3.3112 \(\int (a+b x)^m (c+d x)^{1-m} (e+f x)^3 \, dx\)

Optimal. Leaf size=445 \[ \frac {f (a+b x)^{m+1} (c+d x)^{2-m} \left (a^2 d^2 f^2 \left (m^2-7 m+12\right )-a b d f \left (15 d e (3-m)-c f \left (-2 m^2+2 m+9\right )\right )-3 b d f x (a d f (4-m)-b (7 d e-c f (m+3)))+b^2 \left (c^2 f^2 \left (m^2+5 m+6\right )-15 c d e f (m+2)+48 d^2 e^2\right )\right )}{60 b^3 d^3}-\frac {(b c-a d) (a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \left (a^3 d^3 f^3 \left (-m^3+9 m^2-26 m+24\right )-3 a^2 b d^2 f^2 \left (m^2-5 m+6\right ) (5 d e-c f (m+1))+3 a b^2 d f (2-m) \left (c^2 f^2 \left (m^2+3 m+2\right )-10 c d e f (m+1)+20 d^2 e^2\right )-\left (b^3 \left (-c^3 f^3 \left (m^3+6 m^2+11 m+6\right )+15 c^2 d e f^2 \left (m^2+3 m+2\right )-60 c d^2 e^2 f (m+1)+60 d^3 e^3\right )\right )\right ) \, _2F_1\left (m-1,m+1;m+2;-\frac {d (a+b x)}{b c-a d}\right )}{60 b^5 d^3 (m+1)}+\frac {f (e+f x)^2 (a+b x)^{m+1} (c+d x)^{2-m}}{5 b d} \]

[Out]

1/5*f*(b*x+a)^(1+m)*(d*x+c)^(2-m)*(f*x+e)^2/b/d+1/60*f*(b*x+a)^(1+m)*(d*x+c)^(2-m)*(a^2*d^2*f^2*(m^2-7*m+12)-a
*b*d*f*(15*d*e*(3-m)-c*f*(-2*m^2+2*m+9))+b^2*(48*d^2*e^2-15*c*d*e*f*(2+m)+c^2*f^2*(m^2+5*m+6))-3*b*d*f*(a*d*f*
(4-m)-b*(7*d*e-c*f*(3+m)))*x)/b^3/d^3-1/60*(-a*d+b*c)*(a^3*d^3*f^3*(-m^3+9*m^2-26*m+24)-3*a^2*b*d^2*f^2*(m^2-5
*m+6)*(5*d*e-c*f*(1+m))+3*a*b^2*d*f*(2-m)*(20*d^2*e^2-10*c*d*e*f*(1+m)+c^2*f^2*(m^2+3*m+2))-b^3*(60*d^3*e^3-60
*c*d^2*e^2*f*(1+m)+15*c^2*d*e*f^2*(m^2+3*m+2)-c^3*f^3*(m^3+6*m^2+11*m+6)))*(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^5/d^3/(1+m)/((d*x+c)^m)

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Rubi [A]  time = 0.52, antiderivative size = 444, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {100, 147, 70, 69} \[ -\frac {(b c-a d) (a+b x)^{m+1} (c+d x)^{-m} \left (\frac {b (c+d x)}{b c-a d}\right )^m \left (-3 a^2 b d^2 f^2 \left (m^2-5 m+6\right ) (5 d e-c f (m+1))+a^3 d^3 f^3 \left (-m^3+9 m^2-26 m+24\right )+3 a b^2 d f (2-m) \left (c^2 f^2 \left (m^2+3 m+2\right )-10 c d e f (m+1)+20 d^2 e^2\right )+b^3 \left (-\left (15 c^2 d e f^2 \left (m^2+3 m+2\right )-c^3 f^3 \left (m^3+6 m^2+11 m+6\right )-60 c d^2 e^2 f (m+1)+60 d^3 e^3\right )\right )\right ) \, _2F_1\left (m-1,m+1;m+2;-\frac {d (a+b x)}{b c-a d}\right )}{60 b^5 d^3 (m+1)}+\frac {f (a+b x)^{m+1} (c+d x)^{2-m} \left (a^2 d^2 f^2 \left (m^2-7 m+12\right )-a b d f \left (15 d e (3-m)-c f \left (-2 m^2+2 m+9\right )\right )+3 b d f x (-a d f (4-m)-b c f (m+3)+7 b d e)+b^2 \left (c^2 f^2 \left (m^2+5 m+6\right )-15 c d e f (m+2)+48 d^2 e^2\right )\right )}{60 b^3 d^3}+\frac {f (e+f x)^2 (a+b x)^{m+1} (c+d x)^{2-m}}{5 b d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(f*(a + b*x)^(1 + m)*(c + d*x)^(2 - m)*(e + f*x)^2)/(5*b*d) + (f*(a + b*x)^(1 + m)*(c + d*x)^(2 - m)*(a^2*d^2*
f^2*(12 - 7*m + m^2) - a*b*d*f*(15*d*e*(3 - m) - c*f*(9 + 2*m - 2*m^2)) + b^2*(48*d^2*e^2 - 15*c*d*e*f*(2 + m)
+ c^2*f^2*(6 + 5*m + m^2)) + 3*b*d*f*(7*b*d*e - a*d*f*(4 - m) - b*c*f*(3 + m))*x))/(60*b^3*d^3) - ((b*c - a*d
)*(a^3*d^3*f^3*(24 - 26*m + 9*m^2 - m^3) - 3*a^2*b*d^2*f^2*(6 - 5*m + m^2)*(5*d*e - c*f*(1 + m)) + 3*a*b^2*d*f
*(2 - m)*(20*d^2*e^2 - 10*c*d*e*f*(1 + m) + c^2*f^2*(2 + 3*m + m^2)) - b^3*(60*d^3*e^3 - 60*c*d^2*e^2*f*(1 + m
) + 15*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*Hypergeometric2F1[-1 + m, 1 + m, 2 + m, -((d*(a + b*x))/(b*c - a*d))])/(60*b^5*d^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 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]

