3.1.52 \(\int \frac {(c+d x)^m}{(a+a \tanh (e+f x))^3} \, dx\) [52]

Optimal. Leaf size=224 \[ \frac {(c+d x)^{1+m}}{8 a^3 d (1+m)}-\frac {3\ 2^{-4-m} e^{-2 e+\frac {2 c f}{d}} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {2 f (c+d x)}{d}\right )}{a^3 f}-\frac {3\ 2^{-5-2 m} e^{-4 e+\frac {4 c f}{d}} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {4 f (c+d x)}{d}\right )}{a^3 f}-\frac {2^{-4-m} 3^{-1-m} e^{-6 e+\frac {6 c f}{d}} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {6 f (c+d x)}{d}\right )}{a^3 f} \]

[Out]

1/8*(d*x+c)^(1+m)/a^3/d/(1+m)-3*2^(-4-m)*exp(-2*e+2*c*f/d)*(d*x+c)^m*GAMMA(1+m,2*f*(d*x+c)/d)/a^3/f/((f*(d*x+c
)/d)^m)-3*2^(-5-2*m)*exp(-4*e+4*c*f/d)*(d*x+c)^m*GAMMA(1+m,4*f*(d*x+c)/d)/a^3/f/((f*(d*x+c)/d)^m)-2^(-4-m)*3^(
-1-m)*exp(-6*e+6*c*f/d)*(d*x+c)^m*GAMMA(1+m,6*f*(d*x+c)/d)/a^3/f/((f*(d*x+c)/d)^m)

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Rubi [A]
time = 0.17, antiderivative size = 224, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 2, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.100, Rules used = {3810, 2212} \begin {gather*} -\frac {3\ 2^{-m-4} 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^3 f}-\frac {3\ 2^{-2 m-5} 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^3 f}-\frac {2^{-m-4} 3^{-m-1} e^{\frac {6 c f}{d}-6 e} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \text {Gamma}\left (m+1,\frac {6 f (c+d x)}{d}\right )}{a^3 f}+\frac {(c+d x)^{m+1}}{8 a^3 d (m+1)} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(c + d*x)^m/(a + a*Tanh[e + f*x])^3,x]

[Out]

(c + d*x)^(1 + m)/(8*a^3*d*(1 + m)) - (3*2^(-4 - m)*E^(-2*e + (2*c*f)/d)*(c + d*x)^m*Gamma[1 + m, (2*f*(c + d*
x))/d])/(a^3*f*((f*(c + d*x))/d)^m) - (3*2^(-5 - 2*m)*E^(-4*e + (4*c*f)/d)*(c + d*x)^m*Gamma[1 + m, (4*f*(c +
d*x))/d])/(a^3*f*((f*(c + d*x))/d)^m) - (2^(-4 - m)*3^(-1 - m)*E^(-6*e + (6*c*f)/d)*(c + d*x)^m*Gamma[1 + m, (
6*f*(c + d*x))/d])/(a^3*f*((f*(c + d*x))/d)^m)

Rule 2212

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

Rule 3810

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/b)*(e + f*x))/(2*a))^(-n), x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && EqQ[a^2
+ b^2, 0] && ILtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {(c+d x)^m}{(a+a \tanh (e+f x))^3} \, dx &=\int \left (\frac {(c+d x)^m}{8 a^3}+\frac {e^{-6 e-6 f x} (c+d x)^m}{8 a^3}+\frac {3 e^{-4 e-4 f x} (c+d x)^m}{8 a^3}+\frac {3 e^{-2 e-2 f x} (c+d x)^m}{8 a^3}\right ) \, dx\\ &=\frac {(c+d x)^{1+m}}{8 a^3 d (1+m)}+\frac {\int e^{-6 e-6 f x} (c+d x)^m \, dx}{8 a^3}+\frac {3 \int e^{-4 e-4 f x} (c+d x)^m \, dx}{8 a^3}+\frac {3 \int e^{-2 e-2 f x} (c+d x)^m \, dx}{8 a^3}\\ &=\frac {(c+d x)^{1+m}}{8 a^3 d (1+m)}-\frac {3\ 2^{-4-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^3 f}-\frac {3\ 2^{-5-2 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^3 f}-\frac {2^{-4-m} 3^{-1-m} e^{-6 e+\frac {6 c f}{d}} (c+d x)^m \left (\frac {f (c+d x)}{d}\right )^{-m} \Gamma \left (1+m,\frac {6 f (c+d x)}{d}\right )}{a^3 f}\\ \end {align*}

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Mathematica [F]
time = 180.00, size = 0, normalized size = 0.00 \begin {gather*} \text {\$Aborted} \end {gather*}

Verification is not applicable to the result.

