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

Optimal. Leaf size=138 \[ \frac {d x^2 \left (a+\frac {b}{x}\right )^{m+1} (6 a c-b d (2-m))}{6 a^2}-\frac {b \left (a+\frac {b}{x}\right )^{m+1} \left (6 a^2 c^2-6 a b c d (1-m)+b^2 d^2 \left (m^2-3 m+2\right )\right ) \, _2F_1\left (2,m+1;m+2;\frac {b}{a x}+1\right )}{6 a^4 (m+1)}+\frac {d^2 x^3 \left (a+\frac {b}{x}\right )^{m+1}}{3 a} \]

[Out]

1/6*d*(6*a*c-b*d*(2-m))*(a+b/x)^(1+m)*x^2/a^2+1/3*d^2*(a+b/x)^(1+m)*x^3/a-1/6*b*(6*a^2*c^2-6*a*b*c*d*(1-m)+b^2
*d^2*(m^2-3*m+2))*(a+b/x)^(1+m)*hypergeom([2, 1+m],[2+m],1+b/a/x)/a^4/(1+m)

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Rubi [A]  time = 0.12, antiderivative size = 138, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.294, Rules used = {434, 446, 89, 78, 65} \[ -\frac {b \left (a+\frac {b}{x}\right )^{m+1} \left (6 a^2 c^2-6 a b c d (1-m)+b^2 d^2 \left (m^2-3 m+2\right )\right ) \, _2F_1\left (2,m+1;m+2;\frac {b}{a x}+1\right )}{6 a^4 (m+1)}+\frac {d x^2 \left (a+\frac {b}{x}\right )^{m+1} (6 a c-b d (2-m))}{6 a^2}+\frac {d^2 x^3 \left (a+\frac {b}{x}\right )^{m+1}}{3 a} \]

Antiderivative was successfully verified.

[In]

Int[(a + b/x)^m*(c + d*x)^2,x]

[Out]

(d*(6*a*c - b*d*(2 - m))*(a + b/x)^(1 + m)*x^2)/(6*a^2) + (d^2*(a + b/x)^(1 + m)*x^3)/(3*a) - (b*(6*a^2*c^2 -
6*a*b*c*d*(1 - m) + b^2*d^2*(2 - 3*m + m^2))*(a + b/x)^(1 + m)*Hypergeometric2F1[2, 1 + m, 2 + m, 1 + b/(a*x)]
)/(6*a^4*(1 + m))

Rule 65

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

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 434

Int[((c_) + (d_.)*(x_)^(mn_.))^(q_.)*((a_) + (b_.)*(x_)^(n_.))^(p_.), x_Symbol] :> Int[((a + b*x^n)^p*(d + c*x
^n)^q)/x^(n*q), x] /; FreeQ[{a, b, c, d, n, p}, x] && EqQ[mn, -n] && IntegerQ[q] && (PosQ[n] ||  !IntegerQ[p])

Rule 446

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Dist[1/n, Subst[Int
[x^(Simplify[(m + 1)/n] - 1)*(a + b*x)^p*(c + d*x)^q, x], x, x^n], x] /; FreeQ[{a, b, c, d, m, n, p, q}, x] &&
NeQ[b*c - a*d, 0] && IntegerQ[Simplify[(m + 1)/n]]

Rubi steps

\begin {align*} \int \left (a+\frac {b}{x}\right )^m (c+d x)^2 \, dx &=\int \left (a+\frac {b}{x}\right )^m \left (d+\frac {c}{x}\right )^2 x^2 \, dx\\ &=-\operatorname {Subst}\left (\int \frac {(a+b x)^m (d+c x)^2}{x^4} \, dx,x,\frac {1}{x}\right )\\ &=\frac {d^2 \left (a+\frac {b}{x}\right )^{1+m} x^3}{3 a}-\frac {\operatorname {Subst}\left (\int \frac {(a+b x)^m \left (d (6 a c-b d (2-m))+3 a c^2 x\right )}{x^3} \, dx,x,\frac {1}{x}\right )}{3 a}\\ &=\frac {d (6 a c-b d (2-m)) \left (a+\frac {b}{x}\right )^{1+m} x^2}{6 a^2}+\frac {d^2 \left (a+\frac {b}{x}\right )^{1+m} x^3}{3 a}-\frac {1}{6} \left (6 c^2-\frac {b d (6 a c-b d (2-m)) (1-m)}{a^2}\right ) \operatorname {Subst}\left (\int \frac {(a+b x)^m}{x^2} \, dx,x,\frac {1}{x}\right )\\ &=\frac {d (6 a c-b d (2-m)) \left (a+\frac {b}{x}\right )^{1+m} x^2}{6 a^2}+\frac {d^2 \left (a+\frac {b}{x}\right )^{1+m} x^3}{3 a}-\frac {b \left (6 a^2 c^2-6 a b c d (1-m)+b^2 d^2 \left (2-3 m+m^2\right )\right ) \left (a+\frac {b}{x}\right )^{1+m} \, _2F_1\left (2,1+m;2+m;1+\frac {b}{a x}\right )}{6 a^4 (1+m)}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 112, normalized size = 0.81 \[ \frac {(a x+b) \left (a+\frac {b}{x}\right )^m \left (a^2 d (m+1) x^2 (2 a (3 c+d x)+b d (m-2))-b \left (6 a^2 c^2+6 a b c d (m-1)+b^2 d^2 \left (m^2-3 m+2\right )\right ) \, _2F_1\left (2,m+1;m+2;\frac {b}{a x}+1\right )\right )}{6 a^4 (m+1) x} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b/x)^m*(c + d*x)^2,x]

[Out]

((a + b/x)^m*(b + a*x)*(a^2*d*(1 + m)*x^2*(b*d*(-2 + m) + 2*a*(3*c + d*x)) - b*(6*a^2*c^2 + 6*a*b*c*d*(-1 + m)
+ b^2*d^2*(2 - 3*m + m^2))*Hypergeometric2F1[2, 1 + m, 2 + m, 1 + b/(a*x)]))/(6*a^4*(1 + m)*x)

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fricas [F]  time = 0.42, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} \left (\frac {a x + b}{x}\right )^{m}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x + c)^2*(a + b/x)^m, x)

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maple [F]  time = 0.05, size = 0, normalized size = 0.00 \[ \int \left (d x +c \right )^{2} \left (a +\frac {b}{x}\right )^{m}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x + c)^2*(a + b/x)^m, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b/x)^m*(c + d*x)^2,x)

[Out]

int((a + b/x)^m*(c + d*x)^2, x)

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sympy [C]  time = 6.14, size = 121, normalized size = 0.88 \[ \frac {b^{m} c^{2} x x^{- m} \Gamma \left (1 - m\right ) {{}_{2}F_{1}\left (\begin {matrix} - m, 1 - m \\ 2 - m \end {matrix}\middle | {\frac {a x e^{i \pi }}{b}} \right )}}{\Gamma \left (2 - m\right )} + \frac {2 b^{m} c d x^{2} x^{- m} \Gamma \left (2 - m\right ) {{}_{2}F_{1}\left (\begin {matrix} - m, 2 - m \\ 3 - m \end {matrix}\middle | {\frac {a x e^{i \pi }}{b}} \right )}}{\Gamma \left (3 - m\right )} + \frac {b^{m} d^{2} x^{3} x^{- m} \Gamma \left (3 - m\right ) {{}_{2}F_{1}\left (\begin {matrix} - m, 3 - m \\ 4 - m \end {matrix}\middle | {\frac {a x e^{i \pi }}{b}} \right )}}{\Gamma \left (4 - m\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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