3.127 \(\int \sqrt {a+b \text {sech}(c+d x)} \tanh (c+d x) \, dx\)

Optimal. Leaf size=51 \[ \frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{d}-\frac {2 \sqrt {a+b \text {sech}(c+d x)}}{d} \]

[Out]

2*arctanh((a+b*sech(d*x+c))^(1/2)/a^(1/2))*a^(1/2)/d-2*(a+b*sech(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.05, antiderivative size = 51, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3885, 50, 63, 207} \[ \frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{d}-\frac {2 \sqrt {a+b \text {sech}(c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sech[c + d*x]]*Tanh[c + d*x],x]

[Out]

(2*Sqrt[a]*ArcTanh[Sqrt[a + b*Sech[c + d*x]]/Sqrt[a]])/d - (2*Sqrt[a + b*Sech[c + d*x]])/d

Rule 50

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^n)/(b*
(m + n + 1)), x] + Dist[(n*(b*c - a*d))/(b*(m + n + 1)), Int[(a + b*x)^m*(c + d*x)^(n - 1), x], x] /; FreeQ[{a
, b, c, d}, x] && NeQ[b*c - a*d, 0] && GtQ[n, 0] && NeQ[m + n + 1, 0] &&  !(IGtQ[m, 0] && ( !IntegerQ[n] || (G
tQ[m, 0] && LtQ[m - n, 0]))) &&  !ILtQ[m + n + 2, 0] && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 207

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

Rule 3885

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> -Dist[(-1)^((m - 1
)/2)/(d*b^(m - 1)), Subst[Int[((b^2 - x^2)^((m - 1)/2)*(a + x)^n)/x, x], x, b*Csc[c + d*x]], x] /; FreeQ[{a, b
, c, d, n}, x] && IntegerQ[(m - 1)/2] && NeQ[a^2 - b^2, 0]

Rubi steps

\begin {align*} \int \sqrt {a+b \text {sech}(c+d x)} \tanh (c+d x) \, dx &=-\frac {\operatorname {Subst}\left (\int \frac {\sqrt {a+x}}{x} \, dx,x,b \text {sech}(c+d x)\right )}{d}\\ &=-\frac {2 \sqrt {a+b \text {sech}(c+d x)}}{d}-\frac {a \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+x}} \, dx,x,b \text {sech}(c+d x)\right )}{d}\\ &=-\frac {2 \sqrt {a+b \text {sech}(c+d x)}}{d}-\frac {(2 a) \operatorname {Subst}\left (\int \frac {1}{-a+x^2} \, dx,x,\sqrt {a+b \text {sech}(c+d x)}\right )}{d}\\ &=\frac {2 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+b \text {sech}(c+d x)}}{\sqrt {a}}\right )}{d}-\frac {2 \sqrt {a+b \text {sech}(c+d x)}}{d}\\ \end {align*}

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Mathematica [A]  time = 0.15, size = 90, normalized size = 1.76 \[ -\frac {2 \sqrt {a+b \text {sech}(c+d x)} \left (\sqrt {a \cosh (c+d x)+b}-\sqrt {a \cosh (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {a \cosh (c+d x)+b}}{\sqrt {a \cosh (c+d x)}}\right )\right )}{d \sqrt {a \cosh (c+d x)+b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Sech[c + d*x]]*Tanh[c + d*x],x]

[Out]

(-2*(-(ArcTanh[Sqrt[b + a*Cosh[c + d*x]]/Sqrt[a*Cosh[c + d*x]]]*Sqrt[a*Cosh[c + d*x]]) + Sqrt[b + a*Cosh[c + d
*x]])*Sqrt[a + b*Sech[c + d*x]])/(d*Sqrt[b + a*Cosh[c + d*x]])

