3.4.80 \(\int \frac {x^m (c+d x)^2}{a+b x} \, dx\) [380]

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

[Out]

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

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Rubi [A]
time = 0.04, antiderivative size = 99, normalized size of antiderivative = 1.00, 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 = {90, 66, 45} \begin {gather*} \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)} \end {gather*}

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 45

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])

Rule 66

Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[c^n*((b*x)^(m + 1)/(b*(m + 1)))*Hypergeometr
ic2F1[-n, m + 1, m + 2, (-d)*(x/c)], 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 90

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]))

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*}

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Mathematica [A]
time = 0.11, size = 77, normalized size = 0.78 \begin {gather*} \frac {x^{1+m} \left (a d (2 b c (2+m)-a d (2+m)+b d (1+m) x)+(b c-a d)^2 (2+m) \, _2F_1\left (1,1+m;2+m;-\frac {b x}{a}\right )\right )}{a b^2 (1+m) (2+m)} \end {gather*}

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))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x^{m} \left (d x +c \right )^{2}}{b x +a}\, dx\]

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)

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

[Out]

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

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Sympy [C] Result contains complex when optimal does not.
time = 1.83, size = 219, normalized size = 2.21 \begin {gather*} \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 {gather*}

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))

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^m*(c + d*x)^2)/(a + b*x),x)

[Out]

int((x^m*(c + d*x)^2)/(a + b*x), x)

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