\(\int (c+d x) (b \tanh (e+f x))^{5/2} \, dx\) [16]

Optimal result

Integrand size = 18, antiderivative size = 1392 \[ \int (c+d x) (b \tanh (e+f x))^{5/2} \, dx =\text {Too large to display} \] Output:

2/3*b^(5/2)*d*arctan((b*tanh(f*x+e))^(1/2)/b^(1/2))/f^2-(-b)^(5/2)*(d*x+c) 
*arctanh((b*tanh(f*x+e))^(1/2)/(-b)^(1/2))/f-1/2*(-b)^(5/2)*d*arctanh((b*t 
anh(f*x+e))^(1/2)/(-b)^(1/2))^2/f^2+2/3*b^(5/2)*d*arctanh((b*tanh(f*x+e))^ 
(1/2)/b^(1/2))/f^2+b^(5/2)*(d*x+c)*arctanh((b*tanh(f*x+e))^(1/2)/b^(1/2))/ 
f+1/2*b^(5/2)*d*arctanh((b*tanh(f*x+e))^(1/2)/b^(1/2))^2/f^2-b^(5/2)*d*arc 
tanh((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^2+b^(5/2)*d*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^2-1/2*b^(5/2)*d*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^2-1/2*b^(5/2)*d*arctanh((b*t 
anh(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^2+(-b)^(5/2)*d*ar 
ctanh((b*tanh(f*x+e))^(1/2)/(-b)^(1/2))*ln(2/(1-(b*tanh(f*x+e))^(1/2)/(-b) 
^(1/2)))/f^2-1/2*(-b)^(5/2)*d*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^2-1/2*(-b)^(5/2)*d*arctanh((b*tanh(f*x+e))^(1/2)/(-b 
)^(1/2))*ln((-2*b^(1/2)-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^2-(-b)^(5/2)*d*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^2-1/2*b^(5 
/2)*d*polylog(2,1-2*b^(1/2)/(b^(1/2)-(b*tanh(f*x+e))^(1/2)))/f^2-1/2*b^...
 

Mathematica [F]

\[ \int (c+d x) (b \tanh (e+f x))^{5/2} \, dx=\int (c+d x) (b \tanh (e+f x))^{5/2} \, dx \] Input:

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

Output:

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

Rubi [A] (warning: unable to verify)

Time = 2.61 (sec) , antiderivative size = 1298, normalized size of antiderivative = 0.93, number of steps used = 18, number of rules used = 17, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.944, Rules used = {3042, 4203, 3042, 3954, 3042, 3957, 25, 266, 756, 216, 219, 4219, 4853, 7267, 27, 7276, 2009}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int (c+d x) (-i b \tan (i e+i f x))^{5/2}dx\)

\(\Big \downarrow \) 4203

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

\(\Big \downarrow \) 3042

\(\displaystyle b^2 \int (c+d x) \sqrt {-i b \tan (i e+i f x)}dx+\frac {2 b d \int (-i b \tan (i e+i f x))^{3/2}dx}{3 f}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}\)

\(\Big \downarrow \) 3954

\(\displaystyle b^2 \int (c+d x) \sqrt {-i b \tan (i e+i f x)}dx+\frac {2 b d \left (b^2 \int \frac {1}{\sqrt {b \tanh (e+f x)}}dx-\frac {2 b \sqrt {b \tanh (e+f x)}}{f}\right )}{3 f}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}\)

\(\Big \downarrow \) 3042

\(\displaystyle b^2 \int (c+d x) \sqrt {-i b \tan (i e+i f x)}dx+\frac {2 b d \left (-\frac {2 b \sqrt {b \tanh (e+f x)}}{f}+b^2 \int \frac {1}{\sqrt {-i b \tan (i e+i f x)}}dx\right )}{3 f}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}\)