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fricas [B]  time = 1.04, size = 605, normalized size = 11.86 \[ \left [\frac {\sqrt {a} \log \left (-\frac {2 \, a^{2} \cosh \left (d x + c\right )^{4} + 2 \, a^{2} \sinh \left (d x + c\right )^{4} + 4 \, a b \cosh \left (d x + c\right )^{3} + 4 \, {\left (2 \, a^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right )^{3} + 4 \, a b \cosh \left (d x + c\right ) + {\left (4 \, a^{2} + b^{2}\right )} \cosh \left (d x + c\right )^{2} + {\left (12 \, a^{2} \cosh \left (d x + c\right )^{2} + 12 \, a b \cosh \left (d x + c\right ) + 4 \, a^{2} + b^{2}\right )} \sinh \left (d x + c\right )^{2} + 2 \, a^{2} + 2 \, {\left (a \cosh \left (d x + c\right )^{4} + a \sinh \left (d x + c\right )^{4} + b \cosh \left (d x + c\right )^{3} + {\left (4 \, a \cosh \left (d x + c\right ) + b\right )} \sinh \left (d x + c\right )^{3} + 2 \, a \cosh \left (d x + c\right )^{2} + {\left (6 \, a \cosh \left (d x + c\right )^{2} + 3 \, b \cosh \left (d x + c\right ) + 2 \, a\right )} \sinh \left (d x + c\right )^{2} + b \cosh \left (d x + c\right ) + {\left (4 \, a \cosh \left (d x + c\right )^{3} + 3 \, b \cosh \left (d x + c\right )^{2} + 4 \, a \cosh \left (d x + c\right ) + b\right )} \sinh \left (d x + c\right ) + a\right )} \sqrt {a} \sqrt {\frac {a \cosh \left (d x + c\right ) + b}{\cosh \left (d x + c\right )}} + 2 \, {\left (4 \, a^{2} \cosh \left (d x + c\right )^{3} + 6 \, a b \cosh \left (d x + c\right )^{2} + 2 \, a b + {\left (4 \, a^{2} + b^{2}\right )} \cosh \left (d x + c\right )\right )} \sinh \left (d x + c\right )}{\cosh \left (d x + c\right )^{2} + 2 \, \cosh \left (d x + c\right ) \sinh \left (d x + c\right ) + \sinh \left (d x + c\right )^{2}}\right ) - 4 \, \sqrt {\frac {a \cosh \left (d x + c\right ) + b}{\cosh \left (d x + c\right )}}}{2 \, d}, -\frac {\sqrt {-a} \arctan \left (\frac {{\left (a \cosh \left (d x + c\right )^{2} + a \sinh \left (d x + c\right )^{2} + b \cosh \left (d x + c\right ) + {\left (2 \, a \cosh \left (d x + c\right ) + b\right )} \sinh \left (d x + c\right ) + a\right )} \sqrt {-a} \sqrt {\frac {a \cosh \left (d x + c\right ) + b}{\cosh \left (d x + c\right )}}}{a^{2} \cosh \left (d x + c\right )^{2} + a^{2} \sinh \left (d x + c\right )^{2} + 2 \, a b \cosh \left (d x + c\right ) + a^{2} + 2 \, {\left (a^{2} \cosh \left (d x + c\right ) + a b\right )} \sinh \left (d x + c\right )}\right ) + 2 \, \sqrt {\frac {a \cosh \left (d x + c\right ) + b}{\cosh \left (d x + c\right )}}}{d}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sech(d*x+c))^(1/2)*tanh(d*x+c),x, algorithm="fricas")

[Out]

