3.33 \(\int (a+b x)^m (c+d x)^n (A+B x+C x^2) \, dx\)

Optimal. Leaf size=268 \[ -\frac{(a+b x)^{m+1} (c+d x)^n \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} \, _2F_1\left (m+1,-n;m+2;-\frac{d (a+b x)}{b c-a d}\right ) \left (d (m+n+2) \left (a^2 C d (n+1)+a b c C (m+2)-A b^2 d (m+n+3)\right )-(a d (n+1)+b c (m+1)) (a C d (m+2 n+4)+b (c C (m+2)-B d (m+n+3)))\right )}{b^3 d^2 (m+1) (m+n+2) (m+n+3)}-\frac{(a+b x)^{m+1} (c+d x)^{n+1} (a C d (m+2 n+4)+b (c C (m+2)-B d (m+n+3)))}{b^2 d^2 (m+n+2) (m+n+3)}+\frac{C (a+b x)^{m+2} (c+d x)^{n+1}}{b^2 d (m+n+3)} \]

[Out]

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

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Rubi [A]  time = 0.30949, antiderivative size = 266, normalized size of antiderivative = 0.99, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {951, 80, 70, 69} \[ -\frac{(a+b x)^{m+1} (c+d x)^n \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} \, _2F_1\left (m+1,-n;m+2;-\frac{d (a+b x)}{b c-a d}\right ) \left (d (m+n+2) \left (a^2 C d (n+1)+a b c C (m+2)-A b^2 d (m+n+3)\right )-(a d (n+1)+b c (m+1)) (a C d (m+2 n+4)-b B d (m+n+3)+b c C (m+2))\right )}{b^3 d^2 (m+1) (m+n+2) (m+n+3)}-\frac{(a+b x)^{m+1} (c+d x)^{n+1} (a C d (m+2 n+4)-b B d (m+n+3)+b c C (m+2))}{b^2 d^2 (m+n+2) (m+n+3)}+\frac{C (a+b x)^{m+2} (c+d x)^{n+1}}{b^2 d (m+n+3)} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*x)^m*(c + d*x)^n*(A + B*x + C*x^2),x]

[Out]

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

Rule 951

Int[((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_.) + (b_.)*(x_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :
> Simp[(c^p*(d + e*x)^(m + 2*p)*(f + g*x)^(n + 1))/(g*e^(2*p)*(m + n + 2*p + 1)), x] + Dist[1/(g*e^(2*p)*(m +
n + 2*p + 1)), Int[(d + e*x)^m*(f + g*x)^n*ExpandToSum[g*(m + n + 2*p + 1)*(e^(2*p)*(a + b*x + c*x^2)^p - c^p*
(d + e*x)^(2*p)) - c^p*(e*f - d*g)*(m + 2*p)*(d + e*x)^(2*p - 1), x], x], x] /; FreeQ[{a, b, c, d, e, f, g}, x
] && NeQ[e*f - d*g, 0] && NeQ[b^2 - 4*a*c, 0] && NeQ[c*d^2 - b*d*e + a*e^2, 0] && IGtQ[p, 0] && NeQ[m + n + 2*
p + 1, 0] && (IntegerQ[n] ||  !IntegerQ[m])

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]

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

Mathematica [A]  time = 0.204403, size = 187, normalized size = 0.7 \[ \frac{(a+b x)^{m+1} (c+d x)^n \left (\frac{b (c+d x)}{b c-a d}\right )^{-n} \left (b \left (b \left (A d^2-B c d+c^2 C\right ) \, _2F_1\left (m+1,-n;m+2;\frac{d (a+b x)}{a d-b c}\right )-(b c-a d) (2 c C-B d) \, _2F_1\left (m+1,-n-1;m+2;\frac{d (a+b x)}{a d-b c}\right )\right )+C (b c-a d)^2 \, _2F_1\left (m+1,-n-2;m+2;\frac{d (a+b x)}{a d-b c}\right )\right )}{b^3 d^2 (m+1)} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*x)^m*(c + d*x)^n*(A + B*x + C*x^2),x]

[Out]

((a + b*x)^(1 + m)*(c + d*x)^n*(C*(b*c - a*d)^2*Hypergeometric2F1[1 + m, -2 - n, 2 + m, (d*(a + b*x))/(-(b*c)
+ a*d)] + b*(-((b*c - a*d)*(2*c*C - B*d)*Hypergeometric2F1[1 + m, -1 - n, 2 + m, (d*(a + b*x))/(-(b*c) + a*d)]
) + b*(c^2*C - B*c*d + A*d^2)*Hypergeometric2F1[1 + m, -n, 2 + m, (d*(a + b*x))/(-(b*c) + a*d)])))/(b^3*d^2*(1
+ m)*((b*(c + d*x))/(b*c - a*d))^n)

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Maple [F]  time = 0.064, size = 0, normalized size = 0. \begin{align*} \int \left ( bx+a \right ) ^{m} \left ( dx+c \right ) ^{n} \left ( C{x}^{2}+Bx+A \right ) \, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((b*x+a)^m*(d*x+c)^n*(C*x^2+B*x+A),x)

[Out]

int((b*x+a)^m*(d*x+c)^n*(C*x^2+B*x+A),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C x^{2} + B x + A\right )}{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^n*(C*x^2+B*x+A),x, algorithm="maxima")

[Out]

integrate((C*x^2 + B*x + A)*(b*x + a)^m*(d*x + c)^n, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^n*(C*x^2+B*x+A),x, algorithm="fricas")

[Out]

integral((C*x^2 + B*x + A)*(b*x + a)^m*(d*x + c)^n, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (a + b x\right )^{m} \left (c + d x\right )^{n} \left (A + B x + C x^{2}\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)**m*(d*x+c)**n*(C*x**2+B*x+A),x)

[Out]

Integral((a + b*x)**m*(c + d*x)**n*(A + B*x + C*x**2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (C x^{2} + B x + A\right )}{\left (b x + a\right )}^{m}{\left (d x + c\right )}^{n}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((b*x+a)^m*(d*x+c)^n*(C*x^2+B*x+A),x, algorithm="giac")

[Out]

integrate((C*x^2 + B*x + A)*(b*x + a)^m*(d*x + c)^n, x)