3.1.21 \(\int (c+d x)^2 (b \tanh (e+f x))^{3/2} \, dx\) [21]

Optimal. Leaf size=1341 \[ \frac {4 (-b)^{3/2} d (c+d x) \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )}{f^2}+\frac {2 (-b)^{3/2} d^2 \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )^2}{f^3}+\frac {4 b^{3/2} d (c+d x) \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )}{f^2}+\frac {2 b^{3/2} d^2 \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )^2}{f^3}-\frac {4 b^{3/2} d^2 \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right ) \log \left (\frac {2 \sqrt {b}}{\sqrt {b}-\sqrt {b \tanh (e+f x)}}\right )}{f^3}+\frac {4 b^{3/2} d^2 \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right ) \log \left (\frac {2 \sqrt {b}}{\sqrt {b}+\sqrt {b \tanh (e+f x)}}\right )}{f^3}-\frac {2 b^{3/2} d^2 \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right ) \log \left (\frac {2 \sqrt {b} \left (\sqrt {-b}-\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}-\sqrt {b}\right ) \left (\sqrt {b}+\sqrt {b \tanh (e+f x)}\right )}\right )}{f^3}-\frac {2 b^{3/2} d^2 \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right ) \log \left (\frac {2 \sqrt {b} \left (\sqrt {-b}+\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}+\sqrt {b}\right ) \left (\sqrt {b}+\sqrt {b \tanh (e+f x)}\right )}\right )}{f^3}-\frac {4 (-b)^{3/2} d^2 \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right ) \log \left (\frac {2}{1-\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}}\right )}{f^3}+\frac {2 (-b)^{3/2} d^2 \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right ) \log \left (\frac {2 \left (\sqrt {b}-\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}+\sqrt {b}\right ) \left (1-\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )}\right )}{f^3}+\frac {2 (-b)^{3/2} d^2 \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right ) \log \left (-\frac {2 \left (\sqrt {b}+\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}-\sqrt {b}\right ) \left (1-\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )}\right )}{f^3}+\frac {4 (-b)^{3/2} d^2 \tanh ^{-1}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right ) \log \left (\frac {2}{1+\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}}\right )}{f^3}-\frac {2 b^{3/2} d^2 \text {PolyLog}\left (2,1-\frac {2 \sqrt {b}}{\sqrt {b}-\sqrt {b \tanh (e+f x)}}\right )}{f^3}-\frac {2 b^{3/2} d^2 \text {PolyLog}\left (2,1-\frac {2 \sqrt {b}}{\sqrt {b}+\sqrt {b \tanh (e+f x)}}\right )}{f^3}+\frac {b^{3/2} d^2 \text {PolyLog}\left (2,1-\frac {2 \sqrt {b} \left (\sqrt {-b}-\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}-\sqrt {b}\right ) \left (\sqrt {b}+\sqrt {b \tanh (e+f x)}\right )}\right )}{f^3}+\frac {b^{3/2} d^2 \text {PolyLog}\left (2,1-\frac {2 \sqrt {b} \left (\sqrt {-b}+\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}+\sqrt {b}\right ) \left (\sqrt {b}+\sqrt {b \tanh (e+f x)}\right )}\right )}{f^3}-\frac {2 (-b)^{3/2} d^2 \text {PolyLog}\left (2,1-\frac {2}{1-\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}}\right )}{f^3}+\frac {(-b)^{3/2} d^2 \text {PolyLog}\left (2,1-\frac {2 \left (\sqrt {b}-\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}+\sqrt {b}\right ) \left (1-\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )}\right )}{f^3}+\frac {(-b)^{3/2} d^2 \text {PolyLog}\left (2,1+\frac {2 \left (\sqrt {b}+\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}-\sqrt {b}\right ) \left (1-\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )}\right )}{f^3}-\frac {2 (-b)^{3/2} d^2 \text {PolyLog}\left (2,1-\frac {2}{1+\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}}\right )}{f^3}-\frac {2 b (c+d x)^2 \sqrt {b \tanh (e+f x)}}{f}+b^2 \text {Int}\left (\frac {(c+d x)^2}{\sqrt {b \tanh (e+f x)}},x\right ) \]

[Out]

