### 3.991 $$\int \frac{\text{sech}^2(x) (a+b \tanh (x))^2}{c+d \tanh (x)} \, dx$$

Optimal. Leaf size=53 $-\frac{b \tanh (x) (b c-a d)}{d^2}+\frac{(b c-a d)^2 \log (c+d \tanh (x))}{d^3}+\frac{(a+b \tanh (x))^2}{2 d}$

[Out]

((b*c - a*d)^2*Log[c + d*Tanh[x]])/d^3 - (b*(b*c - a*d)*Tanh[x])/d^2 + (a + b*Tanh[x])^2/(2*d)

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Rubi [A]  time = 0.157541, antiderivative size = 53, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 21, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.095, Rules used = {4342, 43} $-\frac{b \tanh (x) (b c-a d)}{d^2}+\frac{(b c-a d)^2 \log (c+d \tanh (x))}{d^3}+\frac{(a+b \tanh (x))^2}{2 d}$

Antiderivative was successfully veriﬁed.

[In]

Int[(Sech[x]^2*(a + b*Tanh[x])^2)/(c + d*Tanh[x]),x]

[Out]

((b*c - a*d)^2*Log[c + d*Tanh[x]])/d^3 - (b*(b*c - a*d)*Tanh[x])/d^2 + (a + b*Tanh[x])^2/(2*d)

Rule 4342

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))]^2, x_Symbol] :> With[{d = FreeFactors[Tan[c*(a + b*x)], x]}, Dist[d/
(b*c), Subst[Int[SubstFor[1, Tan[c*(a + b*x)]/d, u, x], x], x, Tan[c*(a + b*x)]/d], x] /; FunctionOfQ[Tan[c*(a
+ b*x)]/d, u, x, True]] /; FreeQ[{a, b, c}, x] && NonsumQ[u] && (EqQ[F, Sec] || EqQ[F, sec])

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rubi steps

\begin{align*} \int \frac{\text{sech}^2(x) (a+b \tanh (x))^2}{c+d \tanh (x)} \, dx &=\operatorname{Subst}\left (\int \frac{(a+b x)^2}{c+d x} \, dx,x,\tanh (x)\right )\\ &=\operatorname{Subst}\left (\int \left (-\frac{b (b c-a d)}{d^2}+\frac{b (a+b x)}{d}+\frac{(-b c+a d)^2}{d^2 (c+d x)}\right ) \, dx,x,\tanh (x)\right )\\ &=\frac{(b c-a d)^2 \log (c+d \tanh (x))}{d^3}-\frac{b (b c-a d) \tanh (x)}{d^2}+\frac{(a+b \tanh (x))^2}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.565627, size = 61, normalized size = 1.15 $-\frac{2 b d \tanh (x) (b c-2 a d)+2 (b c-a d)^2 (\log (\cosh (x))-\log (c \cosh (x)+d \sinh (x)))+b^2 d^2 \text{sech}^2(x)}{2 d^3}$

Antiderivative was successfully veriﬁed.

[In]

Integrate[(Sech[x]^2*(a + b*Tanh[x])^2)/(c + d*Tanh[x]),x]

[Out]

-(2*(b*c - a*d)^2*(Log[Cosh[x]] - Log[c*Cosh[x] + d*Sinh[x]]) + b^2*d^2*Sech[x]^2 + 2*b*d*(b*c - 2*a*d)*Tanh[x
])/(2*d^3)

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Maple [B]  time = 0.063, size = 251, normalized size = 4.7 \begin{align*} 4\,{\frac{ \left ( \tanh \left ( x/2 \right ) \right ) ^{3}ab}{d \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-2\,{\frac{ \left ( \tanh \left ( x/2 \right ) \right ) ^{3}{b}^{2}c}{{d}^{2} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+2\,{\frac{{b}^{2} \left ( \tanh \left ( x/2 \right ) \right ) ^{2}}{d \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}+4\,{\frac{a\tanh \left ( x/2 \right ) b}{d \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-2\,{\frac{\tanh \left ( x/2 \right ){b}^{2}c}{{d}^{2} \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) ^{2}}}-{\frac{{a}^{2}}{d}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) }+2\,{\frac{\ln \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}+1 \right ) cba}{{d}^{2}}}-{\frac{{c}^{2}{b}^{2}}{{d}^{3}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}+1 \right ) }+{\frac{{a}^{2}}{d}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}c+2\,\tanh \left ( x/2 \right ) d+c \right ) }-2\,{\frac{\ln \left ( \left ( \tanh \left ( x/2 \right ) \right ) ^{2}c+2\,\tanh \left ( x/2 \right ) d+c \right ) cba}{{d}^{2}}}+{\frac{{c}^{2}{b}^{2}}{{d}^{3}}\ln \left ( \left ( \tanh \left ({\frac{x}{2}} \right ) \right ) ^{2}c+2\,\tanh \left ( x/2 \right ) d+c \right ) } \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sech(x)^2*(a+b*tanh(x))^2/(c+d*tanh(x)),x)