\(\Big \downarrow \) 3957

\(\displaystyle b^2 \int (c+d x) \sqrt {-i b \tan (i e+i f x)}dx+\frac {2 b d \left (-\frac {b^3 \int -\frac {1}{\sqrt {b \tanh (e+f x)} \left (b^2-b^2 \tanh ^2(e+f x)\right )}d(b \tanh (e+f x))}{f}-\frac {2 b \sqrt {b \tanh (e+f x)}}{f}\right )}{3 f}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}\)

\(\Big \downarrow \) 25

\(\displaystyle b^2 \int (c+d x) \sqrt {-i b \tan (i e+i f x)}dx+\frac {2 b d \left (\frac {b^3 \int \frac {1}{\sqrt {b \tanh (e+f x)} \left (b^2-b^2 \tanh ^2(e+f x)\right )}d(b \tanh (e+f x))}{f}-\frac {2 b \sqrt {b \tanh (e+f x)}}{f}\right )}{3 f}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}\)

\(\Big \downarrow \) 266

\(\displaystyle b^2 \int (c+d x) \sqrt {-i b \tan (i e+i f x)}dx+\frac {2 b d \left (\frac {2 b^3 \int \frac {1}{b^2-b^4 \tanh ^4(e+f x)}d\sqrt {b \tanh (e+f x)}}{f}-\frac {2 b \sqrt {b \tanh (e+f x)}}{f}\right )}{3 f}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}\)

\(\Big \downarrow \) 756

\(\displaystyle b^2 \int (c+d x) \sqrt {-i b \tan (i e+i f x)}dx+\frac {2 b d \left (\frac {2 b^3 \left (\frac {\int \frac {1}{b-b^2 \tanh ^2(e+f x)}d\sqrt {b \tanh (e+f x)}}{2 b}+\frac {\int \frac {1}{b^2 \tanh ^2(e+f x)+b}d\sqrt {b \tanh (e+f x)}}{2 b}\right )}{f}-\frac {2 b \sqrt {b \tanh (e+f x)}}{f}\right )}{3 f}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}\)

\(\Big \downarrow \) 216

\(\displaystyle \frac {2 b d \left (\frac {2 b^3 \left (\frac {\int \frac {1}{b-b^2 \tanh ^2(e+f x)}d\sqrt {b \tanh (e+f x)}}{2 b}+\frac {\arctan \left (\sqrt {b} \tanh (e+f x)\right )}{2 b^{3/2}}\right )}{f}-\frac {2 b \sqrt {b \tanh (e+f x)}}{f}\right )}{3 f}+b^2 \int (c+d x) \sqrt {-i b \tan (i e+i f x)}dx-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}\)

\(\Big \downarrow \) 219

\(\displaystyle b^2 \int (c+d x) \sqrt {-i b \tan (i e+i f x)}dx+\frac {2 b d \left (\frac {2 b^3 \left (\frac {\arctan \left (\sqrt {b} \tanh (e+f x)\right )}{2 b^{3/2}}+\frac {\text {arctanh}\left (\sqrt {b} \tanh (e+f x)\right )}{2 b^{3/2}}\right )}{f}-\frac {2 b \sqrt {b \tanh (e+f x)}}{f}\right )}{3 f}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}\)

\(\Big \downarrow \) 4219

\(\displaystyle b^2 \left (\frac {\sqrt {-b} d \int \text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )dx}{f}-\frac {\sqrt {b} d \int \text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )dx}{f}-\frac {\sqrt {-b} (c+d x) \text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )}{f}+\frac {\sqrt {b} (c+d x) \text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )}{f}\right )+\frac {2 b d \left (\frac {2 b^3 \left (\frac {\arctan \left (\sqrt {b} \tanh (e+f x)\right )}{2 b^{3/2}}+\frac {\text {arctanh}\left (\sqrt {b} \tanh (e+f x)\right )}{2 b^{3/2}}\right )}{f}-\frac {2 b \sqrt {b \tanh (e+f x)}}{f}\right )}{3 f}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}\)

