\(\int (a+b x)^{\frac {-2 b c+a d}{b c-a d}} (c+d x)^{\frac {b c-2 a d}{-b c+a d}} \, dx\) [1884]

   Optimal result
   Rubi [A] (verified)
   Mathematica [A] (verified)
   Maple [A] (verified)
   Fricas [A] (verification not implemented)
   Sympy [F(-1)]
   Maxima [F]
   Giac [F]
   Mupad [B] (verification not implemented)

Optimal result

Integrand size = 51, antiderivative size = 97 \[ \int (a+b x)^{\frac {-2 b c+a d}{b c-a d}} (c+d x)^{\frac {b c-2 a d}{-b c+a d}} \, dx=-\frac {(a+b x)^{-\frac {b c}{b c-a d}} (c+d x)^{\frac {a d}{b c-a d}}}{b c}+\frac {(a+b x)^{-\frac {a d}{b c-a d}} (c+d x)^{\frac {a d}{b c-a d}}}{a b c} \]

[Out]

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

Rubi [A] (verified)

Time = 0.01 (sec) , antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.039, Rules used = {47, 37} \[ \int (a+b x)^{\frac {-2 b c+a d}{b c-a d}} (c+d x)^{\frac {b c-2 a d}{-b c+a d}} \, dx=\frac {(a+b x)^{-\frac {a d}{b c-a d}} (c+d x)^{\frac {a d}{b c-a d}}}{a b c}-\frac {(a+b x)^{-\frac {b c}{b c-a d}} (c+d x)^{\frac {a d}{b c-a d}}}{b c} \]

[In]

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

[Out]

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

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 {(a+b x)^{-\frac {b c}{b c-a d}} (c+d x)^{\frac {a d}{b c-a d}}}{b c}-\frac {d \int (a+b x)^{\frac {b c}{-b c+a d}} (c+d x)^{\frac {b c-2 a d}{-b c+a d}} \, dx}{b c} \\ & = -\frac {(a+b x)^{-\frac {b c}{b c-a d}} (c+d x)^{\frac {a d}{b c-a d}}}{b c}+\frac {(a+b x)^{-\frac {a d}{b c-a d}} (c+d x)^{\frac {a d}{b c-a d}}}{a b c} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.13 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.47 \[ \int (a+b x)^{\frac {-2 b c+a d}{b c-a d}} (c+d x)^{\frac {b c-2 a d}{-b c+a d}} \, dx=\frac {x (a+b x)^{\frac {b c}{-b c+a d}} (c+d x)^{\frac {a d}{b c-a d}}}{a c} \]

[In]

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

[Out]

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

Maple [A] (verified)

Time = 0.61 (sec) , antiderivative size = 66, normalized size of antiderivative = 0.68

method result size
gosper \(\frac {x \left (b x +a \right )^{1-\frac {a d -2 b c}{a d -b c}} \left (d x +c \right )^{1-\frac {2 a d -b c}{a d -b c}}}{a c}\) \(66\)
parallelrisch \(\frac {x^{3} \left (b x +a \right )^{-\frac {a d -2 b c}{a d -b c}} \left (d x +c \right )^{-\frac {2 a d -b c}{a d -b c}} b^{2} d^{2}+x^{2} \left (b x +a \right )^{-\frac {a d -2 b c}{a d -b c}} \left (d x +c \right )^{-\frac {2 a d -b c}{a d -b c}} a b \,d^{2}+x^{2} \left (b x +a \right )^{-\frac {a d -2 b c}{a d -b c}} \left (d x +c \right )^{-\frac {2 a d -b c}{a d -b c}} b^{2} c d +x \left (b x +a \right )^{-\frac {a d -2 b c}{a d -b c}} \left (d x +c \right )^{-\frac {2 a d -b c}{a d -b c}} a b c d}{a b c d}\) \(261\)

[In]

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

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.24 (sec) , antiderivative size = 84, normalized size of antiderivative = 0.87 \[ \int (a+b x)^{\frac {-2 b c+a d}{b c-a d}} (c+d x)^{\frac {b c-2 a d}{-b c+a d}} \, dx=\frac {b d x^{3} + a c x + {\left (b c + a d\right )} x^{2}}{{\left (b x + a\right )}^{\frac {2 \, b c - a d}{b c - a d}} {\left (d x + c\right )}^{\frac {b c - 2 \, a d}{b c - a d}} a c} \]

[In]

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

[Out]

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

Sympy [F(-1)]

Timed out. \[ \int (a+b x)^{\frac {-2 b c+a d}{b c-a d}} (c+d x)^{\frac {b c-2 a d}{-b c+a d}} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 0.90 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.46 \[ \int (a+b x)^{\frac {-2 b c+a d}{b c-a d}} (c+d x)^{\frac {b c-2 a d}{-b c+a d}} \, dx=\frac {\frac {x}{{\left (a+b\,x\right )}^{\frac {a\,d-2\,b\,c}{a\,d-b\,c}}}+\frac {x^2\,\left (a\,d+b\,c\right )}{a\,c\,{\left (a+b\,x\right )}^{\frac {a\,d-2\,b\,c}{a\,d-b\,c}}}+\frac {b\,d\,x^3}{a\,c\,{\left (a+b\,x\right )}^{\frac {a\,d-2\,b\,c}{a\,d-b\,c}}}}{{\left (c+d\,x\right )}^{\frac {2\,a\,d-b\,c}{a\,d-b\,c}}} \]

[In]

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

[Out]

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