3.50 \(\int \frac{(c+d x)^m}{a+a \tanh (e+f x)} \, dx\)

Optimal. Leaf size=89 \[ \frac{(c+d x)^{m+1}}{2 a d (m+1)}-\frac{2^{-m-2} e^{\frac{2 c f}{d}-2 e} (c+d x)^m \left (\frac{f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{2 f (c+d x)}{d}\right )}{a f} \]

[Out]

(c + d*x)^(1 + m)/(2*a*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*f*((f*(c + d*x))/d)^m)

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Rubi [A]  time = 0.114647, antiderivative size = 89, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.1, Rules used = {3727, 2181} \[ \frac{(c+d x)^{m+1}}{2 a d (m+1)}-\frac{2^{-m-2} e^{\frac{2 c f}{d}-2 e} (c+d x)^m \left (\frac{f (c+d x)}{d}\right )^{-m} \text{Gamma}\left (m+1,\frac{2 f (c+d x)}{d}\right )}{a f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(c + d*x)^(1 + m)/(2*a*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*f*((f*(c + d*x))/d)^m)

Rule 3727

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

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 \tanh (e+f x)} \, dx &=\frac{(c+d x)^{1+m}}{2 a d (1+m)}+\frac{\int e^{2 i (i e+i f x)} (c+d x)^m \, dx}{2 a}\\ &=\frac{(c+d x)^{1+m}}{2 a 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 f}\\ \end{align*}

Mathematica [B]  time = 1.1318, size = 186, normalized size = 2.09 \[ \frac{2^{-m-2} (c+d x)^m \text{sech}(e+f x) \left (-\frac{f (c+d x)}{d}\right )^m \left (-\frac{f^2 (c+d x)^2}{d^2}\right )^{-m} \left (\sinh \left (f \left (\frac{c}{d}+x\right )\right )+\cosh \left (f \left (\frac{c}{d}+x\right )\right )\right ) \left (d (m+1) \left (\sinh \left (e-\frac{c f}{d}\right )-\cosh \left (e-\frac{c f}{d}\right )\right ) \text{Gamma}\left (m+1,\frac{2 f (c+d x)}{d}\right )+f 2^{m+1} (c+d x) \left (f \left (\frac{c}{d}+x\right )\right )^m \left (\sinh \left (e-\frac{c f}{d}\right )+\cosh \left (e-\frac{c f}{d}\right )\right )\right )}{a d f (m+1) (\tanh (e+f x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2^(-2 - m)*(c + d*x)^m*(-((f*(c + d*x))/d))^m*Sech[e + f*x]*(d*(1 + m)*Gamma[1 + m, (2*f*(c + d*x))/d]*(-Cosh
[e - (c*f)/d] + Sinh[e - (c*f)/d]) + 2^(1 + m)*f*(f*(c/d + x))^m*(c + d*x)*(Cosh[e - (c*f)/d] + Sinh[e - (c*f)
/d]))*(Cosh[f*(c/d + x)] + Sinh[f*(c/d + x)]))/(a*d*f*(1 + m)*(-((f^2*(c + d*x)^2)/d^2))^m*(1 + Tanh[e + f*x])
)

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Maple [F]  time = 0.109, size = 0, normalized size = 0. \begin{align*} \int{\frac{ \left ( dx+c \right ) ^{m}}{a+a\tanh \left ( fx+e \right ) }}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.33017, size = 367, normalized size = 4.12 \begin{align*} -\frac{{\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{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 ) - 2 \,{\left (d f x + c f\right )} \cosh \left (m \log \left (d x + c\right )\right ) - 2 \,{\left (d f x + c f\right )} \sinh \left (m \log \left (d x + c\right )\right )}{4 \,{\left (a d f m + a d f\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/4*((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, 2*(d*f*x + c*f)/d)*sinh((d*m*log(2*f/d) + 2*d*e - 2*c*f)/d) - 2*(d*f*x + c*f)*cosh(m*log(d*x + c)) - 2*(d*
f*x + c*f)*sinh(m*log(d*x + c)))/(a*d*f*m + a*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}}{\tanh{\left (e + f x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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