3.431 \(\int \sqrt{a+a \sinh ^2(e+f x)} \tanh ^6(e+f x) \, dx\)
Optimal. Leaf size=120 \[ -\frac{\tanh ^5(e+f x) \sqrt{a \cosh ^2(e+f x)}}{4 f}-\frac{5 \tanh ^3(e+f x) \sqrt{a \cosh ^2(e+f x)}}{8 f}+\frac{15 \tanh (e+f x) \sqrt{a \cosh ^2(e+f x)}}{8 f}-\frac{15 \text{sech}(e+f x) \sqrt{a \cosh ^2(e+f x)} \tan ^{-1}(\sinh (e+f x))}{8 f} \]
[Out]
(-15*ArcTan[Sinh[e + f*x]]*Sqrt[a*Cosh[e + f*x]^2]*Sech[e + f*x])/(8*f) + (15*Sqrt[a*Cosh[e + f*x]^2]*Tanh[e +
f*x])/(8*f) - (5*Sqrt[a*Cosh[e + f*x]^2]*Tanh[e + f*x]^3)/(8*f) - (Sqrt[a*Cosh[e + f*x]^2]*Tanh[e + f*x]^5)/(
4*f)
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Rubi [A] time = 0.129995, antiderivative size = 120, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used =
{3176, 3207, 2592, 288, 321, 203} \[ -\frac{\tanh ^5(e+f x) \sqrt{a \cosh ^2(e+f x)}}{4 f}-\frac{5 \tanh ^3(e+f x) \sqrt{a \cosh ^2(e+f x)}}{8 f}+\frac{15 \tanh (e+f x) \sqrt{a \cosh ^2(e+f x)}}{8 f}-\frac{15 \text{sech}(e+f x) \sqrt{a \cosh ^2(e+f x)} \tan ^{-1}(\sinh (e+f x))}{8 f} \]
Antiderivative was successfully verified.
[In]
Int[Sqrt[a + a*Sinh[e + f*x]^2]*Tanh[e + f*x]^6,x]
[Out]
(-15*ArcTan[Sinh[e + f*x]]*Sqrt[a*Cosh[e + f*x]^2]*Sech[e + f*x])/(8*f) + (15*Sqrt[a*Cosh[e + f*x]^2]*Tanh[e +
f*x])/(8*f) - (5*Sqrt[a*Cosh[e + f*x]^2]*Tanh[e + f*x]^3)/(8*f) - (Sqrt[a*Cosh[e + f*x]^2]*Tanh[e + f*x]^5)/(
4*f)
Rule 3176
Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Int[ActivateTrig[u*(a*cos[e + f*x]^2)^p]
, x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0]
Rule 3207
Int[(u_.)*((b_.)*sin[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Sin[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Sin[e + f*x]^n)^FracPart[p])/(Sin[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Sin[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] && !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])
Rule 2592
Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> With[{ff = FreeFactors[S
in[e + f*x], x]}, Dist[ff/f, Subst[Int[(ff*x)^(m + n)/(a^2 - ff^2*x^2)^((n + 1)/2), x], x, (a*Sin[e + f*x])/ff
], x]] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n + 1)/2]
Rule 288
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] && !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]
Rule 321
Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]
Rule 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rubi steps
\begin{align*} \int \sqrt{a+a \sinh ^2(e+f x)} \tanh ^6(e+f x) \, dx &=\int \sqrt{a \cosh ^2(e+f x)} \tanh ^6(e+f x) \, dx\\ &=\left (\sqrt{a \cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \int \sinh (e+f x) \tanh ^5(e+f x) \, dx\\ &=\frac{\left (\sqrt{a \cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{x^6}{\left (1+x^2\right )^3} \, dx,x,\sinh (e+f x)\right )}{f}\\ &=-\frac{\sqrt{a \cosh ^2(e+f x)} \tanh ^5(e+f x)}{4 f}+\frac{\left (5 \sqrt{a \cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right )^2} \, dx,x,\sinh (e+f x)\right )}{4 f}\\ &=-\frac{5 \sqrt{a \cosh ^2(e+f x)} \tanh ^3(e+f x)}{8 f}-\frac{\sqrt{a \cosh ^2(e+f x)} \tanh ^5(e+f x)}{4 f}+\frac{\left (15 \sqrt{a \cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{x^2}{1+x^2} \, dx,x,\sinh (e+f x)\right )}{8 f}\\ &=\frac{15 \sqrt{a \cosh ^2(e+f x)} \tanh (e+f x)}{8 f}-\frac{5 \sqrt{a \cosh ^2(e+f x)} \tanh ^3(e+f x)}{8 f}-\frac{\sqrt{a \cosh ^2(e+f x)} \tanh ^5(e+f x)}{4 f}-\frac{\left (15 \sqrt{a \cosh ^2(e+f x)} \text{sech}(e+f x)\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\sinh (e+f x)\right )}{8 f}\\ &=-\frac{15 \tan ^{-1}(\sinh (e+f x)) \sqrt{a \cosh ^2(e+f x)} \text{sech}(e+f x)}{8 f}+\frac{15 \sqrt{a \cosh ^2(e+f x)} \tanh (e+f x)}{8 f}-\frac{5 \sqrt{a \cosh ^2(e+f x)} \tanh ^3(e+f x)}{8 f}-\frac{\sqrt{a \cosh ^2(e+f x)} \tanh ^5(e+f x)}{4 f}\\ \end{align*}
Mathematica [A] time = 0.34292, size = 75, normalized size = 0.62 \[ -\frac{\text{sech}^5(e+f x) \sqrt{a \cosh ^2(e+f x)} \left (-5 \sinh (e+f x)-15 \sinh (3 (e+f x))-2 \sinh (5 (e+f x))+60 \cosh ^4(e+f x) \tan ^{-1}(\sinh (e+f x))\right )}{32 f} \]
Antiderivative was successfully verified.
