3.16.75 \(\int (a+b x)^{-1-\frac {b c}{b c-a d}} (c+d x)^{-1+\frac {a d}{b c-a d}} \, dx\)

Optimal. Leaf size=97 \[ \frac {(a+b x)^{-\frac {a d}{b c-a d}} (c+d x)^{\frac {a d}{b c-a d}}}{a b c}-\frac {(a+b x)^{-\frac {b c}{b c-a d}} (c+d x)^{\frac {a d}{b c-a d}}}{b c} \]

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Rubi [A]  time = 0.02, antiderivative size = 97, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 44, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.045, Rules used = {45, 37} \begin {gather*} \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} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^(-1 - (b*c)/(b*c - a*d))*(c + d*x)^(-1 + (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 45

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*} \int (a+b x)^{-1-\frac {b c}{b c-a d}} (c+d x)^{-1+\frac {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 {d \int (a+b x)^{\frac {b c}{-b c+a d}} (c+d x)^{-1+\frac {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*}

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Mathematica [A]  time = 0.04, size = 46, normalized size = 0.47 \begin {gather*} \frac {x (a+b x)^{\frac {b c}{a d-b c}} (c+d x)^{\frac {a d}{b c-a d}}}{a c} \end {gather*}

Antiderivative was successfully verified.

[In]

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

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IntegrateAlgebraic [F]  time = 0.37, size = 0, normalized size = 0.00 \begin {gather*} \int (a+b x)^{-1-\frac {b c}{b c-a d}} (c+d x)^{-1+\frac {a d}{b c-a d}} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

IntegrateAlgebraic[(a + b*x)^(-1 - (b*c)/(b*c - a*d))*(c + d*x)^(-1 + (a*d)/(b*c - a*d)),x]

[Out]

Defer[IntegrateAlgebraic][(a + b*x)^(-1 - (b*c)/(b*c - a*d))*(c + d*x)^(-1 + (a*d)/(b*c - a*d)), x]

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fricas [A]  time = 1.39, size = 84, normalized size = 0.87 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (b x + a\right )}^{-\frac {b c}{b c - a d} - 1} {\left (d x + c\right )}^{\frac {a d}{b c - a d} - 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.00, size = 66, normalized size = 0.68 \begin {gather*} \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} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(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))/a/c*x

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int {\left (b x + a\right )}^{-\frac {b c}{b c - a d} - 1} {\left (d x + c\right )}^{\frac {a d}{b c - a d} - 1}\,{d x} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 2.14, size = 119, normalized size = 1.23 \begin {gather*} \frac {x\,{\left (a+b\,x\right )}^{\frac {b\,c}{a\,d-b\,c}-1}+\frac {x^2\,\left (a\,d+b\,c\right )\,{\left (a+b\,x\right )}^{\frac {b\,c}{a\,d-b\,c}-1}}{a\,c}+\frac {b\,d\,x^3\,{\left (a+b\,x\right )}^{\frac {b\,c}{a\,d-b\,c}-1}}{a\,c}}{{\left (c+d\,x\right )}^{\frac {a\,d}{a\,d-b\,c}+1}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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