Rubi steps

\begin {align*} \int (a+b x)^m (c+d x)^{1-m} (e+f x)^3 \, dx &=\frac {f (a+b x)^{1+m} (c+d x)^{2-m} (e+f x)^2}{5 b d}+\frac {\int (a+b x)^m (c+d x)^{1-m} (e+f x) (-a f (2 c f+d e (2-m))+b e (5 d e-c f (1+m))+f (7 b d e-a d f (4-m)-b c f (3+m)) x) \, dx}{5 b d}\\ &=\frac {f (a+b x)^{1+m} (c+d x)^{2-m} (e+f x)^2}{5 b d}+\frac {f (a+b x)^{1+m} (c+d x)^{2-m} \left (a^2 d^2 f^2 \left (12-7 m+m^2\right )-a b d f \left (15 d e (3-m)-c f \left (9+2 m-2 m^2\right )\right )+b^2 \left (48 d^2 e^2-15 c d e f (2+m)+c^2 f^2 \left (6+5 m+m^2\right )\right )+3 b d f (7 b d e-a d f (4-m)-b c f (3+m)) x\right )}{60 b^3 d^3}-\frac {\left (a^3 d^3 f^3 \left (24-26 m+9 m^2-m^3\right )-3 a^2 b d^2 f^2 \left (6-5 m+m^2\right ) (5 d e-c f (1+m))+3 a b^2 d f (2-m) \left (20 d^2 e^2-10 c d e f (1+m)+c^2 f^2 \left (2+3 m+m^2\right )\right )-b^3 \left (60 d^3 e^3-60 c d^2 e^2 f (1+m)+15 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)^{1-m} \, dx}{60 b^3 d^3}\\ &=\frac {f (a+b x)^{1+m} (c+d x)^{2-m} (e+f x)^2}{5 b d}+\frac {f (a+b x)^{1+m} (c+d x)^{2-m} \left (a^2 d^2 f^2 \left (12-7 m+m^2\right )-a b d f \left (15 d e (3-m)-c f \left (9+2 m-2 m^2\right )\right )+b^2 \left (48 d^2 e^2-15 c d e f (2+m)+c^2 f^2 \left (6+5 m+m^2\right )\right )+3 b d f (7 b d e-a d f (4-m)-b c f (3+m)) x\right )}{60 b^3 d^3}-\frac {\left ((b c-a d) \left (a^3 d^3 f^3 \left (24-26 m+9 m^2-m^3\right )-3 a^2 b d^2 f^2 \left (6-5 m+m^2\right ) (5 d e-c f (1+m))+3 a b^2 d f (2-m) \left (20 d^2 e^2-10 c d e f (1+m)+c^2 f^2 \left (2+3 m+m^2\right )\right )-b^3 \left (60 d^3 e^3-60 c d^2 e^2 f (1+m)+15 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 )^{1-m} \, dx}{60 b^4 d^3}\\ &=\frac {f (a+b x)^{1+m} (c+d x)^{2-m} (e+f x)^2}{5 b d}+\frac {f (a+b x)^{1+m} (c+d x)^{2-m} \left (a^2 d^2 f^2 \left (12-7 m+m^2\right )-a b d f \left (15 d e (3-m)-c f \left (9+2 m-2 m^2\right )\right )+b^2 \left (48 d^2 e^2-15 c d e f (2+m)+c^2 f^2 \left (6+5 m+m^2\right )\right )+3 b d f (7 b d e-a d f (4-m)-b c f (3+m)) x\right )}{60 b^3 d^3}-\frac {(b c-a d) \left (a^3 d^3 f^3 \left (24-26 m+9 m^2-m^3\right )-3 a^2 b d^2 f^2 \left (6-5 m+m^2\right ) (5 d e-c f (1+m))+3 a b^2 d f (2-m) \left (20 d^2 e^2-10 c d e f (1+m)+c^2 f^2 \left (2+3 m+m^2\right )\right )-b^3 \left (60 d^3 e^3-60 c d^2 e^2 f (1+m)+15 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 (-1+m,1+m;2+m;-\frac {d (a+b x)}{b c-a d}\right )}{60 b^5 d^3 (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.76, size = 317, normalized size = 0.71 \[ \frac {(a+b x)^{m+1} (c+d x)^{-m} \left (b^3 (c+d x) (d e-c f)^2 \left (\frac {b (c+d x)}{b c-a d}\right )^{m-1} (a d f (m-2)-b c f (m+3)+5 b d e) \, _2F_1\left (m-1,m+1;m+2;\frac {d (a+b x)}{a d-b c}\right )+f^2 (b c-a d)^3 \left (\frac {b (c+d x)}{b c-a d}\right )^m (a d f (m-4)-b c f (m+3)+7 b d e) \, _2F_1\left (m-3,m+1;m+2;\frac {d (a+b x)}{a d-b c}\right )+2 b f (b c-a d)^2 (d e-c f) \left (\frac {b (c+d x)}{b c-a d}\right )^m (a d f (m-3)-b c f (m+3)+6 b d e) \, _2F_1\left (m-2,m+1;m+2;\frac {d (a+b x)}{a d-b c}\right )+b^4 d^2 f (m+1) (c+d x)^2 (e+f x)^2\right )}{5 b^5 d^3 (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b*x)^(1 + m)*(b^4*d^2*f*(1 + m)*(c + d*x)^2*(e + f*x)^2 + (b*c - a*d)^3*f^2*(7*b*d*e + a*d*f*(-4 + m) -
b*c*f*(3 + m))*((b*(c + d*x))/(b*c - a*d))^m*Hypergeometric2F1[-3 + m, 1 + m, 2 + m, (d*(a + b*x))/(-(b*c) + a
*d)] + 2*b*(b*c - a*d)^2*f*(d*e - c*f)*(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)] + b^3*(d*e - c*f)^2*(5*b*d*e + a*d*f*(
-2 + m) - b*c*f*(3 + m))*(c + d*x)*((b*(c + d*x))/(b*c - a*d))^(-1 + m)*Hypergeometric2F1[-1 + m, 1 + m, 2 + m
, (d*(a + b*x))/(-(b*c) + a*d)]))/(5*b^5*d^3*(1 + m)*(c + d*x)^m)