[In]
Integrate[Sqrt[a + a*Sinh[e + f*x]^2]*Tanh[e + f*x]^6,x]
[Out]
-(Sqrt[a*Cosh[e + f*x]^2]*Sech[e + f*x]^5*(60*ArcTan[Sinh[e + f*x]]*Cosh[e + f*x]^4 - 5*Sinh[e + f*x] - 15*Sin
h[3*(e + f*x)] - 2*Sinh[5*(e + f*x)]))/(32*f)
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Maple [A] time = 0.109, size = 85, normalized size = 0.7 \begin{align*} -{\frac{a \left ( 15\,\arctan \left ( \sinh \left ( fx+e \right ) \right ) \left ( \cosh \left ( fx+e \right ) \right ) ^{4}-8\, \left ( \cosh \left ( fx+e \right ) \right ) ^{4}\sinh \left ( fx+e \right ) -9\, \left ( \cosh \left ( fx+e \right ) \right ) ^{2}\sinh \left ( fx+e \right ) +2\,\sinh \left ( fx+e \right ) \right ) }{8\, \left ( \cosh \left ( fx+e \right ) \right ) ^{3}f}{\frac{1}{\sqrt{a \left ( \cosh \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sinh(f*x+e)^2)^(1/2)*tanh(f*x+e)^6,x)
[Out]
-1/8*a*(15*arctan(sinh(f*x+e))*cosh(f*x+e)^4-8*cosh(f*x+e)^4*sinh(f*x+e)-9*cosh(f*x+e)^2*sinh(f*x+e)+2*sinh(f*
x+e))/cosh(f*x+e)^3/(a*cosh(f*x+e)^2)^(1/2)/f
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Maxima [B] time = 1.90429, size = 1203, normalized size = 10.02 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sinh(f*x+e)^2)^(1/2)*tanh(f*x+e)^6,x, algorithm="maxima")
[Out]
315/128*sqrt(a)*arctan(e^(-f*x - e))/f + 1/128*(105*sqrt(a)*arctan(e^(-f*x - e)) + (279*sqrt(a)*e^(-f*x - e) +
511*sqrt(a)*e^(-3*f*x - 3*e) + 385*sqrt(a)*e^(-5*f*x - 5*e) + 105*sqrt(a)*e^(-7*f*x - 7*e))/(4*e^(-2*f*x - 2*
e) + 6*e^(-4*f*x - 4*e) + 4*e^(-6*f*x - 6*e) + e^(-8*f*x - 8*e) + 1))/f + 1/128*(105*sqrt(a)*arctan(e^(-f*x -
e)) - (105*sqrt(a)*e^(-f*x - e) + 385*sqrt(a)*e^(-3*f*x - 3*e) + 511*sqrt(a)*e^(-5*f*x - 5*e) + 279*sqrt(a)*e^
(-7*f*x - 7*e))/(4*e^(-2*f*x - 2*e) + 6*e^(-4*f*x - 4*e) + 4*e^(-6*f*x - 6*e) + e^(-8*f*x - 8*e) + 1))/f - 5/2
56*(15*sqrt(a)*arctan(e^(-f*x - e)) - (15*sqrt(a)*e^(-f*x - e) + 55*sqrt(a)*e^(-3*f*x - 3*e) + 73*sqrt(a)*e^(-
5*f*x - 5*e) - 15*sqrt(a)*e^(-7*f*x - 7*e))/(4*e^(-2*f*x - 2*e) + 6*e^(-4*f*x - 4*e) + 4*e^(-6*f*x - 6*e) + e^
(-8*f*x - 8*e) + 1))/f - 5/256*(15*sqrt(a)*arctan(e^(-f*x - e)) - (15*sqrt(a)*e^(-f*x - e) - 73*sqrt(a)*e^(-3*
f*x - 3*e) - 55*sqrt(a)*e^(-5*f*x - 5*e) - 15*sqrt(a)*e^(-7*f*x - 7*e))/(4*e^(-2*f*x - 2*e) + 6*e^(-4*f*x - 4*
e) + 4*e^(-6*f*x - 6*e) + e^(-8*f*x - 8*e) + 1))/f + 5/64*(3*sqrt(a)*arctan(e^(-f*x - e)) - (3*sqrt(a)*e^(-f*x
- e) + 11*sqrt(a)*e^(-3*f*x - 3*e) - 11*sqrt(a)*e^(-5*f*x - 5*e) - 3*sqrt(a)*e^(-7*f*x - 7*e))/(4*e^(-2*f*x -