\(\Big \downarrow \) 4853

\(\displaystyle b^2 \left (\frac {\sqrt {-b} d \int \frac {\text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )}{1-\tanh ^2(e+f x)}d\tanh (e+f x)}{f^2}-\frac {\sqrt {b} d \int \frac {\text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )}{1-\tanh ^2(e+f x)}d\tanh (e+f x)}{f^2}-\frac {\sqrt {-b} (c+d x) \text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )}{f}+\frac {\sqrt {b} (c+d x) \text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )}{f}\right )+\frac {2 b d \left (\frac {2 b^3 \left (\frac {\arctan \left (\sqrt {b} \tanh (e+f x)\right )}{2 b^{3/2}}+\frac {\text {arctanh}\left (\sqrt {b} \tanh (e+f x)\right )}{2 b^{3/2}}\right )}{f}-\frac {2 b \sqrt {b \tanh (e+f x)}}{f}\right )}{3 f}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}\)

\(\Big \downarrow \) 7267

\(\displaystyle b^2 \left (\frac {2 \sqrt {-b} d \int \frac {b^2 \text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right ) \sqrt {b \tanh (e+f x)}}{b^2-b^2 \tanh ^2(e+f x)}d\sqrt {b \tanh (e+f x)}}{b f^2}-\frac {2 d \int \frac {b^2 \text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tanh (e+f x)}}{b^2-b^2 \tanh ^2(e+f x)}d\sqrt {b \tanh (e+f x)}}{\sqrt {b} f^2}-\frac {\sqrt {-b} (c+d x) \text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )}{f}+\frac {\sqrt {b} (c+d x) \text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )}{f}\right )+\frac {2 b d \left (\frac {2 b^3 \left (\frac {\arctan \left (\sqrt {b} \tanh (e+f x)\right )}{2 b^{3/2}}+\frac {\text {arctanh}\left (\sqrt {b} \tanh (e+f x)\right )}{2 b^{3/2}}\right )}{f}-\frac {2 b \sqrt {b \tanh (e+f x)}}{f}\right )}{3 f}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}\)

\(\Big \downarrow \) 27

\(\displaystyle b^2 \left (\frac {2 \sqrt {-b} b d \int \frac {\text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right ) \sqrt {b \tanh (e+f x)}}{b^2-b^2 \tanh ^2(e+f x)}d\sqrt {b \tanh (e+f x)}}{f^2}-\frac {2 b^{3/2} d \int \frac {\text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tanh (e+f x)}}{b^2-b^2 \tanh ^2(e+f x)}d\sqrt {b \tanh (e+f x)}}{f^2}+\frac {\sqrt {b} (c+d x) \text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )}{f}-\frac {\sqrt {-b} (c+d x) \text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )}{f}\right )+\frac {2 b d \left (\frac {2 b^3 \left (\frac {\arctan \left (\sqrt {b} \tanh (e+f x)\right )}{2 b^{3/2}}+\frac {\text {arctanh}\left (\sqrt {b} \tanh (e+f x)\right )}{2 b^{3/2}}\right )}{f}-\frac {2 b \sqrt {b \tanh (e+f x)}}{f}\right )}{3 f}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}\)

\(\Big \downarrow \) 7276

\(\displaystyle b^2 \left (-\frac {2 b^{3/2} d \int \left (\frac {\text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tanh (e+f x)}}{2 b (\tanh (e+f x) b+b)}-\frac {\text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right ) \sqrt {b \tanh (e+f x)}}{2 b (b \tanh (e+f x)-b)}\right )d\sqrt {b \tanh (e+f x)}}{f^2}+\frac {2 \sqrt {-b} b d \int \left (\frac {\text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right ) \sqrt {b \tanh (e+f x)}}{2 b (\tanh (e+f x) b+b)}-\frac {\text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right ) \sqrt {b \tanh (e+f x)}}{2 b (b \tanh (e+f x)-b)}\right )d\sqrt {b \tanh (e+f x)}}{f^2}+\frac {\sqrt {b} (c+d x) \text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )}{f}-\frac {\sqrt {-b} (c+d x) \text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )}{f}\right )+\frac {2 b d \left (\frac {2 b^3 \left (\frac {\arctan \left (\sqrt {b} \tanh (e+f x)\right )}{2 b^{3/2}}+\frac {\text {arctanh}\left (\sqrt {b} \tanh (e+f x)\right )}{2 b^{3/2}}\right )}{f}-\frac {2 b \sqrt {b \tanh (e+f x)}}{f}\right )}{3 f}-\frac {2 b (c+d x) (b \tanh (e+f x))^{3/2}}{3 f}\)

