3.3061 \(\int (a+b x)^m (c+d x)^{-m} (e+f x)^3 \, dx\)

Optimal. Leaf size=432 \[ -\frac{(a+b x)^{m+1} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \left (-3 a^2 b d^2 f^2 \left (m^2-3 m+2\right ) (4 d e-c f (m+1))+a^3 d^3 f^3 \left (-m^3+6 m^2-11 m+6\right )+3 a b^2 d f (1-m) \left (c^2 f^2 \left (m^2+3 m+2\right )-8 c d e f (m+1)+12 d^2 e^2\right )+b^3 \left (-\left (12 c^2 d e f^2 \left (m^2+3 m+2\right )-c^3 f^3 \left (m^3+6 m^2+11 m+6\right )-36 c d^2 e^2 f (m+1)+24 d^3 e^3\right )\right )\right ) \, _2F_1\left (m,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{24 b^4 d^3 (m+1)}+\frac{f (a+b x)^{m+1} (c+d x)^{1-m} \left (a^2 d^2 f^2 \left (m^2-5 m+6\right )-2 a b d f \left (6 d e (2-m)-c f \left (3-m^2\right )\right )-2 b d f x (a d f (3-m)-b (6 d e-c f (m+3)))+b^2 \left (c^2 f^2 \left (m^2+5 m+6\right )-12 c d e f (m+2)+30 d^2 e^2\right )\right )}{24 b^3 d^3}+\frac{f (e+f x)^2 (a+b x)^{m+1} (c+d x)^{1-m}}{4 b d} \]

[Out]

(f*(a + b*x)^(1 + m)*(c + d*x)^(1 - m)*(e + f*x)^2)/(4*b*d) + (f*(a + b*x)^(1 + m)*(c + d*x)^(1 - m)*(a^2*d^2*
f^2*(6 - 5*m + m^2) - 2*a*b*d*f*(6*d*e*(2 - m) - c*f*(3 - m^2)) + b^2*(30*d^2*e^2 - 12*c*d*e*f*(2 + m) + c^2*f
^2*(6 + 5*m + m^2)) - 2*b*d*f*(a*d*f*(3 - m) - b*(6*d*e - c*f*(3 + m)))*x))/(24*b^3*d^3) - ((a^3*d^3*f^3*(6 -
11*m + 6*m^2 - m^3) - 3*a^2*b*d^2*f^2*(2 - 3*m + m^2)*(4*d*e - c*f*(1 + m)) + 3*a*b^2*d*f*(1 - m)*(12*d^2*e^2
- 8*c*d*e*f*(1 + m) + c^2*f^2*(2 + 3*m + m^2)) - b^3*(24*d^3*e^3 - 36*c*d^2*e^2*f*(1 + m) + 12*c^2*d*e*f^2*(2
+ 3*m + m^2) - c^3*f^3*(6 + 11*m + 6*m^2 + m^3)))*(a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*Hypergeometr
ic2F1[m, 1 + m, 2 + m, -((d*(a + b*x))/(b*c - a*d))])/(24*b^4*d^3*(1 + m)*(c + d*x)^m)

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Rubi [A]  time = 0.486399, antiderivative size = 431, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {100, 147, 70, 69} \[ -\frac{(a+b x)^{m+1} (c+d x)^{-m} \left (\frac{b (c+d x)}{b c-a d}\right )^m \left (-3 a^2 b d^2 f^2 \left (m^2-3 m+2\right ) (4 d e-c f (m+1))+a^3 d^3 f^3 \left (-m^3+6 m^2-11 m+6\right )+3 a b^2 d f (1-m) \left (c^2 f^2 \left (m^2+3 m+2\right )-8 c d e f (m+1)+12 d^2 e^2\right )+b^3 \left (-\left (12 c^2 d e f^2 \left (m^2+3 m+2\right )-c^3 f^3 \left (m^3+6 m^2+11 m+6\right )-36 c d^2 e^2 f (m+1)+24 d^3 e^3\right )\right )\right ) \, _2F_1\left (m,m+1;m+2;-\frac{d (a+b x)}{b c-a d}\right )}{24 b^4 d^3 (m+1)}+\frac{f (a+b x)^{m+1} (c+d x)^{1-m} \left (a^2 d^2 f^2 \left (m^2-5 m+6\right )-2 a b d f \left (6 d e (2-m)-c f \left (3-m^2\right )\right )+2 b d f x (-a d f (3-m)-b c f (m+3)+6 b d e)+b^2 \left (c^2 f^2 \left (m^2+5 m+6\right )-12 c d e f (m+2)+30 d^2 e^2\right )\right )}{24 b^3 d^3}+\frac{f (e+f x)^2 (a+b x)^{m+1} (c+d x)^{1-m}}{4 b d} \]