4*(-b)^(3/2)*d*(d*x+c)*arctanh((b*tanh(f*x+e))^(1/2)/(-b)^(1/2))/f^2+2*(-b)^(3/2)*d^2*arctanh((b*tanh(f*x+e))^
(1/2)/(-b)^(1/2))^2/f^3+4*b^(3/2)*d*(d*x+c)*arctanh((b*tanh(f*x+e))^(1/2)/b^(1/2))/f^2+2*b^(3/2)*d^2*arctanh((
b*tanh(f*x+e))^(1/2)/b^(1/2))^2/f^3-4*b^(3/2)*d^2*arctanh((b*tanh(f*x+e))^(1/2)/b^(1/2))*ln(2*b^(1/2)/(b^(1/2)
-(b*tanh(f*x+e))^(1/2)))/f^3+4*b^(3/2)*d^2*arctanh((b*tanh(f*x+e))^(1/2)/b^(1/2))*ln(2*b^(1/2)/(b^(1/2)+(b*tan
h(f*x+e))^(1/2)))/f^3-2*b^(3/2)*d^2*arctanh((b*tanh(f*x+e))^(1/2)/b^(1/2))*ln(2*b^(1/2)*((-b)^(1/2)-(b*tanh(f*
x+e))^(1/2))/((-b)^(1/2)-b^(1/2))/(b^(1/2)+(b*tanh(f*x+e))^(1/2)))/f^3-2*b^(3/2)*d^2*arctanh((b*tanh(f*x+e))^(
1/2)/b^(1/2))*ln(2*b^(1/2)*((-b)^(1/2)+(b*tanh(f*x+e))^(1/2))/((-b)^(1/2)+b^(1/2))/(b^(1/2)+(b*tanh(f*x+e))^(1
/2)))/f^3-4*(-b)^(3/2)*d^2*arctanh((b*tanh(f*x+e))^(1/2)/(-b)^(1/2))*ln(2/(1-(b*tanh(f*x+e))^(1/2)/(-b)^(1/2))
)/f^3+2*(-b)^(3/2)*d^2*arctanh((b*tanh(f*x+e))^(1/2)/(-b)^(1/2))*ln(2*(b^(1/2)-(b*tanh(f*x+e))^(1/2))/((-b)^(1
/2)+b^(1/2))/(1-(b*tanh(f*x+e))^(1/2)/(-b)^(1/2)))/f^3+2*(-b)^(3/2)*d^2*arctanh((b*tanh(f*x+e))^(1/2)/(-b)^(1/
2))*ln(-2*(b^(1/2)+(b*tanh(f*x+e))^(1/2))/((-b)^(1/2)-b^(1/2))/(1-(b*tanh(f*x+e))^(1/2)/(-b)^(1/2)))/f^3+4*(-b
)^(3/2)*d^2*arctanh((b*tanh(f*x+e))^(1/2)/(-b)^(1/2))*ln(2/(1+(b*tanh(f*x+e))^(1/2)/(-b)^(1/2)))/f^3-2*b^(3/2)
*d^2*polylog(2,1-2*b^(1/2)/(b^(1/2)-(b*tanh(f*x+e))^(1/2)))/f^3-2*b^(3/2)*d^2*polylog(2,1-2*b^(1/2)/(b^(1/2)+(
b*tanh(f*x+e))^(1/2)))/f^3+b^(3/2)*d^2*polylog(2,1-2*b^(1/2)*((-b)^(1/2)-(b*tanh(f*x+e))^(1/2))/((-b)^(1/2)-b^
(1/2))/(b^(1/2)+(b*tanh(f*x+e))^(1/2)))/f^3+b^(3/2)*d^2*polylog(2,1-2*b^(1/2)*((-b)^(1/2)+(b*tanh(f*x+e))^(1/2
))/((-b)^(1/2)+b^(1/2))/(b^(1/2)+(b*tanh(f*x+e))^(1/2)))/f^3-2*(-b)^(3/2)*d^2*polylog(2,1-2/(1-(b*tanh(f*x+e))
^(1/2)/(-b)^(1/2)))/f^3+(-b)^(3/2)*d^2*polylog(2,1-2*(b^(1/2)-(b*tanh(f*x+e))^(1/2))/((-b)^(1/2)+b^(1/2))/(1-(
b*tanh(f*x+e))^(1/2)/(-b)^(1/2)))/f^3+(-b)^(3/2)*d^2*polylog(2,1+2*(b^(1/2)+(b*tanh(f*x+e))^(1/2))/((-b)^(1/2)
-b^(1/2))/(1-(b*tanh(f*x+e))^(1/2)/(-b)^(1/2)))/f^3-2*(-b)^(3/2)*d^2*polylog(2,1-2/(1+(b*tanh(f*x+e))^(1/2)/(-
b)^(1/2)))/f^3-2*b*(d*x+c)^2*(b*tanh(f*x+e))^(1/2)/f+b^2*Unintegrable((d*x+c)^2/(b*tanh(f*x+e))^(1/2),x)

