\(\int \frac {1}{(a+b x)^{7/6} (c+d x)^{11/6}} \, dx\) [1835]

Optimal result

Integrand size = 19, antiderivative size = 64 \[ \int \frac {1}{(a+b x)^{7/6} (c+d x)^{11/6}} \, dx=-\frac {6}{(b c-a d) \sqrt [6]{a+b x} (c+d x)^{5/6}}-\frac {36 d (a+b x)^{5/6}}{5 (b c-a d)^2 (c+d x)^{5/6}} \]

[Out]

-6/(-a*d+b*c)/(b*x+a)^(1/6)/(d*x+c)^(5/6)-36/5*d*(b*x+a)^(5/6)/(-a*d+b*c)^2/(d*x+c)^(5/6)

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {47, 37} \[ \int \frac {1}{(a+b x)^{7/6} (c+d x)^{11/6}} \, dx=-\frac {36 d (a+b x)^{5/6}}{5 (c+d x)^{5/6} (b c-a d)^2}-\frac {6}{\sqrt [6]{a+b x} (c+d x)^{5/6} (b c-a d)} \]

[In]

Int[1/((a + b*x)^(7/6)*(c + d*x)^(11/6)),x]

[Out]

-6/((b*c - a*d)*(a + b*x)^(1/6)*(c + d*x)^(5/6)) - (36*d*(a + b*x)^(5/6))/(5*(b*c - a*d)^2*(c + d*x)^(5/6))

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]

Rule 47

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] - Dist[d*(Simplify[m + n + 2]/((b*c - a*d)*(m + 1))), Int[(a + b*x)^Simplify[m +
1]*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && ILtQ[Simplify[m + n + 2], 0] &&
NeQ[m, -1] &&  !(LtQ[m, -1] && LtQ[n, -1] && (EqQ[a, 0] || (NeQ[c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && (
SumSimplerQ[m, 1] ||  !SumSimplerQ[n, 1])

Rubi steps \begin{align*} \text {integral}& = -\frac {6}{(b c-a d) \sqrt [6]{a+b x} (c+d x)^{5/6}}-\frac {(6 d) \int \frac {1}{\sqrt [6]{a+b x} (c+d x)^{11/6}} \, dx}{b c-a d} \\ & = -\frac {6}{(b c-a d) \sqrt [6]{a+b x} (c+d x)^{5/6}}-\frac {36 d (a+b x)^{5/6}}{5 (b c-a d)^2 (c+d x)^{5/6}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.70 \[ \int \frac {1}{(a+b x)^{7/6} (c+d x)^{11/6}} \, dx=-\frac {6 (5 b c+a d+6 b d x)}{5 (b c-a d)^2 \sqrt [6]{a+b x} (c+d x)^{5/6}} \]

[In]

Integrate[1/((a + b*x)^(7/6)*(c + d*x)^(11/6)),x]

[Out]

(-6*(5*b*c + a*d + 6*b*d*x))/(5*(b*c - a*d)^2*(a + b*x)^(1/6)*(c + d*x)^(5/6))

Maple [A] (verified)

Time = 0.79 (sec) , antiderivative size = 53, normalized size of antiderivative = 0.83

method	result	size
gosper	\(-\frac {6 \left (6 b d x +a d +5 b c \right )}{5 \left (b x +a \right )^{\frac {1}{6}} \left (d x +c \right )^{\frac {5}{6}} \left (a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}\)	\(53\)

[In]

int(1/(b*x+a)^(7/6)/(d*x+c)^(11/6),x,method=_RETURNVERBOSE)

[Out]

-6/5*(6*b*d*x+a*d+5*b*c)/(b*x+a)^(1/6)/(d*x+c)^(5/6)/(a^2*d^2-2*a*b*c*d+b^2*c^2)

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 126 vs. \(2 (54) = 108\).

Time = 0.23 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.97 \[ \int \frac {1}{(a+b x)^{7/6} (c+d x)^{11/6}} \, dx=-\frac {6 \, {\left (6 \, b d x + 5 \, b c + a d\right )} {\left (b x + a\right )}^{\frac {5}{6}} {\left (d x + c\right )}^{\frac {1}{6}}}{5 \, {\left (a b^{2} c^{3} - 2 \, a^{2} b c^{2} d + a^{3} c d^{2} + {\left (b^{3} c^{2} d - 2 \, a b^{2} c d^{2} + a^{2} b d^{3}\right )} x^{2} + {\left (b^{3} c^{3} - a b^{2} c^{2} d - a^{2} b c d^{2} + a^{3} d^{3}\right )} x\right )}} \]

[In]

integrate(1/(b*x+a)^(7/6)/(d*x+c)^(11/6),x, algorithm="fricas")

[Out]

-6/5*(6*b*d*x + 5*b*c + a*d)*(b*x + a)^(5/6)*(d*x + c)^(1/6)/(a*b^2*c^3 - 2*a^2*b*c^2*d + a^3*c*d^2 + (b^3*c^2
*d - 2*a*b^2*c*d^2 + a^2*b*d^3)*x^2 + (b^3*c^3 - a*b^2*c^2*d - a^2*b*c*d^2 + a^3*d^3)*x)

Sympy [F]

\[ \int \frac {1}{(a+b x)^{7/6} (c+d x)^{11/6}} \, dx=\int \frac {1}{\left (a + b x\right )^{\frac {7}{6}} \left (c + d x\right )^{\frac {11}{6}}}\, dx \]

[In]

integrate(1/(b*x+a)**(7/6)/(d*x+c)**(11/6),x)

[Out]

Integral(1/((a + b*x)**(7/6)*(c + d*x)**(11/6)), x)

Maxima [F]

\[ \int \frac {1}{(a+b x)^{7/6} (c+d x)^{11/6}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {7}{6}} {\left (d x + c\right )}^{\frac {11}{6}}} \,d x } \]

[In]

integrate(1/(b*x+a)^(7/6)/(d*x+c)^(11/6),x, algorithm="maxima")

[Out]

integrate(1/((b*x + a)^(7/6)*(d*x + c)^(11/6)), x)

Giac [F]

\[ \int \frac {1}{(a+b x)^{7/6} (c+d x)^{11/6}} \, dx=\int { \frac {1}{{\left (b x + a\right )}^{\frac {7}{6}} {\left (d x + c\right )}^{\frac {11}{6}}} \,d x } \]

[In]

integrate(1/(b*x+a)^(7/6)/(d*x+c)^(11/6),x, algorithm="giac")

[Out]

integrate(1/((b*x + a)^(7/6)*(d*x + c)^(11/6)), x)

Mupad [B] (verification not implemented)

Time = 0.98 (sec) , antiderivative size = 72, normalized size of antiderivative = 1.12 \[ \int \frac {1}{(a+b x)^{7/6} (c+d x)^{11/6}} \, dx=-\frac {\left (\frac {36\,b\,x}{5\,{\left (a\,d-b\,c\right )}^2}+\frac {6\,a\,d+30\,b\,c}{5\,d\,{\left (a\,d-b\,c\right )}^2}\right )\,{\left (c+d\,x\right )}^{1/6}}{x\,{\left (a+b\,x\right )}^{1/6}+\frac {c\,{\left (a+b\,x\right )}^{1/6}}{d}} \]

[In]

int(1/((a + b*x)^(7/6)*(c + d*x)^(11/6)),x)

[Out]

-(((36*b*x)/(5*(a*d - b*c)^2) + (6*a*d + 30*b*c)/(5*d*(a*d - b*c)^2))*(c + d*x)^(1/6))/(x*(a + b*x)^(1/6) + (c
*(a + b*x)^(1/6))/d)