[Out]

4/d/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)^3*a*b-2/d^2/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)^3*b^2*c+2/d/(tanh(1/2*x)^2+1)^
2*b^2*tanh(1/2*x)^2+4/d/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)*a*b-2/d^2/(tanh(1/2*x)^2+1)^2*tanh(1/2*x)*b^2*c-1/d*ln
(tanh(1/2*x)^2+1)*a^2+2/d^2*ln(tanh(1/2*x)^2+1)*c*b*a-1/d^3*ln(tanh(1/2*x)^2+1)*c^2*b^2+1/d*ln(tanh(1/2*x)^2*c
+2*tanh(1/2*x)*d+c)*a^2-2/d^2*ln(tanh(1/2*x)^2*c+2*tanh(1/2*x)*d+c)*c*b*a+1/d^3*ln(tanh(1/2*x)^2*c+2*tanh(1/2*
x)*d+c)*c^2*b^2

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Maxima [B]  time = 1.58313, size = 204, normalized size = 3.85 \begin{align*} -b^{2}{\left (\frac{2 \,{\left ({\left (c + d\right )} e^{\left (-2 \, x\right )} + c\right )}}{2 \, d^{2} e^{\left (-2 \, x\right )} + d^{2} e^{\left (-4 \, x\right )} + d^{2}} - \frac{c^{2} \log \left (-{\left (c - d\right )} e^{\left (-2 \, x\right )} - c - d\right )}{d^{3}} + \frac{c^{2} \log \left (e^{\left (-2 \, x\right )} + 1\right )}{d^{3}}\right )} - 2 \, a b{\left (\frac{c \log \left (-{\left (c - d\right )} e^{\left (-2 \, x\right )} - c - d\right )}{d^{2}} - \frac{c \log \left (e^{\left (-2 \, x\right )} + 1\right )}{d^{2}} - \frac{2}{d e^{\left (-2 \, x\right )} + d}\right )} + \frac{a^{2} \log \left (d \tanh \left (x\right ) + c\right )}{d} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2*(a+b*tanh(x))^2/(c+d*tanh(x)),x, algorithm="maxima")

[Out]

-b^2*(2*((c + d)*e^(-2*x) + c)/(2*d^2*e^(-2*x) + d^2*e^(-4*x) + d^2) - c^2*log(-(c - d)*e^(-2*x) - c - d)/d^3
+ c^2*log(e^(-2*x) + 1)/d^3) - 2*a*b*(c*log(-(c - d)*e^(-2*x) - c - d)/d^2 - c*log(e^(-2*x) + 1)/d^2 - 2/(d*e^
(-2*x) + d)) + a^2*log(d*tanh(x) + c)/d

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Fricas [B]  time = 2.37237, size = 1667, normalized size = 31.45 \begin{align*} \text{result too large to display} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2*(a+b*tanh(x))^2/(c+d*tanh(x)),x, algorithm="fricas")

[Out]

