3.16 \(\int (c+d x) (b \tanh (e+f x))^{5/2} \, dx\)

Optimal. Leaf size=1392 \[ \text{result too large to display} \]

[Out]

(2*b^(5/2)*d*ArcTan[Sqrt[b*Tanh[e + f*x]]/Sqrt[b]])/(3*f^2) - ((-b)^(5/2)*(c + d*x)*ArcTanh[Sqrt[b*Tanh[e + f*
x]]/Sqrt[-b]])/f - ((-b)^(5/2)*d*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[-b]]^2)/(2*f^2) + (2*b^(5/2)*d*ArcTanh[Sqr
t[b*Tanh[e + f*x]]/Sqrt[b]])/(3*f^2) + (b^(5/2)*(c + d*x)*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[b]])/f + (b^(5/2)
*d*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[b]]^2)/(2*f^2) - (b^(5/2)*d*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[b]]*Log[(
2*Sqrt[b])/(Sqrt[b] - Sqrt[b*Tanh[e + f*x]])])/f^2 + (b^(5/2)*d*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[b]]*Log[(2*
Sqrt[b])/(Sqrt[b] + Sqrt[b*Tanh[e + f*x]])])/f^2 - (b^(5/2)*d*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[b]]*Log[(2*Sq
rt[b]*(Sqrt[-b] - Sqrt[b*Tanh[e + f*x]]))/((Sqrt[-b] - Sqrt[b])*(Sqrt[b] + Sqrt[b*Tanh[e + f*x]]))])/(2*f^2) -
(b^(5/2)*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*f^2) + ((-b)^(5/2)*d*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[
-b]]*Log[2/(1 - Sqrt[b*Tanh[e + f*x]]/Sqrt[-b])])/f^2 - ((-b)^(5/2)*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*f^2
) - ((-b)^(5/2)*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*f^2) - ((-b)^(5/2)*d*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqr
t[-b]]*Log[2/(1 + Sqrt[b*Tanh[e + f*x]]/Sqrt[-b])])/f^2 - (b^(5/2)*d*PolyLog[2, 1 - (2*Sqrt[b])/(Sqrt[b] - Sqr
t[b*Tanh[e + f*x]])])/(2*f^2) - (b^(5/2)*d*PolyLog[2, 1 - (2*Sqrt[b])/(Sqrt[b] + Sqrt[b*Tanh[e + f*x]])])/(2*f
^2) + (b^(5/2)*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*f^2) + (b^(5/2)*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*f^2) + ((-b)^(5/2)*d*PolyLog[2, 1 - 2/(1 - Sqrt
[b*Tanh[e + f*x]]/Sqrt[-b])])/(2*f^2) - ((-b)^(5/2)*d*PolyLog[2, 1 - (2*(Sqrt[b] - Sqrt[b*Tanh[e + f*x]]))/((S
qrt[-b] + Sqrt[b])*(1 - Sqrt[b*Tanh[e + f*x]]/Sqrt[-b]))])/(4*f^2) - ((-b)^(5/2)*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*f^2) + ((-b)^(5/2)*
d*PolyLog[2, 1 - 2/(1 + Sqrt[b*Tanh[e + f*x]]/Sqrt[-b])])/(2*f^2) - (4*b^2*d*Sqrt[b*Tanh[e + f*x]])/(3*f^2) -
(2*b*(c + d*x)*(b*Tanh[e + f*x])^(3/2))/(3*f)

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Rubi [F]  time = 0.131579, 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 = {} \[ \int (c+d x) (b \tanh (e+f x))^{5/2} \, dx \]

Verification is Not applicable to the result.