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fricas [F]  time = 1.09, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (f^{3} x^{3} + 3 \, e f^{2} x^{2} + 3 \, e^{2} f x + e^{3}\right )} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 1}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(1-m)*(f*x+e)^3,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 + 1), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (f x + e\right )}^{3} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(1-m)*(f*x+e)^3,x, algorithm="giac")

[Out]

integrate((f*x + e)^3*(b*x + a)^m*(d*x + c)^(-m + 1), x)

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maple [F]  time = 0.24, size = 0, normalized size = 0.00 \[ \int \left (f x +e \right )^{3} \left (b x +a \right )^{m} \left (d x +c \right )^{-m +1}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m*(d*x+c)^(-m+1)*(f*x+e)^3,x)

[Out]

int((b*x+a)^m*(d*x+c)^(-m+1)*(f*x+e)^3,x)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (f x + e\right )}^{3} {\left (b x + a\right )}^{m} {\left (d x + c\right )}^{-m + 1}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^(1-m)*(f*x+e)^3,x, algorithm="maxima")

[Out]

integrate((f*x + e)^3*(b*x + a)^m*(d*x + c)^(-m + 1), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\left (e+f\,x\right )}^3\,{\left (a+b\,x\right )}^m\,{\left (c+d\,x\right )}^{1-m} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e + f*x)^3*(a + b*x)^m*(c + d*x)^(1 - m),x)

[Out]

int((e + f*x)^3*(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)**3,x)

[Out]

Exception raised: HeuristicGCDFailed

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