3.35 \(\int \frac{(c+d x)^m}{(a+a \coth (e+f x))^2} \, dx\)

Optimal. Leaf size=152 \[ \frac{2^{-m-2} e^{\frac{2 c f}{d}-2 e} (c+d x)^m \left (\frac{f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{2 f (c+d x)}{d}\right )}{a^2 f}-\frac{4^{-m-2} e^{\frac{4 c f}{d}-4 e} (c+d x)^m \left (\frac{f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{4 f (c+d x)}{d}\right )}{a^2 f}+\frac{(c+d x)^{m+1}}{4 a^2 d (m+1)} \]

[Out]

(c + d*x)^(1 + m)/(4*a^2*d*(1 + m)) + (2^(-2 - m)*E^(-2*e + (2*c*f)/d)*(c + d*x)^m*Gamma[1 + m, (2*f*(c + d*x)
)/d])/(a^2*f*((f*(c + d*x))/d)^m) - (4^(-2 - m)*E^(-4*e + (4*c*f)/d)*(c + d*x)^m*Gamma[1 + m, (4*f*(c + d*x))/
d])/(a^2*f*((f*(c + d*x))/d)^m)

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Rubi [A]  time = 0.172427, antiderivative size = 152, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3729, 2181} \[ \frac{2^{-m-2} e^{\frac{2 c f}{d}-2 e} (c+d x)^m \left (\frac{f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{2 f (c+d x)}{d}\right )}{a^2 f}-\frac{4^{-m-2} e^{\frac{4 c f}{d}-4 e} (c+d x)^m \left (\frac{f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{4 f (c+d x)}{d}\right )}{a^2 f}+\frac{(c+d x)^{m+1}}{4 a^2 d (m+1)} \]

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^m/(a + a*Coth[e + f*x])^2,x]

[Out]

(c + d*x)^(1 + m)/(4*a^2*d*(1 + m)) + (2^(-2 - m)*E^(-2*e + (2*c*f)/d)*(c + d*x)^m*Gamma[1 + m, (2*f*(c + d*x)
)/d])/(a^2*f*((f*(c + d*x))/d)^m) - (4^(-2 - m)*E^(-4*e + (4*c*f)/d)*(c + d*x)^m*Gamma[1 + m, (4*f*(c + d*x))/
d])/(a^2*f*((f*(c + d*x))/d)^m)

Rule 3729

Int[((c_.) + (d_.)*(x_))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Int[ExpandIntegrand[(c
+ d*x)^m, (1/(2*a) + E^((2*a*(e + f*x))/b)/(2*a))^(-n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2
+ b^2, 0] && ILtQ[n, 0]

Rule 2181

Int[(F_)^((g_.)*((e_.) + (f_.)*(x_)))*((c_.) + (d_.)*(x_))^(m_), x_Symbol] :> -Simp[(F^(g*(e - (c*f)/d))*(c +
d*x)^FracPart[m]*Gamma[m + 1, (-((f*g*Log[F])/d))*(c + d*x)])/(d*(-((f*g*Log[F])/d))^(IntPart[m] + 1)*(-((f*g*
Log[F]*(c + d*x))/d))^FracPart[m]), x] /; FreeQ[{F, c, d, e, f, g, m}, x] &&  !IntegerQ[m]

Rubi steps

\begin{align*} \int \frac{(c+d x)^m}{(a+a \coth (e+f x))^2} \, dx &=\int \left (\frac{(c+d x)^m}{4 a^2}+\frac{e^{-4 e-4 f x} (c+d x)^m}{4 a^2}-\frac{e^{-2 e-2 f x} (c+d x)^m}{2 a^2}\right ) \, dx\\ &=\frac{(c+d x)^{1+m}}{4 a^2 d (1+m)}+\frac{\int e^{-4 e-4 f x} (c+d x)^m \, dx}{4 a^2}-\frac{\int e^{-2 e-2 f x} (c+d x)^m \, dx}{2 a^2}\\ &=\frac{(c+d x)^{1+m}}{4 a^2 d (1+m)}+\frac{2^{-2-m} e^{-2 e+\frac{2 c f}{d}} (c+d x)^m \left (\frac{f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac{2 f (c+d x)}{d}\right )}{a^2 f}-\frac{4^{-2-m} e^{-4 e+\frac{4 c f}{d}} (c+d x)^m \left (\frac{f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac{4 f (c+d x)}{d}\right )}{a^2 f}\\ \end{align*}