\(\Big \downarrow \) 2009

\(\displaystyle \left (-\frac {2 d \left (-\frac {\text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )^2}{4 b}+\frac {\log \left (\frac {2 \sqrt {b}}{\sqrt {b}-\sqrt {b \tanh (e+f x)}}\right ) \text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )}{2 b}-\frac {\log \left (\frac {2 \sqrt {b}}{\sqrt {b}+\sqrt {b \tanh (e+f x)}}\right ) \text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )}{2 b}+\frac {\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 ) \text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )}{4 b}+\frac {\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 ) \text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right )}{4 b}+\frac {\operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {b}}{\sqrt {b}-\sqrt {b \tanh (e+f x)}}\right )}{4 b}+\frac {\operatorname {PolyLog}\left (2,1-\frac {2 \sqrt {b}}{\sqrt {b}+\sqrt {b \tanh (e+f x)}}\right )}{4 b}-\frac {\operatorname {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 )}{8 b}-\frac {\operatorname {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 )}{8 b}\right ) b^{3/2}}{f^2}+\frac {2 \sqrt {-b} d \left (\frac {\text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )^2}{4 b}-\frac {\log \left (\frac {2}{1-\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}}\right ) \text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )}{2 b}+\frac {\log \left (\frac {2}{\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}+1}\right ) \text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )}{2 b}-\frac {\log \left (-\frac {2 \left (\sqrt {b}-\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}-\sqrt {b}\right ) \left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}+1\right )}\right ) \text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )}{4 b}-\frac {\log \left (\frac {2 \left (\sqrt {b}+\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}+\sqrt {b}\right ) \left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}+1\right )}\right ) \text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )}{4 b}-\frac {\operatorname {PolyLog}\left (2,1-\frac {2}{1-\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}}\right )}{4 b}-\frac {\operatorname {PolyLog}\left (2,1-\frac {2}{\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}+1}\right )}{4 b}+\frac {\operatorname {PolyLog}\left (2,\frac {2 \left (\sqrt {b}-\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}-\sqrt {b}\right ) \left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}+1\right )}+1\right )}{8 b}+\frac {\operatorname {PolyLog}\left (2,1-\frac {2 \left (\sqrt {b}+\sqrt {b \tanh (e+f x)}\right )}{\left (\sqrt {-b}+\sqrt {b}\right ) \left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}+1\right )}\right )}{8 b}\right ) b}{f^2}+\frac {(c+d x) \text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {b}}\right ) \sqrt {b}}{f}-\frac {\sqrt {-b} (c+d x) \text {arctanh}\left (\frac {\sqrt {b \tanh (e+f x)}}{\sqrt {-b}}\right )}{f}\right ) b^2-\frac {2 (c+d x) (b \tanh (e+f x))^{3/2} b}{3 f}+\frac {2 d \left (\frac {2 b^3 \left (\frac {\arctan \left (\sqrt {b} \tanh (e+f x)\right )}{2 b^{3/2}}+\frac {\text {arctanh}\left (\sqrt {b} \tanh (e+f x)\right )}{2 b^{3/2}}\right )}{f}-\frac {2 b \sqrt {b \tanh (e+f x)}}{f}\right ) b}{3 f}\)

Input:

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

Output:

b^2*(-((Sqrt[-b]*(c + d*x)*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[-b]])/f) + ( 
Sqrt[b]*(c + d*x)*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[b]])/f - (2*b^(3/2)*d 
*(-1/4*ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[b]]^2/b + (ArcTanh[Sqrt[b*Tanh[e 
 + f*x]]/Sqrt[b]]*Log[(2*Sqrt[b])/(Sqrt[b] - Sqrt[b*Tanh[e + f*x]])])/(2*b 
) - (ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[b]]*Log[(2*Sqrt[b])/(Sqrt[b] + Sqr 
t[b*Tanh[e + f*x]])])/(2*b) + (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]]))])/(4*b) + (ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqr 
t[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]]))])/(4*b) + PolyLog[2, 1 - (2*Sqrt[b 
])/(Sqrt[b] - Sqrt[b*Tanh[e + f*x]])]/(4*b) + PolyLog[2, 1 - (2*Sqrt[b])/( 
Sqrt[b] + Sqrt[b*Tanh[e + f*x]])]/(4*b) - 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]]))]/(8*b) - 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]]))]/(8*b)) 
)/f^2 + (2*Sqrt[-b]*b*d*(ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[-b]]^2/(4*b) - 
 (ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[-b]]*Log[2/(1 - Sqrt[b*Tanh[e + f*x]] 
/Sqrt[-b])])/(2*b) + (ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sqrt[-b]]*Log[2/(1 + S 
qrt[b*Tanh[e + f*x]]/Sqrt[-b])])/(2*b) - (ArcTanh[Sqrt[b*Tanh[e + f*x]]/Sq 
rt[-b]]*Log[(-2*(Sqrt[b] - Sqrt[b*Tanh[e + f*x]]))/((Sqrt[-b] - Sqrt[b]...
 

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 216

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A 
rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a 
, 0] || GtQ[b, 0])
 

rule 219

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* 
ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 266

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> With[{k = De 
nominator[m]}, Simp[k/c   Subst[Int[x^(k*(m + 1) - 1)*(a + b*(x^(2*k)/c^2)) 
^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && FractionQ[m] && I 
ntBinomialQ[a, b, c, 2, m, p, x]
 

rule 756

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[-a/b, 2 
]], s = Denominator[Rt[-a/b, 2]]}, Simp[r/(2*a)   Int[1/(r - s*x^2), x], x] 
 + Simp[r/(2*a)   Int[1/(r + s*x^2), x], x]] /; FreeQ[{a, b}, x] &&  !GtQ[a 
/b, 0]
 

rule 2009

Int[u_, x_Symbol] :> Simp[IntSum[u, x], x] /; SumQ[u]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 3954

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b*((b*Tan[c + d 
*x])^(n - 1)/(d*(n - 1))), x] - Simp[b^2   Int[(b*Tan[c + d*x])^(n - 2), x] 
, x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]
 

rule 3957

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[b/d   Subst[Int 
[x^n/(b^2 + x^2), x], x, b*Tan[c + d*x]], x] /; FreeQ[{b, c, d, n}, x] && 
!IntegerQ[n]
 

rule 4203

Int[((c_.) + (d_.)*(x_))^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symb 
ol] :> Simp[b*(c + d*x)^m*((b*Tan[e + f*x])^(n - 1)/(f*(n - 1))), x] + (-Si 
mp[b*d*(m/(f*(n - 1)))   Int[(c + d*x)^(m - 1)*(b*Tan[e + f*x])^(n - 1), x] 
, x] - Simp[b^2   Int[(c + d*x)^m*(b*Tan[e + f*x])^(n - 2), x], x]) /; Free 
Q[{b, c, d, e, f}, x] && GtQ[n, 1] && GtQ[m, 0]
 

rule 4219

Int[((c_.) + (d_.)*(x_))*Sqrt[(a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Sym 
bol] :> Simp[(-I)*Rt[a - I*b, 2]*((c + d*x)/f)*ArcTanh[Sqrt[a + b*Tan[e + f 
*x]]/Rt[a - I*b, 2]], x] + (Simp[I*Rt[a + I*b, 2]*((c + d*x)/f)*ArcTanh[Sqr 
t[a + b*Tan[e + f*x]]/Rt[a + I*b, 2]], x] + Simp[I*d*(Rt[a - I*b, 2]/f)   I 
nt[ArcTanh[Sqrt[a + b*Tan[e + f*x]]/Rt[a - I*b, 2]], x], x] - Simp[I*d*(Rt[ 
a + I*b, 2]/f)   Int[ArcTanh[Sqrt[a + b*Tan[e + f*x]]/Rt[a + I*b, 2]], x], 
x]) /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[a^2 + b^2, 0]
 

