3.3141 \(\int \frac {(a+b x)^m (c+d x)^{2-m}}{b c+a d+2 b d x} \, dx\)

Optimal. Leaf size=231 \[ -\frac {\left (2 m^2-4 m+1\right ) (b c-a d) (a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,m+1;m+2;-\frac {d (a+b x)}{b c-a d}\right )}{8 b^3 m (m+1)}+\frac {(b c-a d)^2 (a+b x)^m (c+d x)^{-m} \, _2F_1\left (1,-m;1-m;-\frac {b (c+d x)}{d (a+b x)}\right )}{8 b^3 d m}+\frac {(2 m+1) (b c-a d) (a+b x)^{m+1} (c+d x)^{-m}}{8 b^3 m}+\frac {d (a+b x)^{m+2} (c+d x)^{-m}}{4 b^3} \]

[Out]

1/8*(-a*d+b*c)*(1+2*m)*(b*x+a)^(1+m)/b^3/m/((d*x+c)^m)+1/4*d*(b*x+a)^(2+m)/b^3/((d*x+c)^m)+1/8*(-a*d+b*c)^2*(b
*x+a)^m*hypergeom([1, -m],[1-m],-b*(d*x+c)/d/(b*x+a))/b^3/d/m/((d*x+c)^m)-1/8*(-a*d+b*c)*(2*m^2-4*m+1)*(b*x+a)
^(1+m)*(b*(d*x+c)/(-a*d+b*c))^m*hypergeom([m, 1+m],[2+m],-d*(b*x+a)/(-a*d+b*c))/b^3/m/(1+m)/((d*x+c)^m)

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Rubi [A]  time = 0.27, antiderivative size = 314, normalized size of antiderivative = 1.36, number of steps used = 10, number of rules used = 4, integrand size = 33, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.121, Rules used = {105, 70, 69, 131} \[ -\frac {(b c-a d)^2 (a+b x)^m (c+d x)^{-m} \, _2F_1\left (1,m;m+1;-\frac {d (a+b x)}{b (c+d x)}\right )}{8 b^3 d m}+\frac {(b c-a d) (a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m-1,m+1;m+2;-\frac {d (a+b x)}{b c-a d}\right )}{2 b^3 (m+1)}+\frac {(b c-a d)^2 (a+b x)^m (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,m;m+1;-\frac {d (a+b x)}{b c-a d}\right )}{8 b^3 d m}+\frac {(b c-a d) (a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,m+1;m+2;-\frac {d (a+b x)}{b c-a d}\right )}{4 b^3 (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((b*c - a*d)^2*(a + b*x)^m*Hypergeometric2F1[1, m, 1 + m, -((d*(a + b*x))/(b*(c + d*x)))])/(8*b^3*d*m*(c + d*
x)^m) + ((b*c - a*d)*(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))])/(2*b^3*(1 + m)*(c + d*x)^m) + ((b*c - a*d)^2*(a + b*x)^m*((b*(c + d*x))/(b*c - a
*d))^m*Hypergeometric2F1[m, m, 1 + m, -((d*(a + b*x))/(b*c - a*d))])/(8*b^3*d*m*(c + d*x)^m) + ((b*c - a*d)*(a
+ b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*Hypergeometric2F1[m, 1 + m, 2 + m, -((d*(a + b*x))/(b*c - a*d))]
)/(4*b^3*(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 105

Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> Dist[b/f, Int[(a
+ b*x)^(m - 1)*(c + d*x)^n, x], x] - Dist[(b*e - a*f)/f, Int[((a + b*x)^(m - 1)*(c + d*x)^n)/(e + f*x), x], x]
/; FreeQ[{a, b, c, d, e, f, m, n}, x] && IGtQ[Simplify[m + n + 1], 0] && (GtQ[m, 0] || ( !RationalQ[m] && (Su
mSimplerQ[m, -1] ||  !SumSimplerQ[n, -1])))

