3.26.24 \(\int \frac {(a+b x)^{4/3}}{(c+d x)^{4/3}} \, dx\)

Optimal. Leaf size=195 \[ \frac {2 \sqrt [3]{b} (b c-a d) \log (a+b x)}{3 d^{7/3}}+\frac {2 \sqrt [3]{b} (b c-a d) \log \left (\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{d^{7/3}}+\frac {4 \sqrt [3]{b} (b c-a d) \tan ^{-1}\left (\frac {2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt {3} \sqrt [3]{d} \sqrt [3]{a+b x}}+\frac {1}{\sqrt {3}}\right )}{\sqrt {3} d^{7/3}}+\frac {4 b \sqrt [3]{a+b x} (c+d x)^{2/3}}{d^2}-\frac {3 (a+b x)^{4/3}}{d \sqrt [3]{c+d x}} \]

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Rubi [A]  time = 0.07, antiderivative size = 195, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {47, 50, 59} \begin {gather*} \frac {4 b \sqrt [3]{a+b x} (c+d x)^{2/3}}{d^2}+\frac {2 \sqrt [3]{b} (b c-a d) \log (a+b x)}{3 d^{7/3}}+\frac {2 \sqrt [3]{b} (b c-a d) \log \left (\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{d^{7/3}}+\frac {4 \sqrt [3]{b} (b c-a d) \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 )}{\sqrt {3} d^{7/3}}-\frac {3 (a+b x)^{4/3}}{d \sqrt [3]{c+d x}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(4/3)/(c + d*x)^(4/3),x]

[Out]

(-3*(a + b*x)^(4/3))/(d*(c + d*x)^(1/3)) + (4*b*(a + b*x)^(1/3)*(c + d*x)^(2/3))/d^2 + (4*b^(1/3)*(b*c - a*d)*
ArcTan[1/Sqrt[3] + (2*b^(1/3)*(c + d*x)^(1/3))/(Sqrt[3]*d^(1/3)*(a + b*x)^(1/3))])/(Sqrt[3]*d^(7/3)) + (2*b^(1
/3)*(b*c - a*d)*Log[a + b*x])/(3*d^(7/3)) + (2*b^(1/3)*(b*c - a*d)*Log[-1 + (b^(1/3)*(c + d*x)^(1/3))/(d^(1/3)
*(a + b*x)^(1/3))])/d^(7/3)

Rule 47

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

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)^{4/3}}{(c+d x)^{4/3}} \, dx &=-\frac {3 (a+b x)^{4/3}}{d \sqrt [3]{c+d x}}+\frac {(4 b) \int \frac {\sqrt [3]{a+b x}}{\sqrt [3]{c+d x}} \, dx}{d}\\ &=-\frac {3 (a+b x)^{4/3}}{d \sqrt [3]{c+d x}}+\frac {4 b \sqrt [3]{a+b x} (c+d x)^{2/3}}{d^2}-\frac {(4 b (b c-a d)) \int \frac {1}{(a+b x)^{2/3} \sqrt [3]{c+d x}} \, dx}{3 d^2}\\ &=-\frac {3 (a+b x)^{4/3}}{d \sqrt [3]{c+d x}}+\frac {4 b \sqrt [3]{a+b x} (c+d x)^{2/3}}{d^2}+\frac {4 \sqrt [3]{b} (b c-a d) \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 )}{\sqrt {3} d^{7/3}}+\frac {2 \sqrt [3]{b} (b c-a d) \log (a+b x)}{3 d^{7/3}}+\frac {2 \sqrt [3]{b} (b c-a d) \log \left (-1+\frac {\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}\right )}{d^{7/3}}\\ \end {align*}

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Mathematica [C]  time = 0.05, size = 73, normalized size = 0.37 \begin {gather*} \frac {3 (a+b x)^{7/3} \left (\frac {b (c+d x)}{b c-a d}\right )^{4/3} \, _2F_1\left (\frac {4}{3},\frac {7}{3};\frac {10}{3};\frac {d (a+b x)}{a d-b c}\right )}{7 b (c+d x)^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^(4/3)/(c + d*x)^(4/3),x]

[Out]

(3*(a + b*x)^(7/3)*((b*(c + d*x))/(b*c - a*d))^(4/3)*Hypergeometric2F1[4/3, 7/3, 10/3, (d*(a + b*x))/(-(b*c) +
a*d)])/(7*b*(c + d*x)^(4/3))

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IntegrateAlgebraic [A]  time = 0.00, size = 326, normalized size = 1.67 \begin {gather*} \frac {d^{4/3} (a+b x)^{4/3} \left (\frac {4 \left (b^{4/3} c-a \sqrt [3]{b} d\right ) \log \left (\sqrt [3]{a d+b (c+d x)-b c}-\sqrt [3]{b} \sqrt [3]{c+d x}\right )}{3 d^{7/3}}-\frac {2 \left (b^{4/3} c-a \sqrt [3]{b} d\right ) \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 d^{7/3}}+\frac {4 \left (b^{4/3} c-a \sqrt [3]{b} d\right ) \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} d^{7/3}}+\frac {\sqrt [3]{a d+b (c+d x)-b c} (-3 a d+b (c+d x)+3 b c)}{d^{7/3} \sqrt [3]{c+d x}}\right )}{(a d+b d x)^{4/3}} \end {gather*}

Antiderivative was successfully verified.