Mathematica [A]  time = 8.10184, size = 194, normalized size = 1.28 \[ \frac{4^{-m-2} (c+d x)^m \text{csch}^2(e+f x) (\sinh (2 f x)+\cosh (2 f x)) \left (-\frac{f (c+d x)}{d}\right )^m \left (-\frac{f^2 (c+d x)^2}{d^2}\right )^{-m} \left (-d (m+1) e^{\frac{4 c f}{d}} (\cosh (2 e)-\sinh (2 e)) \text{Gamma}\left (m+1,\frac{4 f (c+d x)}{d}\right )+d 2^{m+2} (m+1) e^{\frac{2 c f}{d}} \text{Gamma}\left (m+1,\frac{2 f (c+d x)}{d}\right )+d 4^{m+1} (\sinh (e)+\cosh (e))^2 \left (\frac{f (c+d x)}{d}\right )^{m+1}\right )}{a^2 d f (m+1) (\coth (e+f x)+1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[(c + d*x)^m/(a + a*Coth[e + f*x])^2,x]

[Out]

(4^(-2 - m)*(c + d*x)^m*(-((f*(c + d*x))/d))^m*Csch[e + f*x]^2*(2^(2 + m)*d*E^((2*c*f)/d)*(1 + m)*Gamma[1 + m,
(2*f*(c + d*x))/d] + 4^(1 + m)*d*((f*(c + d*x))/d)^(1 + m)*(Cosh[e] + Sinh[e])^2 - d*E^((4*c*f)/d)*(1 + m)*Ga
mma[1 + m, (4*f*(c + d*x))/d]*(Cosh[2*e] - Sinh[2*e]))*(Cosh[2*f*x] + Sinh[2*f*x]))/(a^2*d*f*(1 + m)*(-((f^2*(
c + d*x)^2)/d^2))^m*(1 + Coth[e + f*x])^2)

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Maple [F]  time = 0.128, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( dx+c \right ) ^{m}}{ \left ( a+a{\rm coth} \left (fx+e\right ) \right ) ^{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^m/(a+a*coth(f*x+e))^2,x)

[Out]

int((d*x+c)^m/(a+a*coth(f*x+e))^2,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{m}}{{\left (a \coth \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x + c)^m/(a*coth(f*x + e) + a)^2, x)

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Fricas [A]  time = 2.34152, size = 609, normalized size = 4.01 \begin{align*} -\frac{{\left (d m + d\right )} \cosh \left (\frac{d m \log \left (\frac{4 \, f}{d}\right ) + 4 \, d e - 4 \, c f}{d}\right ) \Gamma \left (m + 1, \frac{4 \,{\left (d f x + c f\right )}}{d}\right ) - 4 \,{\left (d m + d\right )} \cosh \left (\frac{d m \log \left (\frac{2 \, f}{d}\right ) + 2 \, d e - 2 \, c f}{d}\right ) \Gamma \left (m + 1, \frac{2 \,{\left (d f x + c f\right )}}{d}\right ) -{\left (d m + d\right )} \Gamma \left (m + 1, \frac{4 \,{\left (d f x + c f\right )}}{d}\right ) \sinh \left (\frac{d m \log \left (\frac{4 \, f}{d}\right ) + 4 \, d e - 4 \, c f}{d}\right ) + 4 \,{\left (d m + d\right )} \Gamma \left (m + 1, \frac{2 \,{\left (d f x + c f\right )}}{d}\right ) \sinh \left (\frac{d m \log \left (\frac{2 \, f}{d}\right ) + 2 \, d e - 2 \, c f}{d}\right ) - 4 \,{\left (d f x + c f\right )} \cosh \left (m \log \left (d x + c\right )\right ) - 4 \,{\left (d f x + c f\right )} \sinh \left (m \log \left (d x + c\right )\right )}{16 \,{\left (a^{2} d f m + a^{2} d f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*((d*m + d)*cosh((d*m*log(4*f/d) + 4*d*e - 4*c*f)/d)*gamma(m + 1, 4*(d*f*x + c*f)/d) - 4*(d*m + d)*cosh((
d*m*log(2*f/d) + 2*d*e - 2*c*f)/d)*gamma(m + 1, 2*(d*f*x + c*f)/d) - (d*m + d)*gamma(m + 1, 4*(d*f*x + c*f)/d)
*sinh((d*m*log(4*f/d) + 4*d*e - 4*c*f)/d) + 4*(d*m + d)*gamma(m + 1, 2*(d*f*x + c*f)/d)*sinh((d*m*log(2*f/d) +
2*d*e - 2*c*f)/d) - 4*(d*f*x + c*f)*cosh(m*log(d*x + c)) - 4*(d*f*x + c*f)*sinh(m*log(d*x + c)))/(a^2*d*f*m +
a^2*d*f)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\left (c + d x\right )^{m}}{\coth ^{2}{\left (e + f x \right )} + 2 \coth{\left (e + f x \right )} + 1}\, dx}{a^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**m/(a+a*coth(f*x+e))**2,x)

[Out]

Integral((c + d*x)**m/(coth(e + f*x)**2 + 2*coth(e + f*x) + 1), x)/a**2

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (d x + c\right )}^{m}}{{\left (a \coth \left (f x + e\right ) + a\right )}^{2}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((d*x + c)^m/(a*coth(f*x + e) + a)^2, x)