3.1615 \(\int \frac{(a+b x)^{7/3}}{(c+d x)^{4/3}} \, dx\)

Optimal. Leaf size=241 \[ \frac{7 b (a+b x)^{4/3} (c+d x)^{2/3}}{2 d^2}-\frac{14 b \sqrt [3]{a+b x} (c+d x)^{2/3} (b c-a d)}{3 d^3}-\frac{7 \sqrt [3]{b} (b c-a d)^2 \log (a+b x)}{9 d^{10/3}}-\frac{7 \sqrt [3]{b} (b c-a d)^2 \log \left (\frac{\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{3 d^{10/3}}-\frac{14 \sqrt [3]{b} (b c-a d)^2 \tan ^{-1}\left (\frac{2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt{3} \sqrt [3]{d} \sqrt [3]{a+b x}}+\frac{1}{\sqrt{3}}\right )}{3 \sqrt{3} d^{10/3}}-\frac{3 (a+b x)^{7/3}}{d \sqrt [3]{c+d x}} \]

[Out]

(-3*(a + b*x)^(7/3))/(d*(c + d*x)^(1/3)) - (14*b*(b*c - a*d)*(a + b*x)^(1/3)*(c + d*x)^(2/3))/(3*d^3) + (7*b*(
a + b*x)^(4/3)*(c + d*x)^(2/3))/(2*d^2) - (14*b^(1/3)*(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]*d^(10/3)) - (7*b^(1/3)*(b*c - a*d)^2*Log[a + b*x])/(9*d^(1
0/3)) - (7*b^(1/3)*(b*c - a*d)^2*Log[-1 + (b^(1/3)*(c + d*x)^(1/3))/(d^(1/3)*(a + b*x)^(1/3))])/(3*d^(10/3))

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Rubi [A]  time = 0.106996, antiderivative size = 241, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {47, 50, 59} \[ \frac{7 b (a+b x)^{4/3} (c+d x)^{2/3}}{2 d^2}-\frac{14 b \sqrt [3]{a+b x} (c+d x)^{2/3} (b c-a d)}{3 d^3}-\frac{7 \sqrt [3]{b} (b c-a d)^2 \log (a+b x)}{9 d^{10/3}}-\frac{7 \sqrt [3]{b} (b c-a d)^2 \log \left (\frac{\sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt [3]{d} \sqrt [3]{a+b x}}-1\right )}{3 d^{10/3}}-\frac{14 \sqrt [3]{b} (b c-a d)^2 \tan ^{-1}\left (\frac{2 \sqrt [3]{b} \sqrt [3]{c+d x}}{\sqrt{3} \sqrt [3]{d} \sqrt [3]{a+b x}}+\frac{1}{\sqrt{3}}\right )}{3 \sqrt{3} d^{10/3}}-\frac{3 (a+b x)^{7/3}}{d \sqrt [3]{c+d x}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*(a + b*x)^(7/3))/(d*(c + d*x)^(1/3)) - (14*b*(b*c - a*d)*(a + b*x)^(1/3)*(c + d*x)^(2/3))/(3*d^3) + (7*b*(
a + b*x)^(4/3)*(c + d*x)^(2/3))/(2*d^2) - (14*b^(1/3)*(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]*d^(10/3)) - (7*b^(1/3)*(b*c - a*d)^2*Log[a + b*x])/(9*d^(1
0/3)) - (7*b^(1/3)*(b*c - a*d)^2*Log[-1 + (b^(1/3)*(c + d*x)^(1/3))/(d^(1/3)*(a + b*x)^(1/3))])/(3*d^(10/3))

