3.3114 \(\int (a+b x)^m (c+d x)^{1-m} (e+f x) \, dx\)

Optimal. Leaf size=145 \[ \frac {f (a+b x)^{m+1} (c+d x)^{2-m}}{3 b d}-\frac {(b c-a d) (a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m (a d f (2-m)-b (3 d e-c f (m+1))) \, _2F_1\left (m-1,m+1;m+2;-\frac {d (a+b x)}{b c-a d}\right )}{3 b^3 d (m+1)} \]

[Out]

1/3*f*(b*x+a)^(1+m)*(d*x+c)^(2-m)/b/d-1/3*(-a*d+b*c)*(a*d*f*(2-m)-b*(3*d*e-c*f*(1+m)))*(b*x+a)^(1+m)*(b*(d*x+c
)/(-a*d+b*c))^m*hypergeom([-1+m, 1+m],[2+m],-d*(b*x+a)/(-a*d+b*c))/b^3/d/(1+m)/((d*x+c)^m)

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Rubi [A]  time = 0.08, antiderivative size = 144, normalized size of antiderivative = 0.99, number of steps used = 3, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {80, 70, 69} \[ \frac {(b c-a d) (a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m (-a d f (2-m)-b c f (m+1)+3 b d e) \, _2F_1\left (m-1,m+1;m+2;-\frac {d (a+b x)}{b c-a d}\right )}{3 b^3 d (m+1)}+\frac {f (a+b x)^{m+1} (c+d x)^{2-m}}{3 b d} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^m*(c + d*x)^(1 - m)*(e + f*x),x]

[Out]

(f*(a + b*x)^(1 + m)*(c + d*x)^(2 - m))/(3*b*d) + ((b*c - a*d)*(3*b*d*e - a*d*f*(2 - m) - b*c*f*(1 + m))*(a +
b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*Hypergeometric2F1[-1 + m, 1 + m, 2 + m, -((d*(a + b*x))/(b*c - a*d)
)])/(3*b^3*d*(1 + m)*(c + d*x)^m)

Rule 69

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*Hypergeometric2F1[
-n, m + 1, m + 2, -((d*(a + b*x))/(b*c - a*d))])/(b*(m + 1)*(b/(b*c - a*d))^n), x] /; FreeQ[{a, b, c, d, m, n}
, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-(d/(b*c - a*d)), 0]))

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(c + d*x)^FracPart[n]/((b/(b*c - a*d)
)^IntPart[n]*((b*(c + d*x))/(b*c - a*d))^FracPart[n]), Int[(a + b*x)^m*Simp[(b*c)/(b*c - a*d) + (b*d*x)/(b*c -
a*d), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 80

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[(b*(c + d*x)
^(n + 1)*(e + f*x)^(p + 1))/(d*f*(n + p + 2)), x] + Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1) + c*f*(p + 1)))/(
d*f*(n + p + 2)), Int[(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && NeQ[n + p + 2,
0]

Rubi steps

\begin {align*} \int (a+b x)^m (c+d x)^{1-m} (e+f x) \, dx &=\frac {f (a+b x)^{1+m} (c+d x)^{2-m}}{3 b d}+\frac {(3 b d e-f (a d (2-m)+b c (1+m))) \int (a+b x)^m (c+d x)^{1-m} \, dx}{3 b d}\\ &=\frac {f (a+b x)^{1+m} (c+d x)^{2-m}}{3 b d}+\frac {\left ((b c-a d) (3 b d e-f (a d (2-m)+b c (1+m))) (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m\right ) \int (a+b x)^m \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{1-m} \, dx}{3 b^2 d}\\ &=\frac {f (a+b x)^{1+m} (c+d x)^{2-m}}{3 b d}+\frac {(b c-a d) (3 b d e-a d f (2-m)-b c f (1+m)) (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (-1+m,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{3 b^3 d (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.13, size = 126, normalized size = 0.87 \[ \frac {(a+b x)^{m+1} (c+d x)^{-m} \left (b^2 f (m+1) (c+d x)^2-(b c-a d) \left (\frac {b (c+d x)}{b c-a d}\right )^m (-a d f (m-2)+b c f (m+1)-3 b d e) \, _2F_1\left (m-1,m+1;m+2;\frac {d (a+b x)}{a d-b c}\right )\right )}{3 b^3 d (m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^m*(c + d*x)^(1 - m)*(e + f*x),x]

[Out]

((a + b*x)^(1 + m)*(b^2*f*(1 + m)*(c + d*x)^2 - (b*c - a*d)*(-3*b*d*e - a*d*f*(-2 + m) + b*c*f*(1 + m))*((b*(c
+ d*x))/(b*c - a*d))^m*Hypergeometric2F1[-1 + m, 1 + m, 2 + m, (d*(a + b*x))/(-(b*c) + a*d)]))/(3*b^3*d*(1 +
m)*(c + d*x)^m)

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fricas [F]  time = 1.08, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (f x + e\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(1-m)*(f*x+e),x, algorithm="fricas")

[Out]

integral((f*x + e)*(b*x + a)^m*(d*x + c)^(-m + 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (f x + e\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((f*x + e)*(b*x + a)^m*(d*x + c)^(-m + 1), x)

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maple [F]  time = 0.23, size = 0, normalized size = 0.00 \[ \int \left (f x +e \right ) \left (b x +a \right )^{m} \left (d x +c \right )^{-m +1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m*(d*x+c)^(-m+1)*(f*x+e),x)

[Out]

int((b*x+a)^m*(d*x+c)^(-m+1)*(f*x+e),x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (f x + e\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((f*x + e)*(b*x + a)^m*(d*x + c)^(-m + 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \left (e+f\,x\right )\,{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^{1-m} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e + f*x)*(a + b*x)^m*(c + d*x)^(1 - m),x)

[Out]

int((e + f*x)*(a + b*x)^m*(c + d*x)^(1 - m), x)

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sympy [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: HeuristicGCDFailed} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**m*(d*x+c)**(1-m)*(f*x+e),x)

[Out]

Exception raised: HeuristicGCDFailed

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