(2*b^2*c*d - 4*a*b*d^2 + 2*(b^2*c*d - (2*a*b + b^2)*d^2)*cosh(x)^2 + 4*(b^2*c*d - (2*a*b + b^2)*d^2)*cosh(x)*s
inh(x) + 2*(b^2*c*d - (2*a*b + b^2)*d^2)*sinh(x)^2 + ((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*cosh(x)^4 + 4*(b^2*c^2 -
2*a*b*c*d + a^2*d^2)*cosh(x)*sinh(x)^3 + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sinh(x)^4 + b^2*c^2 - 2*a*b*c*d + a^
2*d^2 + 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*cosh(x)^2 + 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2 + 3*(b^2*c^2 - 2*a*b*c*
d + a^2*d^2)*cosh(x)^2)*sinh(x)^2 + 4*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*cosh(x)^3 + (b^2*c^2 - 2*a*b*c*d + a^2*
d^2)*cosh(x))*sinh(x))*log(2*(c*cosh(x) + d*sinh(x))/(cosh(x) - sinh(x))) - ((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*c
osh(x)^4 + 4*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*cosh(x)*sinh(x)^3 + (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*sinh(x)^4 + b
^2*c^2 - 2*a*b*c*d + a^2*d^2 + 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*cosh(x)^2 + 2*(b^2*c^2 - 2*a*b*c*d + a^2*d^2
+ 3*(b^2*c^2 - 2*a*b*c*d + a^2*d^2)*cosh(x)^2)*sinh(x)^2 + 4*((b^2*c^2 - 2*a*b*c*d + a^2*d^2)*cosh(x)^3 + (b^2
*c^2 - 2*a*b*c*d + a^2*d^2)*cosh(x))*sinh(x))*log(2*cosh(x)/(cosh(x) - sinh(x))))/(d^3*cosh(x)^4 + 4*d^3*cosh(
x)*sinh(x)^3 + d^3*sinh(x)^4 + 2*d^3*cosh(x)^2 + d^3 + 2*(3*d^3*cosh(x)^2 + d^3)*sinh(x)^2 + 4*(d^3*cosh(x)^3
+ d^3*cosh(x))*sinh(x))

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

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)**2*(a+b*tanh(x))**2/(c+d*tanh(x)),x)

[Out]

Integral((a + b*tanh(x))**2*sech(x)**2/(c + d*tanh(x)), x)

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Giac [B]  time = 1.15526, size = 356, normalized size = 6.72 \begin{align*} \frac{{\left (b^{2} c^{3} - 2 \, a b c^{2} d + b^{2} c^{2} d + a^{2} c d^{2} - 2 \, a b c d^{2} + a^{2} d^{3}\right )} \log \left ({\left | c e^{\left (2 \, x\right )} + d e^{\left (2 \, x\right )} + c - d \right |}\right )}{c d^{3} + d^{4}} - \frac{{\left (b^{2} c^{2} - 2 \, a b c d + a^{2} d^{2}\right )} \log \left (e^{\left (2 \, x\right )} + 1\right )}{d^{3}} + \frac{3 \, b^{2} c^{2} e^{\left (4 \, x\right )} - 6 \, a b c d e^{\left (4 \, x\right )} + 3 \, a^{2} d^{2} e^{\left (4 \, x\right )} + 6 \, b^{2} c^{2} e^{\left (2 \, x\right )} - 12 \, a b c d e^{\left (2 \, x\right )} + 4 \, b^{2} c d e^{\left (2 \, x\right )} + 6 \, a^{2} d^{2} e^{\left (2 \, x\right )} - 8 \, a b d^{2} e^{\left (2 \, x\right )} - 4 \, b^{2} d^{2} e^{\left (2 \, x\right )} + 3 \, b^{2} c^{2} - 6 \, a b c d + 4 \, b^{2} c d + 3 \, a^{2} d^{2} - 8 \, a b d^{2}}{2 \, d^{3}{\left (e^{\left (2 \, x\right )} + 1\right )}^{2}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sech(x)^2*(a+b*tanh(x))^2/(c+d*tanh(x)),x, algorithm="giac")

[Out]

(b^2*c^3 - 2*a*b*c^2*d + b^2*c^2*d + a^2*c*d^2 - 2*a*b*c*d^2 + a^2*d^3)*log(abs(c*e^(2*x) + d*e^(2*x) + c - d)
)/(c*d^3 + d^4) - (b^2*c^2 - 2*a*b*c*d + a^2*d^2)*log(e^(2*x) + 1)/d^3 + 1/2*(3*b^2*c^2*e^(4*x) - 6*a*b*c*d*e^
(4*x) + 3*a^2*d^2*e^(4*x) + 6*b^2*c^2*e^(2*x) - 12*a*b*c*d*e^(2*x) + 4*b^2*c*d*e^(2*x) + 6*a^2*d^2*e^(2*x) - 8
*a*b*d^2*e^(2*x) - 4*b^2*d^2*e^(2*x) + 3*b^2*c^2 - 6*a*b*c*d + 4*b^2*c*d + 3*a^2*d^2 - 8*a*b*d^2)/(d^3*(e^(2*x
) + 1)^2)