Antiderivative was successfully verified.

[In]

Int[((a + b*x)^m*(e + f*x)^3)/(c + d*x)^m,x]

[Out]

(f*(a + b*x)^(1 + m)*(c + d*x)^(1 - m)*(e + f*x)^2)/(4*b*d) + (f*(a + b*x)^(1 + m)*(c + d*x)^(1 - m)*(a^2*d^2*
f^2*(6 - 5*m + m^2) - 2*a*b*d*f*(6*d*e*(2 - m) - c*f*(3 - m^2)) + b^2*(30*d^2*e^2 - 12*c*d*e*f*(2 + m) + c^2*f
^2*(6 + 5*m + m^2)) + 2*b*d*f*(6*b*d*e - a*d*f*(3 - m) - b*c*f*(3 + m))*x))/(24*b^3*d^3) - ((a^3*d^3*f^3*(6 -
11*m + 6*m^2 - m^3) - 3*a^2*b*d^2*f^2*(2 - 3*m + m^2)*(4*d*e - c*f*(1 + m)) + 3*a*b^2*d*f*(1 - m)*(12*d^2*e^2
- 8*c*d*e*f*(1 + m) + c^2*f^2*(2 + 3*m + m^2)) - b^3*(24*d^3*e^3 - 36*c*d^2*e^2*f*(1 + m) + 12*c^2*d*e*f^2*(2
+ 3*m + m^2) - c^3*f^3*(6 + 11*m + 6*m^2 + m^3)))*(a + b*x)^(1 + m)*((b*(c + d*x))/(b*c - a*d))^m*Hypergeometr
ic2F1[m, 1 + m, 2 + m, -((d*(a + b*x))/(b*c - a*d))])/(24*b^4*d^3*(1 + m)*(c + d*x)^m)

Rule 100

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

Rule 147

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_) + (f_.)*(x_))*((g_.) + (h_.)*(x_)), x_Symbol]
:> -Simp[((a*d*f*h*(n + 2) + b*c*f*h*(m + 2) - b*d*(f*g + e*h)*(m + n + 3) - b*d*f*h*(m + n + 2)*x)*(a + b*x)^
(m + 1)*(c + d*x)^(n + 1))/(b^2*d^2*(m + n + 2)*(m + n + 3)), x] + Dist[(a^2*d^2*f*h*(n + 1)*(n + 2) + a*b*d*(
n + 1)*(2*c*f*h*(m + 1) - d*(f*g + e*h)*(m + n + 3)) + b^2*(c^2*f*h*(m + 1)*(m + 2) - c*d*(f*g + e*h)*(m + 1)*
(m + n + 3) + d^2*e*g*(m + n + 2)*(m + n + 3)))/(b^2*d^2*(m + n + 2)*(m + n + 3)), Int[(a + b*x)^m*(c + d*x)^n
, x], x] /; FreeQ[{a, b, c, d, e, f, g, h, m, n}, x] && NeQ[m + n + 2, 0] && NeQ[m + n + 3, 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 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]))

