frac {{\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} b d - \left (-b d^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}}{b x + a}\right ) - \left (-b d^{2}\right )^{\frac {2}{3}} {\left (b c - a d\right )} \log \left (\frac {{\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} b d + \left (-b d^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} - \left (-b d^{2}\right )^{\frac {1}{3}} {\left (b d x + a d\right )}}{b x + a}\right )}{3 \, b d^{3}}, \frac {3 \, {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} b d^{2} - 6 \, \sqrt {\frac {1}{3}} {\left (b^{2} c d - a b d^{2}\right )} \sqrt {-\frac {\left (-b d^{2}\right )^{\frac {1}{3}}}{b}} \arctan \left (\frac {\sqrt {\frac {1}{3}} {\left (2 \, \left (-b d^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} - \left (-b d^{2}\right )^{\frac {1}{3}} {\left (b d x + a d\right )}\right )} \sqrt {-\frac {\left (-b d^{2}\right )^{\frac {1}{3}}}{b}}}{b d^{2} x + a d^{2}}\right ) + 2 \, \left (-b d^{2}\right )^{\frac {2}{3}} {\left (b c - a d\right )} \log \left (\frac {{\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} b d - \left (-b d^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}}{b x + a}\right ) - \left (-b d^{2}\right )^{\frac {2}{3}} {\left (b c - a d\right )} \log \left (\frac {{\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} b d + \left (-b d^{2}\right )^{\frac {2}{3}} {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}} - \left (-b d^{2}\right )^{\frac {1}{3}} {\left (b d x + a d\right )}}{b x + a}\right )}{3 \, b d^{3}}\right ] \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.03, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b x +a \right )^{\frac {2}{3}}}{\left (d x +c \right )^{\frac {2}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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