3.379 \(\int \frac{x^m (c+d x)^3}{a+b x} \, dx\)

Optimal. Leaf size=127 \[ \frac{d x^{m+1} \left (a^2 d^2-3 a b c d+3 b^2 c^2\right )}{b^3 (m+1)}+\frac{d^2 x^{m+2} (3 b c-a d)}{b^2 (m+2)}+\frac{x^{m+1} (b c-a d)^3 \, _2F_1\left (1,1;1-m;\frac{a}{a+b x}\right )}{b^3 m (a+b x)}+\frac{d^3 x^{m+3}}{b (m+3)} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.106496, antiderivative size = 171, normalized size of antiderivative = 1.35, number of steps used = 7, number of rules used = 3, integrand size = 18, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.167, Rules used = {88, 64, 43} \[ \frac{d^2 x^{m+2} (b c-a d)}{b^2 (m+2)}+\frac{x^{m+1} (b c-a d)^3 \, _2F_1\left (1,m+1;m+2;-\frac{b x}{a}\right )}{a b^3 (m+1)}+\frac{d x^{m+1} (b c-a d)^2}{b^3 (m+1)}+\frac{c d x^{m+1} (b c-a d)}{b^2 (m+1)}+\frac{c^2 d x^{m+1}}{b (m+1)}+\frac{2 c d^2 x^{m+2}}{b (m+2)}+\frac{d^3 x^{m+3}}{b (m+3)} \]

Antiderivative was successfully verified.

[In]

Int[(x^m*(c + d*x)^3)/(a + b*x),x]

[Out]

(c^2*d*x^(1 + m))/(b*(1 + m)) + (c*d*(b*c - a*d)*x^(1 + m))/(b^2*(1 + m)) + (d*(b*c - a*d)^2*x^(1 + m))/(b^3*(
1 + m)) + (2*c*d^2*x^(2 + m))/(b*(2 + m)) + (d^2*(b*c - a*d)*x^(2 + m))/(b^2*(2 + m)) + (d^3*x^(3 + m))/(b*(3
+ m)) + ((b*c - a*d)^3*x^(1 + m)*Hypergeometric2F1[1, 1 + m, 2 + m, -((b*x)/a)])/(a*b^3*(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)^3}{a+b x} \, dx &=\int \left (\frac{d (b c-a d)^2 x^m}{b^3}+\frac{(b c-a d)^3 x^m}{b^3 (a+b x)}+\frac{d (b c-a d) x^m (c+d x)}{b^2}+\frac{d x^m (c+d x)^2}{b}\right ) \, dx\\ &=\frac{d (b c-a d)^2 x^{1+m}}{b^3 (1+m)}+\frac{d \int x^m (c+d x)^2 \, dx}{b}+\frac{(d (b c-a d)) \int x^m (c+d x) \, dx}{b^2}+\frac{(b c-a d)^3 \int \frac{x^m}{a+b x} \, dx}{b^3}\\ &=\frac{d (b c-a d)^2 x^{1+m}}{b^3 (1+m)}+\frac{(b c-a d)^3 x^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac{b x}{a}\right )}{a b^3 (1+m)}+\frac{d \int \left (c^2 x^m+2 c d x^{1+m}+d^2 x^{2+m}\right ) \, dx}{b}+\frac{(d (b c-a d)) \int \left (c x^m+d x^{1+m}\right ) \, dx}{b^2}\\ &=\frac{c^2 d x^{1+m}}{b (1+m)}+\frac{c d (b c-a d) x^{1+m}}{b^2 (1+m)}+\frac{d (b c-a d)^2 x^{1+m}}{b^3 (1+m)}+\frac{2 c d^2 x^{2+m}}{b (2+m)}+\frac{d^2 (b c-a d) x^{2+m}}{b^2 (2+m)}+\frac{d^3 x^{3+m}}{b (3+m)}+\frac{(b c-a d)^3 x^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac{b x}{a}\right )}{a b^3 (1+m)}\\ \end{align*}

Mathematica [A]  time = 0.1868, size = 118, normalized size = 0.93 \[ \frac{x^{m+1} \left (d \left (\frac{a^2 d^2}{m+1}+a b d \left (-\frac{3 c}{m+1}-\frac{d x}{m+2}\right )+b^2 \left (\frac{3 c^2}{m+1}+\frac{3 c d x}{m+2}+\frac{d^2 x^2}{m+3}\right )\right )+\frac{(b c-a d)^3 \, _2F_1\left (1,m+1;m+2;-\frac{b x}{a}\right )}{a (m+1)}\right )}{b^3} \]

Antiderivative was successfully verified.

[In]

Integrate[(x^m*(c + d*x)^3)/(a + b*x),x]

[Out]

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

________________________________________________________________________________________

Maple [F]  time = 0.037, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( dx+c \right ) ^{3}{x}^{m}}{bx+a}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(d*x+c)^3/(b*x+a),x)

[Out]

int(x^m*(d*x+c)^3/(b*x+a),x)

________________________________________________________________________________________

Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{3} 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)^3/(b*x+a),x, algorithm="maxima")

[Out]

integrate((d*x + c)^3*x^m/(b*x + a), x)

________________________________________________________________________________________

Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{{\left (d^{3} x^{3} + 3 \, c d^{2} x^{2} + 3 \, c^{2} d x + c^{3}\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)^3/(b*x+a),x, algorithm="fricas")

[Out]

integral((d^3*x^3 + 3*c*d^2*x^2 + 3*c^2*d*x + c^3)*x^m/(b*x + a), x)

________________________________________________________________________________________

Sympy [C]  time = 5.19416, size = 303, normalized size = 2.39 \begin{align*} \frac{c^{3} 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^{3} 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{3 c^{2} 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{6 c^{2} 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{3 c 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{9 c 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 )} + \frac{d^{3} m x^{4} x^{m} \Phi \left (\frac{b x e^{i \pi }}{a}, 1, m + 4\right ) \Gamma \left (m + 4\right )}{a \Gamma \left (m + 5\right )} + \frac{4 d^{3} x^{4} x^{m} \Phi \left (\frac{b x e^{i \pi }}{a}, 1, m + 4\right ) \Gamma \left (m + 4\right )}{a \Gamma \left (m + 5\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*(d*x+c)**3/(b*x+a),x)

[Out]

c**3*m*x*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 1)*gamma(m + 1)/(a*gamma(m + 2)) + c**3*x*x**m*lerchphi(b
*x*exp_polar(I*pi)/a, 1, m + 1)*gamma(m + 1)/(a*gamma(m + 2)) + 3*c**2*d*m*x**2*x**m*lerchphi(b*x*exp_polar(I*
pi)/a, 1, m + 2)*gamma(m + 2)/(a*gamma(m + 3)) + 6*c**2*d*x**2*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 2)*
gamma(m + 2)/(a*gamma(m + 3)) + 3*c*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)) + 9*c*d**2*x**3*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 3)*gamma(m + 3)/(a*gamma(m + 4)) +
d**3*m*x**4*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 4)*gamma(m + 4)/(a*gamma(m + 5)) + 4*d**3*x**4*x**m*le
rchphi(b*x*exp_polar(I*pi)/a, 1, m + 4)*gamma(m + 4)/(a*gamma(m + 5))

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{3} 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)^3/(b*x+a),x, algorithm="giac")

[Out]

integrate((d*x + c)^3*x^m/(b*x + a), x)