3.380 \(\int \frac{x^m (c+d x)^2}{a+b x} \, dx\)

Optimal. Leaf size=99 \[ \frac{x^{m+1} (b c-a d)^2 \, _2F_1\left (1,m+1;m+2;-\frac{b x}{a}\right )}{a b^2 (m+1)}+\frac{d x^{m+1} (b c-a d)}{b^2 (m+1)}+\frac{c d x^{m+1}}{b (m+1)}+\frac{d^2 x^{m+2}}{b (m+2)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.0550933, antiderivative size = 99, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {88, 64, 43} \[ \frac{x^{m+1} (b c-a d)^2 \, _2F_1\left (1,m+1;m+2;-\frac{b x}{a}\right )}{a b^2 (m+1)}+\frac{d x^{m+1} (b c-a d)}{b^2 (m+1)}+\frac{c d x^{m+1}}{b (m+1)}+\frac{d^2 x^{m+2}}{b (m+2)} \]

Antiderivative was successfully verified.

[In]

Int[(x^m*(c + d*x)^2)/(a + b*x),x]

[Out]

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

Rule 88

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

Rule 64

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

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

Mathematica [A]  time = 0.0741327, size = 77, normalized size = 0.78 \[ \frac{x^{m+1} \left ((m+2) (b c-a d)^2 \, _2F_1\left (1,m+1;m+2;-\frac{b x}{a}\right )+a d (-a d (m+2)+2 b c (m+2)+b d (m+1) x)\right )}{a b^2 (m+1) (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.035, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( dx+c \right ) ^{2}{x}^{m}}{bx+a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

Sympy [C]  time = 4.71882, size = 219, normalized size = 2.21 \begin{align*} \frac{c^{2} m x x^{m} \Phi \left (\frac{b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{a \Gamma \left (m + 2\right )} + \frac{c^{2} x x^{m} \Phi \left (\frac{b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{a \Gamma \left (m + 2\right )} + \frac{2 c d m x^{2} x^{m} \Phi \left (\frac{b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a \Gamma \left (m + 3\right )} + \frac{4 c d x^{2} x^{m} \Phi \left (\frac{b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a \Gamma \left (m + 3\right )} + \frac{d^{2} m x^{3} x^{m} \Phi \left (\frac{b x e^{i \pi }}{a}, 1, m + 3\right ) \Gamma \left (m + 3\right )}{a \Gamma \left (m + 4\right )} + \frac{3 d^{2} x^{3} x^{m} \Phi \left (\frac{b x e^{i \pi }}{a}, 1, m + 3\right ) \Gamma \left (m + 3\right )}{a \Gamma \left (m + 4\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

c**2*m*x*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 1)*gamma(m + 1)/(a*gamma(m + 2)) + c**2*x*x**m*lerchphi(b
*x*exp_polar(I*pi)/a, 1, m + 1)*gamma(m + 1)/(a*gamma(m + 2)) + 2*c*d*m*x**2*x**m*lerchphi(b*x*exp_polar(I*pi)
/a, 1, m + 2)*gamma(m + 2)/(a*gamma(m + 3)) + 4*c*d*x**2*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 2)*gamma(
m + 2)/(a*gamma(m + 3)) + d**2*m*x**3*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 3)*gamma(m + 3)/(a*gamma(m +
4)) + 3*d**2*x**3*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 3)*gamma(m + 3)/(a*gamma(m + 4))

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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