\(\int \frac {1}{(a+b x)^{4/3} (c+d x)^{2/3}} \, dx\) [1604]

Optimal result

Integrand size = 19, antiderivative size = 30 \[ \int \frac {1}{(a+b x)^{4/3} (c+d x)^{2/3}} \, dx=-\frac {3 \sqrt [3]{c+d x}}{(b c-a d) \sqrt [3]{a+b x}} \]

[Out]

-3*(d*x+c)^(1/3)/(-a*d+b*c)/(b*x+a)^(1/3)

Rubi [A] (verified)

Time = 0.00 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00, number of steps used = 1, number of rules used = 1, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.053, Rules used = {37} \[ \int \frac {1}{(a+b x)^{4/3} (c+d x)^{2/3}} \, dx=-\frac {3 \sqrt [3]{c+d x}}{\sqrt [3]{a+b x} (b c-a d)} \]

[In]

Int[1/((a + b*x)^(4/3)*(c + d*x)^(2/3)),x]

[Out]

(-3*(c + d*x)^(1/3))/((b*c - a*d)*(a + b*x)^(1/3))

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(a + b*x)^(m + 1)*((c + d*x)^(n +
1)/((b*c - a*d)*(m + 1))), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rubi steps \begin{align*} \text {integral}& = -\frac {3 \sqrt [3]{c+d x}}{(b c-a d) \sqrt [3]{a+b x}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.02 (sec) , antiderivative size = 30, normalized size of antiderivative = 1.00 \[ \int \frac {1}{(a+b x)^{4/3} (c+d x)^{2/3}} \, dx=-\frac {3 \sqrt [3]{c+d x}}{(b c-a d) \sqrt [3]{a+b x}} \]

[In]

Integrate[1/((a + b*x)^(4/3)*(c + d*x)^(2/3)),x]

[Out]

(-3*(c + d*x)^(1/3))/((b*c - a*d)*(a + b*x)^(1/3))

Maple [A] (verified)

Time = 0.31 (sec) , antiderivative size = 27, normalized size of antiderivative = 0.90

method	result	size
gosper	\(\frac {3 \left (d x +c \right )^{\frac {1}{3}}}{\left (b x +a \right )^{\frac {1}{3}} \left (a d -b c \right )}\)	\(27\)

[In]

int(1/(b*x+a)^(4/3)/(d*x+c)^(2/3),x,method=_RETURNVERBOSE)

[Out]

3/(b*x+a)^(1/3)*(d*x+c)^(1/3)/(a*d-b*c)

Fricas [A] (verification not implemented)

none

Time = 0.22 (sec) , antiderivative size = 42, normalized size of antiderivative = 1.40 \[ \int \frac {1}{(a+b x)^{4/3} (c+d x)^{2/3}} \, dx=-\frac {3 \, {\left (b x + a\right )}^{\frac {2}{3}} {\left (d x + c\right )}^{\frac {1}{3}}}{a b c - a^{2} d + {\left (b^{2} c - a b d\right )} x} \]

[In]

integrate(1/(b*x+a)^(4/3)/(d*x+c)^(2/3),x, algorithm="fricas")

[Out]

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

Sympy [F]

\[ \int \frac {1}{(a+b x)^{4/3} (c+d x)^{2/3}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {4}{3}} \left (c + d x\right )^{\frac {2}{3}}}\, dx \]

[In]

integrate(1/(b*x+a)**(4/3)/(d*x+c)**(2/3),x)

[Out]

Integral(1/((a + b*x)**(4/3)*(c + d*x)**(2/3)), x)

Maxima [F]

\[ \int \frac {1}{(a+b x)^{4/3} (c+d x)^{2/3}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {4}{3}} {\left (d x + c\right )}^{\frac {2}{3}}} \,d x } \]

[In]

integrate(1/(b*x+a)^(4/3)/(d*x+c)^(2/3),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(4/3)*(d*x + c)^(2/3)), x)

Giac [F]

\[ \int \frac {1}{(a+b x)^{4/3} (c+d x)^{2/3}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {4}{3}} {\left (d x + c\right )}^{\frac {2}{3}}} \,d x } \]

[In]

integrate(1/(b*x+a)^(4/3)/(d*x+c)^(2/3),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(4/3)*(d*x + c)^(2/3)), x)

Mupad [B] (verification not implemented)

Time = 0.68 (sec) , antiderivative size = 26, normalized size of antiderivative = 0.87 \[ \int \frac {1}{(a+b x)^{4/3} (c+d x)^{2/3}} \, dx=\frac {3\,{\left (c+d\,x\right )}^{1/3}}{\left (a\,d-b\,c\right )\,{\left (a+b\,x\right )}^{1/3}} \]

[In]

int(1/((a + b*x)^(4/3)*(c + d*x)^(2/3)),x)

[Out]

(3*(c + d*x)^(1/3))/((a*d - b*c)*(a + b*x)^(1/3))