3.78 \(\int \frac {(e x)^m}{(a+b x) (a c-b c x)^2} \, dx\)

Optimal. Leaf size=98 \[ \frac {b (e x)^{m+2} \, _2F_1\left (2,\frac {m+2}{2};\frac {m+4}{2};\frac {b^2 x^2}{a^2}\right )}{a^4 c^2 e^2 (m+2)}+\frac {(e x)^{m+1} \, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};\frac {b^2 x^2}{a^2}\right )}{a^3 c^2 e (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 98, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 3, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {82, 73, 364} \[ \frac {b (e x)^{m+2} \, _2F_1\left (2,\frac {m+2}{2};\frac {m+4}{2};\frac {b^2 x^2}{a^2}\right )}{a^4 c^2 e^2 (m+2)}+\frac {(e x)^{m+1} \, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};\frac {b^2 x^2}{a^2}\right )}{a^3 c^2 e (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(e*x)^m/((a + b*x)*(a*c - b*c*x)^2),x]

[Out]

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

Rule 73

Int[((a_) + (b_.)*(x_))^(m_.)*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[(a*c + b*
d*x^2)^m*(e + f*x)^p, x] /; FreeQ[{a, b, c, d, e, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[n, m] && Integer
Q[m]

Rule 82

Int[((f_.)*(x_))^(p_.)*((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Dist[a, Int[(a + b*
x)^n*(c + d*x)^n*(f*x)^p, x], x] + Dist[b/f, Int[(a + b*x)^n*(c + d*x)^n*(f*x)^(p + 1), x], x] /; FreeQ[{a, b,
c, d, f, m, n, p}, x] && EqQ[b*c + a*d, 0] && EqQ[m - n - 1, 0] &&  !RationalQ[p] &&  !IGtQ[m, 0] && NeQ[m +
n + p + 2, 0]

Rule 364

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(a^p*(c*x)^(m + 1)*Hypergeometric2F1[-
p, (m + 1)/n, (m + 1)/n + 1, -((b*x^n)/a)])/(c*(m + 1)), x] /; FreeQ[{a, b, c, m, n, p}, x] &&  !IGtQ[p, 0] &&
(ILtQ[p, 0] || GtQ[a, 0])

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 87, normalized size = 0.89 \[ \frac {x (e x)^m \left (b (m+1) x \, _2F_1\left (2,\frac {m}{2}+1;\frac {m}{2}+2;\frac {b^2 x^2}{a^2}\right )+a (m+2) \, _2F_1\left (2,\frac {m+1}{2};\frac {m+3}{2};\frac {b^2 x^2}{a^2}\right )\right )}{a^4 c^2 (m+1) (m+2)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.93, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\left (e x\right )^{m}}{b^{3} c^{2} x^{3} - a b^{2} c^{2} x^{2} - a^{2} b c^{2} x + a^{3} c^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [F]  time = 0.11, size = 0, normalized size = 0.00 \[ \int \frac {\left (e x \right )^{m}}{\left (b x +a \right ) \left (-b c x +a c \right )^{2}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [C]  time = 3.48, size = 440, normalized size = 4.49 \[ - \frac {2 a e^{m} m^{2} x^{m} \Phi \left (\frac {a}{b x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{- 4 a^{3} b c^{2} \Gamma \left (1 - m\right ) + 4 a^{2} b^{2} c^{2} x \Gamma \left (1 - m\right )} + \frac {a e^{m} m x^{m} \Phi \left (\frac {a}{b x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{- 4 a^{3} b c^{2} \Gamma \left (1 - m\right ) + 4 a^{2} b^{2} c^{2} x \Gamma \left (1 - m\right )} - \frac {a e^{m} m x^{m} \Phi \left (\frac {a e^{i \pi }}{b x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{- 4 a^{3} b c^{2} \Gamma \left (1 - m\right ) + 4 a^{2} b^{2} c^{2} x \Gamma \left (1 - m\right )} + \frac {2 b e^{m} m^{2} x x^{m} \Phi \left (\frac {a}{b x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{- 4 a^{3} b c^{2} \Gamma \left (1 - m\right ) + 4 a^{2} b^{2} c^{2} x \Gamma \left (1 - m\right )} - \frac {b e^{m} m x x^{m} \Phi \left (\frac {a}{b x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{- 4 a^{3} b c^{2} \Gamma \left (1 - m\right ) + 4 a^{2} b^{2} c^{2} x \Gamma \left (1 - m\right )} + \frac {b e^{m} m x x^{m} \Phi \left (\frac {a e^{i \pi }}{b x}, 1, m e^{i \pi }\right ) \Gamma \left (- m\right )}{- 4 a^{3} b c^{2} \Gamma \left (1 - m\right ) + 4 a^{2} b^{2} c^{2} x \Gamma \left (1 - m\right )} + \frac {2 b e^{m} m x x^{m} \Gamma \left (- m\right )}{- 4 a^{3} b c^{2} \Gamma \left (1 - m\right ) + 4 a^{2} b^{2} c^{2} x \Gamma \left (1 - m\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2*a*e**m*m**2*x**m*lerchphi(a/(b*x), 1, m*exp_polar(I*pi))*gamma(-m)/(-4*a**3*b*c**2*gamma(1 - m) + 4*a**2*b*
*2*c**2*x*gamma(1 - m)) + a*e**m*m*x**m*lerchphi(a/(b*x), 1, m*exp_polar(I*pi))*gamma(-m)/(-4*a**3*b*c**2*gamm
a(1 - m) + 4*a**2*b**2*c**2*x*gamma(1 - m)) - a*e**m*m*x**m*lerchphi(a*exp_polar(I*pi)/(b*x), 1, m*exp_polar(I
*pi))*gamma(-m)/(-4*a**3*b*c**2*gamma(1 - m) + 4*a**2*b**2*c**2*x*gamma(1 - m)) + 2*b*e**m*m**2*x*x**m*lerchph
i(a/(b*x), 1, m*exp_polar(I*pi))*gamma(-m)/(-4*a**3*b*c**2*gamma(1 - m) + 4*a**2*b**2*c**2*x*gamma(1 - m)) - b
*e**m*m*x*x**m*lerchphi(a/(b*x), 1, m*exp_polar(I*pi))*gamma(-m)/(-4*a**3*b*c**2*gamma(1 - m) + 4*a**2*b**2*c*
*2*x*gamma(1 - m)) + b*e**m*m*x*x**m*lerchphi(a*exp_polar(I*pi)/(b*x), 1, m*exp_polar(I*pi))*gamma(-m)/(-4*a**
3*b*c**2*gamma(1 - m) + 4*a**2*b**2*c**2*x*gamma(1 - m)) + 2*b*e**m*m*x*x**m*gamma(-m)/(-4*a**3*b*c**2*gamma(1
- m) + 4*a**2*b**2*c**2*x*gamma(1 - m))

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