3.10.22 \(\int \frac {(d+e x)^m (a+b x+c x^2)}{f+g x} \, dx\) [922]

Optimal. Leaf size=129 \[ -\frac {(c e f+c d g-b e g) (d+e x)^{1+m}}{e^2 g^2 (1+m)}+\frac {c (d+e x)^{2+m}}{e^2 g (2+m)}+\frac {\left (c f^2-b f g+a g^2\right ) (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {g (d+e x)}{e f-d g}\right )}{g^2 (e f-d g) (1+m)} \]

[Out]

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

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Rubi [A]
time = 0.10, antiderivative size = 129, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {965, 81, 70} \begin {gather*} \frac {(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right ) \, _2F_1\left (1,m+1;m+2;-\frac {g (d+e x)}{e f-d g}\right )}{g^2 (m+1) (e f-d g)}-\frac {(d+e x)^{m+1} (-b e g+c d g+c e f)}{e^2 g^2 (m+1)}+\frac {c (d+e x)^{m+2}}{e^2 g (m+2)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 70

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

Rule 81

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 965

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])

Rubi steps

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

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Mathematica [A]
time = 0.55, size = 166, normalized size = 1.29 \begin {gather*} \frac {(d+e x)^m \left (\frac {g \left (b e g (2+m) (d+e x)+c \left (d e (-f (2+m)+g m x)+e^2 x (-f (2+m)+g (1+m) x)+d^2 g \left (-1+\left (1+\frac {e x}{d}\right )^{-m}\right )\right )\right )}{e^2 (1+m) (2+m)}+\frac {\left (c f^2+g (-b f+a g)\right ) \left (\frac {g (d+e x)}{e (f+g x)}\right )^{-m} \, _2F_1\left (-m,-m;1-m;\frac {e f-d g}{e f+e g x}\right )}{m}\right )}{g^3} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]
time = 0.03, size = 0, normalized size = 0.00 \[\int \frac {\left (e x +d \right )^{m} \left (c \,x^{2}+b x +a \right )}{g x +f}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (d + e x\right )^{m} \left (a + b x + c x^{2}\right )}{f + g x}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*x+d)**m*(c*x**2+b*x+a)/(g*x+f),x)

[Out]

Integral((d + e*x)**m*(a + b*x + c*x**2)/(f + g*x), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d+e\,x\right )}^m\,\left (c\,x^2+b\,x+a\right )}{f+g\,x} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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