[In]

IntegrateAlgebraic[(a + b*x)^(4/3)/(c + d*x)^(4/3),x]

[Out]

(d^(4/3)*(a + b*x)^(4/3)*(((3*b*c - 3*a*d + b*(c + d*x))*(-(b*c) + a*d + b*(c + d*x))^(1/3))/(d^(7/3)*(c + d*x
)^(1/3)) + (4*(b^(4/3)*c - a*b^(1/3)*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]*d^(7/3)) + (4*(b^(4/3)*c - a*b^(1/3)*d)*Log[-(b^(1/3)*(c + d*x)
^(1/3)) + (-(b*c) + a*d + b*(c + d*x))^(1/3)])/(3*d^(7/3)) - (2*(b^(4/3)*c - a*b^(1/3)*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
*d^(7/3))))/(a*d + b*d*x)^(4/3)

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fricas [A]  time = 1.91, size = 306, normalized size = 1.57 \begin {gather*} \frac {4 \, \sqrt {3} {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x\right )} \left (-\frac {b}{d}\right )^{\frac {1}{3}} \arctan \left (\frac {2 \, \sqrt {3} {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} d \left (-\frac {b}{d}\right )^{\frac {2}{3}} + \sqrt {3} {\left (b d x + b c\right )}}{3 \, {\left (b d x + b c\right )}}\right ) + 2 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x\right )} \left (-\frac {b}{d}\right )^{\frac {1}{3}} \log \left (\frac {{\left (d x + c\right )} \left (-\frac {b}{d}\right )^{\frac {2}{3}} - {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}} \left (-\frac {b}{d}\right )^{\frac {1}{3}} + {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}}}{d x + c}\right ) - 4 \, {\left (b c^{2} - a c d + {\left (b c d - a d^{2}\right )} x\right )} \left (-\frac {b}{d}\right )^{\frac {1}{3}} \log \left (\frac {{\left (d x + c\right )} \left (-\frac {b}{d}\right )^{\frac {1}{3}} + {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{d x + c}\right ) + 3 \, {\left (b d x + 4 \, b c - 3 \, a d\right )} {\left (b x + a\right )}^{\frac {1}{3}} {\left (d x + c\right )}^{\frac {2}{3}}}{3 \, {\left (d^{3} x + c d^{2}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(4/3)/(d*x+c)^(4/3),x, algorithm="fricas")

[Out]

1/3*(4*sqrt(3)*(b*c^2 - a*c*d + (b*c*d - a*d^2)*x)*(-b/d)^(1/3)*arctan(1/3*(2*sqrt(3)*(b*x + a)^(1/3)*(d*x + c
)^(2/3)*d*(-b/d)^(2/3) + sqrt(3)*(b*d*x + b*c))/(b*d*x + b*c)) + 2*(b*c^2 - a*c*d + (b*c*d - a*d^2)*x)*(-b/d)^
(1/3)*log(((d*x + c)*(-b/d)^(2/3) - (b*x + a)^(1/3)*(d*x + c)^(2/3)*(-b/d)^(1/3) + (b*x + a)^(2/3)*(d*x + c)^(
1/3))/(d*x + c)) - 4*(b*c^2 - a*c*d + (b*c*d - a*d^2)*x)*(-b/d)^(1/3)*log(((d*x + c)*(-b/d)^(1/3) + (b*x + a)^
(1/3)*(d*x + c)^(2/3))/(d*x + c)) + 3*(b*d*x + 4*b*c - 3*a*d)*(b*x + a)^(1/3)*(d*x + c)^(2/3))/(d^3*x + c*d^2)

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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 {4}{3}}}{{\left (d x + c\right )}^{\frac {4}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(4/3)/(d*x+c)^(4/3),x, algorithm="giac")

[Out]

integrate((b*x + a)^(4/3)/(d*x + c)^(4/3), x)

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maple [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (b x +a \right )^{\frac {4}{3}}}{\left (d x +c \right )^{\frac {4}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(4/3)/(d*x+c)^(4/3),x)

[Out]

int((b*x+a)^(4/3)/(d*x+c)^(4/3),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 {4}{3}}}{{\left (d x + c\right )}^{\frac {4}{3}}}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(4/3)/(d*x+c)^(4/3),x, algorithm="maxima")

[Out]

integrate((b*x + a)^(4/3)/(d*x + c)^(4/3), 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 )}^{4/3}}{{\left (c+d\,x\right )}^{4/3}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*x)^(4/3)/(c + d*x)^(4/3),x)

[Out]

int((a + b*x)^(4/3)/(c + d*x)^(4/3), 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 {4}{3}}}{\left (c + d x\right )^{\frac {4}{3}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(4/3)/(d*x+c)**(4/3),x)

[Out]

Integral((a + b*x)**(4/3)/(c + d*x)**(4/3), x)

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