[In]

Integrate[(c + d*x)^m/(a + a*Tanh[e + f*x])^3,x]

[Out]

$Aborted

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Maple [F]
time = 180.00, size = 0, normalized size = 0.00 \[\int \frac {\left (d x +c \right )^{m}}{\left (a +a \tanh \left (f x +e \right )\right )^{3}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)^m/(a+a*tanh(f*x+e))^3,x)

[Out]

int((d*x+c)^m/(a+a*tanh(f*x+e))^3,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((d*x+c)^m/(a+a*tanh(f*x+e))^3,x, algorithm="maxima")

[Out]

integrate((d*x + c)^m/(a*tanh(f*x + e) + a)^3, x)

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Fricas [A]
time = 0.10, size = 381, normalized size = 1.70 \begin {gather*} -\frac {2 \, {\left (d m + d\right )} \cosh \left (\frac {d m \log \left (\frac {6 \, f}{d}\right ) - 6 \, c f + 6 \, d \cosh \left (1\right ) + 6 \, d \sinh \left (1\right )}{d}\right ) \Gamma \left (m + 1, \frac {6 \, {\left (d f x + c f\right )}}{d}\right ) + 9 \, {\left (d m + d\right )} \cosh \left (\frac {d m \log \left (\frac {4 \, f}{d}\right ) - 4 \, c f + 4 \, d \cosh \left (1\right ) + 4 \, d \sinh \left (1\right )}{d}\right ) \Gamma \left (m + 1, \frac {4 \, {\left (d f x + c f\right )}}{d}\right ) + 18 \, {\left (d m + d\right )} \cosh \left (\frac {d m \log \left (\frac {2 \, f}{d}\right ) - 2 \, c f + 2 \, d \cosh \left (1\right ) + 2 \, d \sinh \left (1\right )}{d}\right ) \Gamma \left (m + 1, \frac {2 \, {\left (d f x + c f\right )}}{d}\right ) - 2 \, {\left (d m + d\right )} \Gamma \left (m + 1, \frac {6 \, {\left (d f x + c f\right )}}{d}\right ) \sinh \left (\frac {d m \log \left (\frac {6 \, f}{d}\right ) - 6 \, c f + 6 \, d \cosh \left (1\right ) + 6 \, d \sinh \left (1\right )}{d}\right ) - 9 \, {\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 \, c f + 4 \, d \cosh \left (1\right ) + 4 \, d \sinh \left (1\right )}{d}\right ) - 18 \, {\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 \, c f + 2 \, d \cosh \left (1\right ) + 2 \, d \sinh \left (1\right )}{d}\right ) - 12 \, {\left (d f x + c f\right )} \cosh \left (m \log \left (d x + c\right )\right ) - 12 \, {\left (d f x + c f\right )} \sinh \left (m \log \left (d x + c\right )\right )}{96 \, {\left (a^{3} d f m + a^{3} d f\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)^m/(a+a*tanh(f*x+e))^3,x, algorithm="fricas")

[Out]

-1/96*(2*(d*m + d)*cosh((d*m*log(6*f/d) - 6*c*f + 6*d*cosh(1) + 6*d*sinh(1))/d)*gamma(m + 1, 6*(d*f*x + c*f)/d
) + 9*(d*m + d)*cosh((d*m*log(4*f/d) - 4*c*f + 4*d*cosh(1) + 4*d*sinh(1))/d)*gamma(m + 1, 4*(d*f*x + c*f)/d) +
18*(d*m + d)*cosh((d*m*log(2*f/d) - 2*c*f + 2*d*cosh(1) + 2*d*sinh(1))/d)*gamma(m + 1, 2*(d*f*x + c*f)/d) - 2
*(d*m + d)*gamma(m + 1, 6*(d*f*x + c*f)/d)*sinh((d*m*log(6*f/d) - 6*c*f + 6*d*cosh(1) + 6*d*sinh(1))/d) - 9*(d
*m + d)*gamma(m + 1, 4*(d*f*x + c*f)/d)*sinh((d*m*log(4*f/d) - 4*c*f + 4*d*cosh(1) + 4*d*sinh(1))/d) - 18*(d*m
+ d)*gamma(m + 1, 2*(d*f*x + c*f)/d)*sinh((d*m*log(2*f/d) - 2*c*f + 2*d*cosh(1) + 2*d*sinh(1))/d) - 12*(d*f*x
+ c*f)*cosh(m*log(d*x + c)) - 12*(d*f*x + c*f)*sinh(m*log(d*x + c)))/(a^3*d*f*m + a^3*d*f)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\left (c + d x\right )^{m}}{\tanh ^{3}{\left (e + f x \right )} + 3 \tanh ^{2}{\left (e + f x \right )} + 3 \tanh {\left (e + f x \right )} + 1}\, dx}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)**m/(a+a*tanh(f*x+e))**3,x)

[Out]

Integral((c + d*x)**m/(tanh(e + f*x)**3 + 3*tanh(e + f*x)**2 + 3*tanh(e + f*x) + 1), x)/a**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((d*x+c)^m/(a+a*tanh(f*x+e))^3,x, algorithm="giac")

[Out]

integrate((d*x + c)^m/(a*tanh(f*x + e) + a)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {{\left (c+d\,x\right )}^m}{{\left (a+a\,\mathrm {tanh}\left (e+f\,x\right )\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c + d*x)^m/(a + a*tanh(e + f*x))^3,x)

[Out]

int((c + d*x)^m/(a + a*tanh(e + f*x))^3, x)

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