3.4.76 \(\int \frac {x^m (A+B x)}{(a+b x)^2} \, dx\) [376]

Optimal. Leaf size=73 \[ \frac {(A b-a B) x^{1+m}}{a b (a+b x)}-\frac {(A b m-a B (1+m)) x^{1+m} \, _2F_1\left (1,1+m;2+m;-\frac {b x}{a}\right )}{a^2 b (1+m)} \]

[Out]

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

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Rubi [A]
time = 0.02, antiderivative size = 73, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {79, 66} \begin {gather*} \frac {x^{m+1} (A b-a B)}{a b (a+b x)}-\frac {x^{m+1} (A b m-a B (m+1)) \, _2F_1\left (1,m+1;m+2;-\frac {b x}{a}\right )}{a^2 b (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(x^m*(A + B*x))/(a + b*x)^2,x]

[Out]

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

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 79

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] || L
tQ[p, n]))))

Rubi steps

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

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Mathematica [A]
time = 0.06, size = 63, normalized size = 0.86 \begin {gather*} \frac {x^{1+m} \left (\frac {a (A b-a B)}{a+b x}+\frac {(-A b m+a B (1+m)) \, _2F_1\left (1,1+m;2+m;-\frac {b x}{a}\right )}{1+m}\right )}{a^2 b} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(x^m*(A + B*x))/(a + b*x)^2,x]

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^m*(B*x+A)/(b*x+a)^2,x)

[Out]

int(x^m*(B*x+A)/(b*x+a)^2,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*(B*x+A)/(b*x+a)^2,x, algorithm="maxima")

[Out]

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

[Out]

integral((B*x + A)*x^m/(b^2*x^2 + 2*a*b*x + a^2), x)

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Sympy [C] Result contains complex when optimal does not.
time = 2.58, size = 639, normalized size = 8.75 \begin {gather*} A \left (- \frac {a m^{2} x x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{a^{3} \Gamma \left (m + 2\right ) + a^{2} b x \Gamma \left (m + 2\right )} - \frac {a m x x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{a^{3} \Gamma \left (m + 2\right ) + a^{2} b x \Gamma \left (m + 2\right )} + \frac {a m x x^{m} \Gamma \left (m + 1\right )}{a^{3} \Gamma \left (m + 2\right ) + a^{2} b x \Gamma \left (m + 2\right )} + \frac {a x x^{m} \Gamma \left (m + 1\right )}{a^{3} \Gamma \left (m + 2\right ) + a^{2} b x \Gamma \left (m + 2\right )} - \frac {b m^{2} x^{2} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{a^{3} \Gamma \left (m + 2\right ) + a^{2} b x \Gamma \left (m + 2\right )} - \frac {b m x^{2} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 1\right ) \Gamma \left (m + 1\right )}{a^{3} \Gamma \left (m + 2\right ) + a^{2} b x \Gamma \left (m + 2\right )}\right ) + B \left (- \frac {a m^{2} x^{2} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a^{3} \Gamma \left (m + 3\right ) + a^{2} b x \Gamma \left (m + 3\right )} - \frac {3 a m x^{2} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a^{3} \Gamma \left (m + 3\right ) + a^{2} b x \Gamma \left (m + 3\right )} + \frac {a m x^{2} x^{m} \Gamma \left (m + 2\right )}{a^{3} \Gamma \left (m + 3\right ) + a^{2} b x \Gamma \left (m + 3\right )} - \frac {2 a x^{2} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a^{3} \Gamma \left (m + 3\right ) + a^{2} b x \Gamma \left (m + 3\right )} + \frac {2 a x^{2} x^{m} \Gamma \left (m + 2\right )}{a^{3} \Gamma \left (m + 3\right ) + a^{2} b x \Gamma \left (m + 3\right )} - \frac {b m^{2} x^{3} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a^{3} \Gamma \left (m + 3\right ) + a^{2} b x \Gamma \left (m + 3\right )} - \frac {3 b m x^{3} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a^{3} \Gamma \left (m + 3\right ) + a^{2} b x \Gamma \left (m + 3\right )} - \frac {2 b x^{3} x^{m} \Phi \left (\frac {b x e^{i \pi }}{a}, 1, m + 2\right ) \Gamma \left (m + 2\right )}{a^{3} \Gamma \left (m + 3\right ) + a^{2} b x \Gamma \left (m + 3\right )}\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**m*(B*x+A)/(b*x+a)**2,x)

[Out]

A*(-a*m**2*x*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 1)*gamma(m + 1)/(a**3*gamma(m + 2) + a**2*b*x*gamma(m
+ 2)) - a*m*x*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 1)*gamma(m + 1)/(a**3*gamma(m + 2) + a**2*b*x*gamma
(m + 2)) + a*m*x*x**m*gamma(m + 1)/(a**3*gamma(m + 2) + a**2*b*x*gamma(m + 2)) + a*x*x**m*gamma(m + 1)/(a**3*g
amma(m + 2) + a**2*b*x*gamma(m + 2)) - b*m**2*x**2*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 1)*gamma(m + 1)
/(a**3*gamma(m + 2) + a**2*b*x*gamma(m + 2)) - b*m*x**2*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 1)*gamma(m
+ 1)/(a**3*gamma(m + 2) + a**2*b*x*gamma(m + 2))) + B*(-a*m**2*x**2*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m
+ 2)*gamma(m + 2)/(a**3*gamma(m + 3) + a**2*b*x*gamma(m + 3)) - 3*a*m*x**2*x**m*lerchphi(b*x*exp_polar(I*pi)/
a, 1, m + 2)*gamma(m + 2)/(a**3*gamma(m + 3) + a**2*b*x*gamma(m + 3)) + a*m*x**2*x**m*gamma(m + 2)/(a**3*gamma
(m + 3) + a**2*b*x*gamma(m + 3)) - 2*a*x**2*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 2)*gamma(m + 2)/(a**3*
gamma(m + 3) + a**2*b*x*gamma(m + 3)) + 2*a*x**2*x**m*gamma(m + 2)/(a**3*gamma(m + 3) + a**2*b*x*gamma(m + 3))
- b*m**2*x**3*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 2)*gamma(m + 2)/(a**3*gamma(m + 3) + a**2*b*x*gamma
(m + 3)) - 3*b*m*x**3*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 2)*gamma(m + 2)/(a**3*gamma(m + 3) + a**2*b*
x*gamma(m + 3)) - 2*b*x**3*x**m*lerchphi(b*x*exp_polar(I*pi)/a, 1, m + 2)*gamma(m + 2)/(a**3*gamma(m + 3) + a*
*2*b*x*gamma(m + 3)))

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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*(B*x+A)/(b*x+a)^2,x, algorithm="giac")

[Out]

integrate((B*x + A)*x^m/(b*x + a)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((x^m*(A + B*x))/(a + b*x)^2,x)

[Out]

int((x^m*(A + B*x))/(a + b*x)^2, x)

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