[1/2*(sqrt(a)*log(-(2*a^2*cosh(d*x + c)^4 + 2*a^2*sinh(d*x + c)^4 + 4*a*b*cosh(d*x + c)^3 + 4*(2*a^2*cosh(d*x
+ c) + a*b)*sinh(d*x + c)^3 + 4*a*b*cosh(d*x + c) + (4*a^2 + b^2)*cosh(d*x + c)^2 + (12*a^2*cosh(d*x + c)^2 +
12*a*b*cosh(d*x + c) + 4*a^2 + b^2)*sinh(d*x + c)^2 + 2*a^2 + 2*(a*cosh(d*x + c)^4 + a*sinh(d*x + c)^4 + b*cos
h(d*x + c)^3 + (4*a*cosh(d*x + c) + b)*sinh(d*x + c)^3 + 2*a*cosh(d*x + c)^2 + (6*a*cosh(d*x + c)^2 + 3*b*cosh
(d*x + c) + 2*a)*sinh(d*x + c)^2 + b*cosh(d*x + c) + (4*a*cosh(d*x + c)^3 + 3*b*cosh(d*x + c)^2 + 4*a*cosh(d*x
+ c) + b)*sinh(d*x + c) + a)*sqrt(a)*sqrt((a*cosh(d*x + c) + b)/cosh(d*x + c)) + 2*(4*a^2*cosh(d*x + c)^3 + 6
*a*b*cosh(d*x + c)^2 + 2*a*b + (4*a^2 + b^2)*cosh(d*x + c))*sinh(d*x + c))/(cosh(d*x + c)^2 + 2*cosh(d*x + c)*
sinh(d*x + c) + sinh(d*x + c)^2)) - 4*sqrt((a*cosh(d*x + c) + b)/cosh(d*x + c)))/d, -(sqrt(-a)*arctan((a*cosh(
d*x + c)^2 + a*sinh(d*x + c)^2 + b*cosh(d*x + c) + (2*a*cosh(d*x + c) + b)*sinh(d*x + c) + a)*sqrt(-a)*sqrt((a
*cosh(d*x + c) + b)/cosh(d*x + c))/(a^2*cosh(d*x + c)^2 + a^2*sinh(d*x + c)^2 + 2*a*b*cosh(d*x + c) + a^2 + 2*
(a^2*cosh(d*x + c) + a*b)*sinh(d*x + c))) + 2*sqrt((a*cosh(d*x + c) + b)/cosh(d*x + c)))/d]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \operatorname {sech}\left (d x + c\right ) + a} \tanh \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sech(d*x+c))^(1/2)*tanh(d*x+c),x, algorithm="giac")

[Out]

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

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maple [A]  time = 0.11, size = 43, normalized size = 0.84 \[ -\frac {2 \sqrt {a +b \,\mathrm {sech}\left (d x +c \right )}-2 \sqrt {a}\, \arctanh \left (\frac {\sqrt {a +b \,\mathrm {sech}\left (d x +c \right )}}{\sqrt {a}}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*sech(d*x+c))^(1/2)*tanh(d*x+c),x)

[Out]

-1/d*(2*(a+b*sech(d*x+c))^(1/2)-2*a^(1/2)*arctanh((a+b*sech(d*x+c))^(1/2)/a^(1/2)))

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \operatorname {sech}\left (d x + c\right ) + a} \tanh \left (d x + c\right )\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sech(d*x+c))^(1/2)*tanh(d*x+c),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 1.69, size = 47, normalized size = 0.92 \[ \frac {2\,\sqrt {a}\,\mathrm {atanh}\left (\frac {\sqrt {a+\frac {b}{\mathrm {cosh}\left (c+d\,x\right )}}}{\sqrt {a}}\right )}{d}-\frac {2\,\sqrt {a+\frac {b}{\mathrm {cosh}\left (c+d\,x\right )}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tanh(c + d*x)*(a + b/cosh(c + d*x))^(1/2),x)

[Out]

(2*a^(1/2)*atanh((a + b/cosh(c + d*x))^(1/2)/a^(1/2)))/d - (2*(a + b/cosh(c + d*x))^(1/2))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {a + b \operatorname {sech}{\left (c + d x \right )}} \tanh {\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sech(d*x+c))**(1/2)*tanh(d*x+c),x)

[Out]

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

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