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)^{7/3}}{(c+d x)^{4/3}} \, dx &=-\frac{3 (a+b x)^{7/3}}{d \sqrt [3]{c+d x}}+\frac{(7 b) \int \frac{(a+b x)^{4/3}}{\sqrt [3]{c+d x}} \, dx}{d}\\ &=-\frac{3 (a+b x)^{7/3}}{d \sqrt [3]{c+d x}}+\frac{7 b (a+b x)^{4/3} (c+d x)^{2/3}}{2 d^2}-\frac{(14 b (b c-a d)) \int \frac{\sqrt [3]{a+b x}}{\sqrt [3]{c+d x}} \, dx}{3 d^2}\\ &=-\frac{3 (a+b x)^{7/3}}{d \sqrt [3]{c+d x}}-\frac{14 b (b c-a d) \sqrt [3]{a+b x} (c+d x)^{2/3}}{3 d^3}+\frac{7 b (a+b x)^{4/3} (c+d x)^{2/3}}{2 d^2}+\frac{\left (14 b (b c-a d)^2\right ) \int \frac{1}{(a+b x)^{2/3} \sqrt [3]{c+d x}} \, dx}{9 d^3}\\ &=-\frac{3 (a+b x)^{7/3}}{d \sqrt [3]{c+d x}}-\frac{14 b (b c-a d) \sqrt [3]{a+b x} (c+d x)^{2/3}}{3 d^3}+\frac{7 b (a+b x)^{4/3} (c+d x)^{2/3}}{2 d^2}-\frac{14 \sqrt [3]{b} (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} d^{10/3}}-\frac{7 \sqrt [3]{b} (b c-a d)^2 \log (a+b x)}{9 d^{10/3}}-\frac{7 \sqrt [3]{b} (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 )}{3 d^{10/3}}\\ \end{align*}

Mathematica [C]  time = 0.0584074, size = 73, normalized size = 0.3 \[ \frac{3 (a+b x)^{10/3} \left (\frac{b (c+d x)}{b c-a d}\right )^{4/3} \, _2F_1\left (\frac{4}{3},\frac{10}{3};\frac{13}{3};\frac{d (a+b x)}{a d-b c}\right )}{10 b (c+d x)^{4/3}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(3*(a + b*x)^(10/3)*((b*(c + d*x))/(b*c - a*d))^(4/3)*Hypergeometric2F1[4/3, 10/3, 13/3, (d*(a + b*x))/(-(b*c)
+ a*d)])/(10*b*(c + d*x)^(4/3))

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Maple [F]  time = 0.049, size = 0, normalized size = 0. \begin{align*} \int{ \left ( bx+a \right ) ^{{\frac{7}{3}}} \left ( dx+c \right ) ^{-{\frac{4}{3}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^(7/3)/(d*x+c)^(4/3),x)

[Out]

int((b*x+a)^(7/3)/(d*x+c)^(4/3),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{\frac{7}{3}}}{{\left (d x + c\right )}^{\frac{4}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(7/3)/(d*x+c)^(4/3),x, algorithm="maxima")

[Out]

integrate((b*x + a)^(7/3)/(d*x + c)^(4/3), x)

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Fricas [B]  time = 2.45291, size = 995, normalized size = 4.13 \begin{align*} -\frac{28 \, \sqrt{3}{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2} +{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\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 ) + 14 \,{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2} +{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\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 ) - 28 \,{\left (b^{2} c^{3} - 2 \, a b c^{2} d + a^{2} c d^{2} +{\left (b^{2} c^{2} d - 2 \, a b c d^{2} + a^{2} d^{3}\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 (3 \, b^{2} d^{2} x^{2} - 28 \, b^{2} c^{2} + 49 \, a b c d - 18 \, a^{2} d^{2} -{\left (7 \, b^{2} c d - 13 \, a b d^{2}\right )} x\right )}{\left (b x + a\right )}^{\frac{1}{3}}{\left (d x + c\right )}^{\frac{2}{3}}}{18 \,{\left (d^{4} x + c d^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(7/3)/(d*x+c)^(4/3),x, algorithm="fricas")

[Out]

-1/18*(28*sqrt(3)*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2 + (b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*x)*(-b/d)^(1/3)*arc
tan(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)) + 14
*(b^2*c^3 - 2*a*b*c^2*d + a^2*c*d^2 + (b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*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)) - 28*(b^2
*c^3 - 2*a*b*c^2*d + a^2*c*d^2 + (b^2*c^2*d - 2*a*b*c*d^2 + a^2*d^3)*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*(3*b^2*d^2*x^2 - 28*b^2*c^2 + 49*a*b*c*d - 18*a^2*d^2 - (
7*b^2*c*d - 13*a*b*d^2)*x)*(b*x + a)^(1/3)*(d*x + c)^(2/3))/(d^4*x + c*d^3)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (a + b x\right )^{\frac{7}{3}}}{\left (c + d x\right )^{\frac{4}{3}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**(7/3)/(d*x+c)**(4/3),x)

[Out]

Integral((a + b*x)**(7/3)/(c + d*x)**(4/3), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (b x + a\right )}^{\frac{7}{3}}}{{\left (d x + c\right )}^{\frac{4}{3}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^(7/3)/(d*x+c)^(4/3),x, algorithm="giac")

[Out]

integrate((b*x + a)^(7/3)/(d*x + c)^(4/3), x)