3.10.27 \(\int \frac {(d+e x)^m (a+b x+c x^2)^2}{f+g x} \, dx\) [927]

Optimal. Leaf size=287 \[ \frac {(b e g-c (e f+d g)) \left (c \left (e^2 f^2+d^2 g^2\right )+e g (2 a e g-b (e f+d g))\right ) (d+e x)^{1+m}}{e^4 g^4 (1+m)}+\frac {\left (b^2 e^2 g^2+c^2 \left (e^2 f^2+2 d e f g+3 d^2 g^2\right )+2 c e g (a e g-b (e f+2 d g))\right ) (d+e x)^{2+m}}{e^4 g^3 (2+m)}-\frac {c (c e f+3 c d g-2 b e g) (d+e x)^{3+m}}{e^4 g^2 (3+m)}+\frac {c^2 (d+e x)^{4+m}}{e^4 g (4+m)}+\frac {\left (c f^2-b f g+a g^2\right )^2 (d+e x)^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {g (d+e x)}{e f-d g}\right )}{g^4 (e f-d g) (1+m)} \]

[Out]

(b*e*g-c*(d*g+e*f))*(c*(d^2*g^2+e^2*f^2)+e*g*(2*a*e*g-b*(d*g+e*f)))*(e*x+d)^(1+m)/e^4/g^4/(1+m)+(b^2*e^2*g^2+c
^2*(3*d^2*g^2+2*d*e*f*g+e^2*f^2)+2*c*e*g*(a*e*g-b*(2*d*g+e*f)))*(e*x+d)^(2+m)/e^4/g^3/(2+m)-c*(-2*b*e*g+3*c*d*
g+c*e*f)*(e*x+d)^(3+m)/e^4/g^2/(3+m)+c^2*(e*x+d)^(4+m)/e^4/g/(4+m)+(a*g^2-b*f*g+c*f^2)^2*(e*x+d)^(1+m)*hyperge
om([1, 1+m],[2+m],-g*(e*x+d)/(-d*g+e*f))/g^4/(-d*g+e*f)/(1+m)

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Rubi [A]
time = 0.55, antiderivative size = 287, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.111, Rules used = {965, 1634, 70} \begin {gather*} \frac {(d+e x)^{m+2} \left (2 c e g (a e g-b (2 d g+e f))+b^2 e^2 g^2+c^2 \left (3 d^2 g^2+2 d e f g+e^2 f^2\right )\right )}{e^4 g^3 (m+2)}+\frac {(d+e x)^{m+1} (b e g-c (d g+e f)) \left (e g (2 a e g-b (d g+e f))+c \left (d^2 g^2+e^2 f^2\right )\right )}{e^4 g^4 (m+1)}+\frac {(d+e x)^{m+1} \left (a g^2-b f g+c f^2\right )^2 \, _2F_1\left (1,m+1;m+2;-\frac {g (d+e x)}{e f-d g}\right )}{g^4 (m+1) (e f-d g)}-\frac {c (d+e x)^{m+3} (-2 b e g+3 c d g+c e f)}{e^4 g^2 (m+3)}+\frac {c^2 (d+e x)^{m+4}}{e^4 g (m+4)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 1634

Int[(Px_)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[Px*(a + b*x)
^m*(c + d*x)^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && PolyQ[Px, x] && (IntegersQ[m, n] || IGtQ[m, -2]) &&
GtQ[Expon[Px, x], 2]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 6 vs. order 5 in optimal.
time = 0.58, size = 254, normalized size = 0.89 \begin {gather*} \frac {(d+e x)^m \left (30 a b g m x^2 \left (1+\frac {e x}{d}\right )^{-m} F_1\left (2;-m,1;3;-\frac {e x}{d},-\frac {g x}{f}\right )+10 \left (b^2+2 a c\right ) g m x^3 \left (1+\frac {e x}{d}\right )^{-m} F_1\left (3;-m,1;4;-\frac {e x}{d},-\frac {g x}{f}\right )+15 b c g m x^4 \left (1+\frac {e x}{d}\right )^{-m} F_1\left (4;-m,1;5;-\frac {e x}{d},-\frac {g x}{f}\right )+6 c^2 g m x^5 \left (1+\frac {e x}{d}\right )^{-m} F_1\left (5;-m,1;6;-\frac {e x}{d},-\frac {g x}{f}\right )+30 a^2 f \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 )\right )}{30 f g m} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((d + e*x)^m*((30*a*b*g*m*x^2*AppellF1[2, -m, 1, 3, -((e*x)/d), -((g*x)/f)])/(1 + (e*x)/d)^m + (10*(b^2 + 2*a*
c)*g*m*x^3*AppellF1[3, -m, 1, 4, -((e*x)/d), -((g*x)/f)])/(1 + (e*x)/d)^m + (15*b*c*g*m*x^4*AppellF1[4, -m, 1,
5, -((e*x)/d), -((g*x)/f)])/(1 + (e*x)/d)^m + (6*c^2*g*m*x^5*AppellF1[5, -m, 1, 6, -((e*x)/d), -((g*x)/f)])/(
1 + (e*x)/d)^m + (30*a^2*f*Hypergeometric2F1[-m, -m, 1 - m, (e*f - d*g)/(e*f + e*g*x)])/((g*(d + e*x))/(e*(f +
g*x)))^m))/(30*f*g*m)

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Maple [F]
time = 0.06, size = 0, normalized size = 0.00 \[\int \frac {\left (e x +d \right )^{m} \left (c \,x^{2}+b x +a \right )^{2}}{g x +f}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

[Out]

integral((c^2*x^4 + 2*b*c*x^3 + 2*a*b*x + (b^2 + 2*a*c)*x^2 + a^2)*(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 )^{2}}{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)**2/(g*x+f),x)

[Out]

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

[Out]

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

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

[Out]

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

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