Rubi steps

\begin{align*} \int (a+b x)^m (c+d x)^{-m} (e+f x)^3 \, dx &=\frac{f (a+b x)^{1+m} (c+d x)^{1-m} (e+f x)^2}{4 b d}+\frac{\int (a+b x)^m (c+d x)^{-m} (e+f x) (-a f (2 c f+d e (1-m))+b e (4 d e-c f (1+m))+f (6 b d e-a d f (3-m)-b c f (3+m)) x) \, dx}{4 b d}\\ &=\frac{f (a+b x)^{1+m} (c+d x)^{1-m} (e+f x)^2}{4 b d}+\frac{f (a+b x)^{1+m} (c+d x)^{1-m} \left (a^2 d^2 f^2 \left (6-5 m+m^2\right )-2 a b d f \left (6 d e (2-m)-c f \left (3-m^2\right )\right )+b^2 \left (30 d^2 e^2-12 c d e f (2+m)+c^2 f^2 \left (6+5 m+m^2\right )\right )+2 b d f (6 b d e-a d f (3-m)-b c f (3+m)) x\right )}{24 b^3 d^3}-\frac{\left (a^3 d^3 f^3 \left (6-11 m+6 m^2-m^3\right )-3 a^2 b d^2 f^2 \left (2-3 m+m^2\right ) (4 d e-c f (1+m))+3 a b^2 d f (1-m) \left (12 d^2 e^2-8 c d e f (1+m)+c^2 f^2 \left (2+3 m+m^2\right )\right )-b^3 \left (24 d^3 e^3-36 c d^2 e^2 f (1+m)+12 c^2 d e f^2 \left (2+3 m+m^2\right )-c^3 f^3 \left (6+11 m+6 m^2+m^3\right )\right )\right ) \int (a+b x)^m (c+d x)^{-m} \, dx}{24 b^3 d^3}\\ &=\frac{f (a+b x)^{1+m} (c+d x)^{1-m} (e+f x)^2}{4 b d}+\frac{f (a+b x)^{1+m} (c+d x)^{1-m} \left (a^2 d^2 f^2 \left (6-5 m+m^2\right )-2 a b d f \left (6 d e (2-m)-c f \left (3-m^2\right )\right )+b^2 \left (30 d^2 e^2-12 c d e f (2+m)+c^2 f^2 \left (6+5 m+m^2\right )\right )+2 b d f (6 b d e-a d f (3-m)-b c f (3+m)) x\right )}{24 b^3 d^3}-\frac{\left (\left (a^3 d^3 f^3 \left (6-11 m+6 m^2-m^3\right )-3 a^2 b d^2 f^2 \left (2-3 m+m^2\right ) (4 d e-c f (1+m))+3 a b^2 d f (1-m) \left (12 d^2 e^2-8 c d e f (1+m)+c^2 f^2 \left (2+3 m+m^2\right )\right )-b^3 \left (24 d^3 e^3-36 c d^2 e^2 f (1+m)+12 c^2 d e f^2 \left (2+3 m+m^2\right )-c^3 f^3 \left (6+11 m+6 m^2+m^3\right )\right )\right ) (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}{24 b^3 d^3}\\ &=\frac{f (a+b x)^{1+m} (c+d x)^{1-m} (e+f x)^2}{4 b d}+\frac{f (a+b x)^{1+m} (c+d x)^{1-m} \left (a^2 d^2 f^2 \left (6-5 m+m^2\right )-2 a b d f \left (6 d e (2-m)-c f \left (3-m^2\right )\right )+b^2 \left (30 d^2 e^2-12 c d e f (2+m)+c^2 f^2 \left (6+5 m+m^2\right )\right )+2 b d f (6 b d e-a d f (3-m)-b c f (3+m)) x\right )}{24 b^3 d^3}-\frac{\left (a^3 d^3 f^3 \left (6-11 m+6 m^2-m^3\right )-3 a^2 b d^2 f^2 \left (2-3 m+m^2\right ) (4 d e-c f (1+m))+3 a b^2 d f (1-m) \left (12 d^2 e^2-8 c d e f (1+m)+c^2 f^2 \left (2+3 m+m^2\right )\right )-b^3 \left (24 d^3 e^3-36 c d^2 e^2 f (1+m)+12 c^2 d e f^2 \left (2+3 m+m^2\right )-c^3 f^3 \left (6+11 m+6 m^2+m^3\right )\right )\right ) (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 )}{24 b^4 d^3 (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.782234, size = 304, normalized size = 0.7 \[ \frac{(a+b x)^{m+1} (c+d x)^{-m} \left (b^2 (d e-c f)^2 \left (\frac{b (c+d x)}{b c-a d}\right )^m (a d f (m-1)-b c f (m+3)+4 b d e) \, _2F_1\left (m,m+1;m+2;\frac{d (a+b x)}{a d-b c}\right )+f^2 (b c-a d)^2 \left (\frac{b (c+d x)}{b c-a d}\right )^m (a d f (m-3)-b c f (m+3)+6 b d e) \, _2F_1\left (m-2,m+1;m+2;\frac{d (a+b x)}{a d-b c}\right )+2 b f (b c-a d) (d e-c f) \left (\frac{b (c+d x)}{b c-a d}\right )^m (a d f (m-2)-b c f (m+3)+5 b d e) \, _2F_1\left (m-1,m+1;m+2;\frac{d (a+b x)}{a d-b c}\right )+b^3 d^2 f (m+1) (c+d x) (e+f x)^2\right )}{4 b^4 d^3 (m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + b*x)^m*(e + f*x)^3)/(c + d*x)^m,x]