[In]

Int[(c + d*x)*(b*Tanh[e + f*x])^(5/2),x]

[Out]

(2*b^(5/2)*d*ArcTan[Sqrt[b*Tanh[e + f*x]]/Sqrt[b]])/(3*f^2) + (2*b^(5/2)*d*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[
b]])/(3*f^2) - (4*b^2*d*Sqrt[b*Tanh[e + f*x]])/(3*f^2) - (2*b*(c + d*x)*(b*Tanh[e + f*x])^(3/2))/(3*f) + b^2*D
efer[Int][(c + d*x)*Sqrt[b*Tanh[e + f*x]], x]

Rubi steps

\begin{align*} \int (c+d x) (b \tanh (e+f x))^{5/2} \, dx &=-\frac{2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}+b^2 \int (c+d x) \sqrt{b \tanh (e+f x)} \, dx+\frac{(2 b d) \int (b \tanh (e+f x))^{3/2} \, dx}{3 f}\\ &=-\frac{4 b^2 d \sqrt{b \tanh (e+f x)}}{3 f^2}-\frac{2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}+b^2 \int (c+d x) \sqrt{b \tanh (e+f x)} \, dx+\frac{\left (2 b^3 d\right ) \int \frac{1}{\sqrt{b \tanh (e+f x)}} \, dx}{3 f}\\ &=-\frac{4 b^2 d \sqrt{b \tanh (e+f x)}}{3 f^2}-\frac{2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}+b^2 \int (c+d x) \sqrt{b \tanh (e+f x)} \, dx-\frac{\left (2 b^4 d\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (-b^2+x^2\right )} \, dx,x,b \tanh (e+f x)\right )}{3 f^2}\\ &=-\frac{4 b^2 d \sqrt{b \tanh (e+f x)}}{3 f^2}-\frac{2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}+b^2 \int (c+d x) \sqrt{b \tanh (e+f x)} \, dx-\frac{\left (4 b^4 d\right ) \operatorname{Subst}\left (\int \frac{1}{-b^2+x^4} \, dx,x,\sqrt{b \tanh (e+f x)}\right )}{3 f^2}\\ &=-\frac{4 b^2 d \sqrt{b \tanh (e+f x)}}{3 f^2}-\frac{2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}+b^2 \int (c+d x) \sqrt{b \tanh (e+f x)} \, dx+\frac{\left (2 b^3 d\right ) \operatorname{Subst}\left (\int \frac{1}{b-x^2} \, dx,x,\sqrt{b \tanh (e+f x)}\right )}{3 f^2}+\frac{\left (2 b^3 d\right ) \operatorname{Subst}\left (\int \frac{1}{b+x^2} \, dx,x,\sqrt{b \tanh (e+f x)}\right )}{3 f^2}\\ &=\frac{2 b^{5/2} d \tan ^{-1}\left (\frac{\sqrt{b \tanh (e+f x)}}{\sqrt{b}}\right )}{3 f^2}+\frac{2 b^{5/2} d \tanh ^{-1}\left (\frac{\sqrt{b \tanh (e+f x)}}{\sqrt{b}}\right )}{3 f^2}-\frac{4 b^2 d \sqrt{b \tanh (e+f x)}}{3 f^2}-\frac{2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}+b^2 \int (c+d x) \sqrt{b \tanh (e+f x)} \, dx\\ \end{align*}

Mathematica [F]  time = 36.7754, size = 0, normalized size = 0. \[ \int (c+d x) (b \tanh (e+f x))^{5/2} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[(c + d*x)*(b*Tanh[e + f*x])^(5/2),x]

[Out]

Integrate[(c + d*x)*(b*Tanh[e + f*x])^(5/2), x]

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Maple [F]  time = 0.08, size = 0, normalized size = 0. \begin{align*} \int \left ( dx+c \right ) \left ( b\tanh \left ( fx+e \right ) \right ) ^{{\frac{5}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*x+c)*(b*tanh(f*x+e))^(5/2),x)

[Out]

int((d*x+c)*(b*tanh(f*x+e))^(5/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (d x + c\right )} \left (b \tanh \left (f x + e\right )\right )^{\frac{5}{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))^(5/2),x, algorithm="maxima")

[Out]

integrate((d*x + c)*(b*tanh(f*x + e))^(5/2), 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))^(5/2),x, algorithm="fricas")

[Out]

Exception raised: UnboundLocalError

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \tanh{\left (e + f x \right )}\right )^{\frac{5}{2}} \left (c + d x\right )\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*(b*tanh(f*x+e))**(5/2),x)

[Out]

Integral((b*tanh(e + f*x))**(5/2)*(c + d*x), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*x+c)*(b*tanh(f*x+e))^(5/2),x, algorithm="giac")

[Out]

Exception raised: RuntimeError