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Rubi [A]
time = 1.16, antiderivative size = 0, normalized size of antiderivative = 0.00, number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used = {} \begin {gather*} \int (c+d x)^2 (b \tanh (e+f x))^{3/2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

(4*(-b)^(3/2)*d*(c + d*x)*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[-b]])/f^2 + (2*(-b)^(3/2)*d^2*ArcTanh[Sqrt[b*Tanh
[e + f*x]]/Sqrt[-b]]^2)/f^3 + (4*b^(3/2)*d*(c + d*x)*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[b]])/f^2 + (2*b^(3/2)*
d^2*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[b]]^2)/f^3 - (4*b^(3/2)*d^2*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[b]]*Log[
(2*Sqrt[b])/(Sqrt[b] - Sqrt[b*Tanh[e + f*x]])])/f^3 + (4*b^(3/2)*d^2*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[b]]*Lo
g[(2*Sqrt[b])/(Sqrt[b] + Sqrt[b*Tanh[e + f*x]])])/f^3 - (2*b^(3/2)*d^2*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]]))])/
f^3 - (2*b^(3/2)*d^2*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]]))])/f^3 - (4*(-b)^(3/2)*d^2*ArcTanh[Sqrt[b*Tanh[e + f*
x]]/Sqrt[-b]]*Log[2/(1 - Sqrt[b*Tanh[e + f*x]]/Sqrt[-b])])/f^3 + (2*(-b)^(3/2)*d^2*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]))])/f^3 + (2*(-b)^(3/2)*d^2*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]))])/f^3 + (4*(-b)^(3/2)*d^2*ArcTanh[Sqrt[b*Tanh[
e + f*x]]/Sqrt[-b]]*Log[2/(1 + Sqrt[b*Tanh[e + f*x]]/Sqrt[-b])])/f^3 - (2*b^(3/2)*d^2*PolyLog[2, 1 - (2*Sqrt[b
])/(Sqrt[b] - Sqrt[b*Tanh[e + f*x]])])/f^3 - (2*b^(3/2)*d^2*PolyLog[2, 1 - (2*Sqrt[b])/(Sqrt[b] + Sqrt[b*Tanh[
e + f*x]])])/f^3 + (b^(3/2)*d^2*PolyLog[2, 1 - (2*Sqrt[b]*(Sqrt[-b] - Sqrt[b*Tanh[e + f*x]]))/((Sqrt[-b] - Sqr
t[b])*(Sqrt[b] + Sqrt[b*Tanh[e + f*x]]))])/f^3 + (b^(3/2)*d^2*PolyLog[2, 1 - (2*Sqrt[b]*(Sqrt[-b] + Sqrt[b*Tan
h[e + f*x]]))/((Sqrt[-b] + Sqrt[b])*(Sqrt[b] + Sqrt[b*Tanh[e + f*x]]))])/f^3 - (2*(-b)^(3/2)*d^2*PolyLog[2, 1
- 2/(1 - Sqrt[b*Tanh[e + f*x]]/Sqrt[-b])])/f^3 + ((-b)^(3/2)*d^2*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]))])/f^3 + ((-b)^(3/2)*d^2*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]))])/f^3 - (2*(-b)
^(3/2)*d^2*PolyLog[2, 1 - 2/(1 + Sqrt[b*Tanh[e + f*x]]/Sqrt[-b])])/f^3 - (2*b*(c + d*x)^2*Sqrt[b*Tanh[e + f*x]
])/f + b^2*Defer[Int][(c + d*x)^2/Sqrt[b*Tanh[e + f*x]], x]

Rubi steps

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

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Mathematica [A]
time = 19.22, size = 0, normalized size = 0.00 \begin {gather*} \int (c+d x)^2 (b \tanh (e+f x))^{3/2} \, dx \end {gather*}

Verification is not applicable to the result.

[In]

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

[Out]

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

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Maple [A]
time = 180.00, size = 0, normalized size = 0.00 \[\int \left (d x +c \right )^{2} \left (b \tanh \left (f x +e \right )\right )^{\frac {3}{2}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
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)^2*(b*tanh(f*x+e))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: TypeError >>  Error detected within library code:   integrate: implementation incomplete (co
nstant residues)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
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)^2*(b*tanh(f*x+e))^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [A]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int {\left (b\,\mathrm {tanh}\left (e+f\,x\right )\right )}^{3/2}\,{\left (c+d\,x\right )}^2 \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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