[Out]

((a + b*x)^(1 + m)*(b^3*d^2*f*(1 + m)*(c + d*x)*(e + f*x)^2 + (b*c - a*d)^2*f^2*(6*b*d*e + a*d*f*(-3 + m) - b*
c*f*(3 + m))*((b*(c + d*x))/(b*c - a*d))^m*Hypergeometric2F1[-2 + m, 1 + m, 2 + m, (d*(a + b*x))/(-(b*c) + a*d
)] + 2*b*(b*c - a*d)*f*(d*e - c*f)*(5*b*d*e + a*d*f*(-2 + m) - b*c*f*(3 + m))*((b*(c + d*x))/(b*c - a*d))^m*Hy
pergeometric2F1[-1 + m, 1 + m, 2 + m, (d*(a + b*x))/(-(b*c) + a*d)] + b^2*(d*e - c*f)^2*(4*b*d*e + a*d*f*(-1 +
m) - b*c*f*(3 + 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^4*d^3*(1 + m)*(c + d*x)^m)

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Maple [F]  time = 0.069, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( fx+e \right ) ^{3} \left ( bx+a \right ) ^{m}}{ \left ( dx+c \right ) ^{m}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m*(f*x+e)^3/((d*x+c)^m),x)

[Out]

int((b*x+a)^m*(f*x+e)^3/((d*x+c)^m),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{3}{\left (b x + a\right )}^{m}}{{\left (d x + c\right )}^{m}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(f*x+e)^3/((d*x+c)^m),x, algorithm="maxima")

[Out]

integrate((f*x + e)^3*(b*x + a)^m/(d*x + c)^m, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (f^{3} x^{3} + 3 \, e f^{2} x^{2} + 3 \, e^{2} f x + e^{3}\right )}{\left (b x + a\right )}^{m}}{{\left (d x + c\right )}^{m}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(f*x+e)^3/((d*x+c)^m),x, algorithm="fricas")

[Out]

integral((f^3*x^3 + 3*e*f^2*x^2 + 3*e^2*f*x + e^3)*(b*x + a)^m/(d*x + c)^m, x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**m*(f*x+e)**3/((d*x+c)**m),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (f x + e\right )}^{3}{\left (b x + a\right )}^{m}}{{\left (d x + c\right )}^{m}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(f*x+e)^3/((d*x+c)^m),x, algorithm="giac")

[Out]

integrate((f*x + e)^3*(b*x + a)^m/(d*x + c)^m, x)