rule 4853

Int[u_, x_Symbol] :> With[{v = FunctionOfTrig[u, x]}, Simp[With[{d = FreeFa 
ctors[Tan[v], x]}, d/Coefficient[v, x, 1]   Subst[Int[SubstFor[1/(1 + d^2*x 
^2), Tan[v]/d, u, x], x], x, Tan[v]/d]], x] /;  !FalseQ[v] && FunctionOfQ[N 
onfreeFactors[Tan[v], x], u, x, True] && TryPureTanSubst[ActivateTrig[u], x 
]]
 

rule 7267

Int[u_, x_Symbol] :> With[{lst = SubstForFractionalPowerOfLinear[u, x]}, Si 
mp[lst[[2]]*lst[[4]]   Subst[Int[lst[[1]], x], x, lst[[3]]^(1/lst[[2]])], x 
] /;  !FalseQ[lst] && SubstForFractionalPowerQ[u, lst[[3]], x]]
 

rule 7276

Int[(u_)/((a_) + (b_.)*(x_)^(n_)), x_Symbol] :> With[{v = RationalFunctionE 
xpand[u/(a + b*x^n), x]}, Int[v, x] /; SumQ[v]] /; FreeQ[{a, b}, x] && IGtQ 
[n, 0]
 

Maple [F]

\[\int \left (d x +c \right ) \left (b \tanh \left (f x +e \right )\right )^{\frac {5}{2}}d x\]

Input:

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

Output:

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

Fricas [F(-2)]

Exception generated. \[ \int (c+d x) (b \tanh (e+f x))^{5/2} \, dx=\text {Exception raised: TypeError} \] Input:

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

Output:

Exception raised: TypeError >>  Error detected within library code:   inte 
grate: implementation incomplete (constant residues)
 

Sympy [F]

\[ \int (c+d x) (b \tanh (e+f x))^{5/2} \, dx=\int \left (b \tanh {\left (e + f x \right )}\right )^{\frac {5}{2}} \left (c + d x\right )\, dx \] Input:

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

Output:

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

Maxima [F]

\[ \int (c+d x) (b \tanh (e+f x))^{5/2} \, dx=\int { {\left (d x + c\right )} \left (b \tanh \left (f x + e\right )\right )^{\frac {5}{2}} \,d x } \] Input:

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

Output:

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

Giac [F]

\[ \int (c+d x) (b \tanh (e+f x))^{5/2} \, dx=\int { {\left (d x + c\right )} \left (b \tanh \left (f x + e\right )\right )^{\frac {5}{2}} \,d x } \] Input:

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

Output:

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

Mupad [F(-1)]

Timed out. \[ \int (c+d x) (b \tanh (e+f x))^{5/2} \, dx=\int {\left (b\,\mathrm {tanh}\left (e+f\,x\right )\right )}^{5/2}\,\left (c+d\,x\right ) \,d x \] Input:

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

Output:

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

Reduce [F]

\[ \int (c+d x) (b \tanh (e+f x))^{5/2} \, dx=\sqrt {b}\, b^{2} \left (\left (\int \sqrt {\tanh \left (f x +e \right )}\, \tanh \left (f x +e \right )^{2} x d x \right ) d +\left (\int \sqrt {\tanh \left (f x +e \right )}\, \tanh \left (f x +e \right )^{2}d x \right ) c \right ) \] Input:

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

Output:

sqrt(b)*b**2*(int(sqrt(tanh(e + f*x))*tanh(e + f*x)**2*x,x)*d + int(sqrt(t 
anh(e + f*x))*tanh(e + f*x)**2,x)*c)