Rule 131

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[((b*c -
a*d)^n*(a + b*x)^(m + 1)*Hypergeometric2F1[m + 1, -n, m + 2, -(((d*e - c*f)*(a + b*x))/((b*c - a*d)*(e + f*x))
)])/((m + 1)*(b*e - a*f)^(n + 1)*(e + f*x)^(m + 1)), x] /; FreeQ[{a, b, c, d, e, f, m, p}, x] && EqQ[m + n + p
+ 2, 0] && ILtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {(a+b x)^m (c+d x)^{2-m}}{b c+a d+2 b d x} \, dx &=\frac {\int (a+b x)^m (c+d x)^{1-m} \, dx}{2 b}+\frac {(b c-a d) \int \frac {(a+b x)^m (c+d x)^{1-m}}{b c+a d+2 b d x} \, dx}{2 b}\\ &=\frac {(b c-a d) \int (a+b x)^m (c+d x)^{-m} \, dx}{4 b^2}+\frac {(b c-a d)^2 \int \frac {(a+b x)^m (c+d x)^{-m}}{b c+a d+2 b d x} \, dx}{4 b^2}+\frac {\left ((b c-a d) (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}{2 b^2}\\ &=\frac {(b c-a d) (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 )}{2 b^3 (1+m)}+\frac {(b c-a d)^2 \int (a+b x)^{-1+m} (c+d x)^{-m} \, dx}{8 b^2 d}-\frac {(b c-a d)^3 \int \frac {(a+b x)^{-1+m} (c+d x)^{-m}}{b c+a d+2 b d x} \, dx}{8 b^2 d}+\frac {\left ((b c-a d) (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 )^{-m} \, dx}{4 b^2}\\ &=-\frac {(b c-a d)^2 (a+b x)^m (c+d x)^{-m} \, _2F_1\left (1,m;1+m;-\frac {d (a+b x)}{b (c+d x)}\right )}{8 b^3 d m}+\frac {(b c-a d) (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 )}{2 b^3 (1+m)}+\frac {(b c-a d) (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{4 b^3 (1+m)}+\frac {\left ((b c-a d)^2 (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m\right ) \int (a+b x)^{-1+m} \left (\frac {b c}{b c-a d}+\frac {b d x}{b c-a d}\right )^{-m} \, dx}{8 b^2 d}\\ &=-\frac {(b c-a d)^2 (a+b x)^m (c+d x)^{-m} \, _2F_1\left (1,m;1+m;-\frac {d (a+b x)}{b (c+d x)}\right )}{8 b^3 d m}+\frac {(b c-a d) (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 )}{2 b^3 (1+m)}+\frac {(b c-a d)^2 (a+b x)^m (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,m;1+m;-\frac {d (a+b x)}{b c-a d}\right )}{8 b^3 d m}+\frac {(b c-a d) (a+b x)^{1+m} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{4 b^3 (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 243, normalized size = 1.05 \[ \frac {(a+b x)^m (c+d x)^{-m} \left (-4 d m (a+b x) (a d-b c) \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m-1,m+1;m+2;\frac {d (a+b x)}{a d-b c}\right )-(b c-a d) \left ((m+1) (b c-a d) \left (\, _2F_1\left (1,m;m+1;-\frac {d (a+b x)}{b (c+d x)}\right )-\left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,m;m+1;\frac {d (a+b x)}{a d-b c}\right )\right )-2 d m (a+b x) \left (\frac {b (c+d x)}{b c-a d}\right )^m \, _2F_1\left (m,m+1;m+2;\frac {d (a+b x)}{a d-b c}\right )\right )\right )}{8 b^3 d m (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^m*(-4*d*(-(b*c) + a*d)*m*(a + b*x)*((b*(c + d*x))/(b*c - a*d))^m*Hypergeometric2F1[-1 + m, 1 + m, 2
+ m, (d*(a + b*x))/(-(b*c) + a*d)] - (b*c - a*d)*((b*c - a*d)*(1 + m)*(Hypergeometric2F1[1, m, 1 + m, -((d*(a
+ b*x))/(b*(c + d*x)))] - ((b*(c + d*x))/(b*c - a*d))^m*Hypergeometric2F1[m, m, 1 + m, (d*(a + b*x))/(-(b*c)
+ a*d)]) - 2*d*m*(a + b*x)*((b*(c + d*x))/(b*c - a*d))^m*Hypergeometric2F1[m, 1 + m, 2 + m, (d*(a + b*x))/(-(b
*c) + a*d)])))/(8*b^3*d*m*(1 + m)*(c + d*x)^m)

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fricas [F]  time = 1.11, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2}}{2 \, b d x + b c + a d}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*x + a)^m*(d*x + c)^(-m + 2)/(2*b*d*x + b*c + a*d), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2}}{2 \, b d x + b c + a d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.24, size = 0, normalized size = 0.00 \[ \int \frac {\left (b x +a \right )^{m} \left (d x +c \right )^{-m +2}}{2 b d x +a d +b c}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 2}}{2 \, b d x + b c + a d}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^{2-m}}{a\,d+b\,c+2\,b\,d\,x} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Exception raised: HeuristicGCDFailed

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