3.20 \(\int \frac{c+d x}{(b \tanh (e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=1365 \[ \text{result too large to display} \]

[Out]

(2*d*ArcTan[Sqrt[b*Tanh[e + f*x]]/Sqrt[b]])/(b^(3/2)*f^2) - ((c + d*x)*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[-b]]
)/((-b)^(3/2)*f) - (d*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[-b]]^2)/(2*(-b)^(3/2)*f^2) + (2*d*ArcTanh[Sqrt[b*Tanh
[e + f*x]]/Sqrt[b]])/(b^(3/2)*f^2) + ((c + d*x)*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[b]])/(b^(3/2)*f) + (d*ArcTa
nh[Sqrt[b*Tanh[e + f*x]]/Sqrt[b]]^2)/(2*b^(3/2)*f^2) - (d*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[b]]*Log[(2*Sqrt[b
])/(Sqrt[b] - Sqrt[b*Tanh[e + f*x]])])/(b^(3/2)*f^2) + (d*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[b]]*Log[(2*Sqrt[b
])/(Sqrt[b] + Sqrt[b*Tanh[e + f*x]])])/(b^(3/2)*f^2) - (d*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[b]]*Log[(2*Sqrt[b
]*(Sqrt[-b] - Sqrt[b*Tanh[e + f*x]]))/((Sqrt[-b] - Sqrt[b])*(Sqrt[b] + Sqrt[b*Tanh[e + f*x]]))])/(2*b^(3/2)*f^
2) - (d*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[b]]*Log[(2*Sqrt[b]*(Sqrt[-b] + Sqrt[b*Tanh[e + f*x]]))/((Sqrt[-b] +
Sqrt[b])*(Sqrt[b] + Sqrt[b*Tanh[e + f*x]]))])/(2*b^(3/2)*f^2) + (d*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[-b]]*Lo
g[2/(1 - Sqrt[b*Tanh[e + f*x]]/Sqrt[-b])])/((-b)^(3/2)*f^2) - (d*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[-b]]*Log[(
2*(Sqrt[b] - Sqrt[b*Tanh[e + f*x]]))/((Sqrt[-b] + Sqrt[b])*(1 - Sqrt[b*Tanh[e + f*x]]/Sqrt[-b]))])/(2*(-b)^(3/
2)*f^2) - (d*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[-b]]*Log[(-2*(Sqrt[b] + Sqrt[b*Tanh[e + f*x]]))/((Sqrt[-b] - S
qrt[b])*(1 - Sqrt[b*Tanh[e + f*x]]/Sqrt[-b]))])/(2*(-b)^(3/2)*f^2) - (d*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[-b]
]*Log[2/(1 + Sqrt[b*Tanh[e + f*x]]/Sqrt[-b])])/((-b)^(3/2)*f^2) - (d*PolyLog[2, 1 - (2*Sqrt[b])/(Sqrt[b] - Sqr
t[b*Tanh[e + f*x]])])/(2*b^(3/2)*f^2) - (d*PolyLog[2, 1 - (2*Sqrt[b])/(Sqrt[b] + Sqrt[b*Tanh[e + f*x]])])/(2*b
^(3/2)*f^2) + (d*PolyLog[2, 1 - (2*Sqrt[b]*(Sqrt[-b] - Sqrt[b*Tanh[e + f*x]]))/((Sqrt[-b] - Sqrt[b])*(Sqrt[b]
+ Sqrt[b*Tanh[e + f*x]]))])/(4*b^(3/2)*f^2) + (d*PolyLog[2, 1 - (2*Sqrt[b]*(Sqrt[-b] + Sqrt[b*Tanh[e + f*x]]))
/((Sqrt[-b] + Sqrt[b])*(Sqrt[b] + Sqrt[b*Tanh[e + f*x]]))])/(4*b^(3/2)*f^2) + (d*PolyLog[2, 1 - 2/(1 - Sqrt[b*
Tanh[e + f*x]]/Sqrt[-b])])/(2*(-b)^(3/2)*f^2) - (d*PolyLog[2, 1 - (2*(Sqrt[b] - Sqrt[b*Tanh[e + f*x]]))/((Sqrt
[-b] + Sqrt[b])*(1 - Sqrt[b*Tanh[e + f*x]]/Sqrt[-b]))])/(4*(-b)^(3/2)*f^2) - (d*PolyLog[2, 1 + (2*(Sqrt[b] + S
qrt[b*Tanh[e + f*x]]))/((Sqrt[-b] - Sqrt[b])*(1 - Sqrt[b*Tanh[e + f*x]]/Sqrt[-b]))])/(4*(-b)^(3/2)*f^2) + (d*P
olyLog[2, 1 - 2/(1 + Sqrt[b*Tanh[e + f*x]]/Sqrt[-b])])/(2*(-b)^(3/2)*f^2) - (2*(c + d*x))/(b*f*Sqrt[b*Tanh[e +
f*x]])

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Rubi [F]  time = 0.110031, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \frac{c+d x}{(b \tanh (e+f x))^{3/2}} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(c + d*x)/(b*Tanh[e + f*x])^(3/2),x]

[Out]

(2*d*ArcTan[Sqrt[b*Tanh[e + f*x]]/Sqrt[b]])/(b^(3/2)*f^2) + (2*d*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[b]])/(b^(3
/2)*f^2) - (2*(c + d*x))/(b*f*Sqrt[b*Tanh[e + f*x]]) + Defer[Int][(c + d*x)*Sqrt[b*Tanh[e + f*x]], x]/b^2

Rubi steps

\begin{align*} \int \frac{c+d x}{(b \tanh (e+f x))^{3/2}} \, dx &=-\frac{2 (c+d x)}{b f \sqrt{b \tanh (e+f x)}}+\frac{\int (c+d x) \sqrt{b \tanh (e+f x)} \, dx}{b^2}+\frac{(2 d) \int \frac{1}{\sqrt{b \tanh (e+f x)}} \, dx}{b f}\\ &=-\frac{2 (c+d x)}{b f \sqrt{b \tanh (e+f x)}}+\frac{\int (c+d x) \sqrt{b \tanh (e+f x)} \, dx}{b^2}-\frac{(2 d) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (-b^2+x^2\right )} \, dx,x,b \tanh (e+f x)\right )}{f^2}\\ &=-\frac{2 (c+d x)}{b f \sqrt{b \tanh (e+f x)}}+\frac{\int (c+d x) \sqrt{b \tanh (e+f x)} \, dx}{b^2}-\frac{(4 d) \operatorname{Subst}\left (\int \frac{1}{-b^2+x^4} \, dx,x,\sqrt{b \tanh (e+f x)}\right )}{f^2}\\ &=-\frac{2 (c+d x)}{b f \sqrt{b \tanh (e+f x)}}+\frac{\int (c+d x) \sqrt{b \tanh (e+f x)} \, dx}{b^2}+\frac{(2 d) \operatorname{Subst}\left (\int \frac{1}{b-x^2} \, dx,x,\sqrt{b \tanh (e+f x)}\right )}{b f^2}+\frac{(2 d) \operatorname{Subst}\left (\int \frac{1}{b+x^2} \, dx,x,\sqrt{b \tanh (e+f x)}\right )}{b f^2}\\ &=\frac{2 d \tan ^{-1}\left (\frac{\sqrt{b \tanh (e+f x)}}{\sqrt{b}}\right )}{b^{3/2} f^2}+\frac{2 d \tanh ^{-1}\left (\frac{\sqrt{b \tanh (e+f x)}}{\sqrt{b}}\right )}{b^{3/2} f^2}-\frac{2 (c+d x)}{b f \sqrt{b \tanh (e+f x)}}+\frac{\int (c+d x) \sqrt{b \tanh (e+f x)} \, dx}{b^2}\\ \end{align*}

Mathematica [F]  time = 25.7603, size = 0, normalized size = 0. \[ \int \frac{c+d x}{(b \tanh (e+f x))^{3/2}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(c + d*x)/(b*Tanh[e + f*x])^(3/2),x]

[Out]

Integrate[(c + d*x)/(b*Tanh[e + f*x])^(3/2), x]

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Maple [F]  time = 0.055, size = 0, normalized size = 0. \begin{align*} \int{(dx+c) \left ( b\tanh \left ( fx+e \right ) \right ) ^{-{\frac{3}{2}}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)/(b*tanh(f*x+e))^(3/2),x)

[Out]

int((d*x+c)/(b*tanh(f*x+e))^(3/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d x + c}{\left (b \tanh \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(b*tanh(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

integrate((d*x + c)/(b*tanh(f*x + e))^(3/2), x)

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Fricas [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: UnboundLocalError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(b*tanh(f*x+e))^(3/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(b*tanh(f*x+e))**(3/2),x)

[Out]

Integral((c + d*x)/(b*tanh(e + f*x))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d x + c}{\left (b \tanh \left (f x + e\right )\right )^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)/(b*tanh(f*x+e))^(3/2),x, algorithm="giac")

[Out]

integrate((d*x + c)/(b*tanh(f*x + e))^(3/2), x)