2*e) + 6*e^(-4*f*x - 4*e) + 4*e^(-6*f*x - 6*e) + e^(-8*f*x - 8*e) + 1))/f + 1/256*(837*sqrt(a)*e^(-2*f*x - 2*
e) + 1533*sqrt(a)*e^(-4*f*x - 4*e) + 1155*sqrt(a)*e^(-6*f*x - 6*e) + 315*sqrt(a)*e^(-8*f*x - 8*e) + 128*sqrt(a
))/(f*(e^(-f*x - e) + 4*e^(-3*f*x - 3*e) + 6*e^(-5*f*x - 5*e) + 4*e^(-7*f*x - 7*e) + e^(-9*f*x - 9*e))) - 1/25
6*(315*sqrt(a)*e^(-f*x - e) + 1155*sqrt(a)*e^(-3*f*x - 3*e) + 1533*sqrt(a)*e^(-5*f*x - 5*e) + 837*sqrt(a)*e^(-
7*f*x - 7*e) + 128*sqrt(a)*e^(-9*f*x - 9*e))/(f*(4*e^(-2*f*x - 2*e) + 6*e^(-4*f*x - 4*e) + 4*e^(-6*f*x - 6*e)
+ e^(-8*f*x - 8*e) + 1))
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Fricas [B] time = 2.04515, size = 4517, normalized size = 37.64 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sinh(f*x+e)^2)^(1/2)*tanh(f*x+e)^6,x, algorithm="fricas")
[Out]
1/4*(20*cosh(f*x + e)*e^(f*x + e)*sinh(f*x + e)^9 + 2*e^(f*x + e)*sinh(f*x + e)^10 + 15*(6*cosh(f*x + e)^2 + 1
)*e^(f*x + e)*sinh(f*x + e)^8 + 120*(2*cosh(f*x + e)^3 + cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e)^7 + 5*(84*co
sh(f*x + e)^4 + 84*cosh(f*x + e)^2 + 1)*e^(f*x + e)*sinh(f*x + e)^6 + 6*(84*cosh(f*x + e)^5 + 140*cosh(f*x + e
)^3 + 5*cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e)^5 + 5*(84*cosh(f*x + e)^6 + 210*cosh(f*x + e)^4 + 15*cosh(f*x
+ e)^2 - 1)*e^(f*x + e)*sinh(f*x + e)^4 + 20*(12*cosh(f*x + e)^7 + 42*cosh(f*x + e)^5 + 5*cosh(f*x + e)^3 - c
osh(f*x + e))*e^(f*x + e)*sinh(f*x + e)^3 + 15*(6*cosh(f*x + e)^8 + 28*cosh(f*x + e)^6 + 5*cosh(f*x + e)^4 - 2
*cosh(f*x + e)^2 - 1)*e^(f*x + e)*sinh(f*x + e)^2 + 10*(2*cosh(f*x + e)^9 + 12*cosh(f*x + e)^7 + 3*cosh(f*x +
e)^5 - 2*cosh(f*x + e)^3 - 3*cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e) - 15*(9*cosh(f*x + e)*e^(f*x + e)*sinh(f
*x + e)^8 + e^(f*x + e)*sinh(f*x + e)^9 + 4*(9*cosh(f*x + e)^2 + 1)*e^(f*x + e)*sinh(f*x + e)^7 + 28*(3*cosh(f
*x + e)^3 + cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e)^6 + 6*(21*cosh(f*x + e)^4 + 14*cosh(f*x + e)^2 + 1)*e^(f*
x + e)*sinh(f*x + e)^5 + 2*(63*cosh(f*x + e)^5 + 70*cosh(f*x + e)^3 + 15*cosh(f*x + e))*e^(f*x + e)*sinh(f*x +
e)^4 + 4*(21*cosh(f*x + e)^6 + 35*cosh(f*x + e)^4 + 15*cosh(f*x + e)^2 + 1)*e^(f*x + e)*sinh(f*x + e)^3 + 12*
(3*cosh(f*x + e)^7 + 7*cosh(f*x + e)^5 + 5*cosh(f*x + e)^3 + cosh(f*x + e))*e^(f*x + e)*sinh(f*x + e)^2 + (9*c
osh(f*x + e)^8 + 28*cosh(f*x + e)^6 + 30*cosh(f*x + e)^4 + 12*cosh(f*x + e)^2 + 1)*e^(f*x + e)*sinh(f*x + e) +
(cosh(f*x + e)^9 + 4*cosh(f*x + e)^7 + 6*cosh(f*x + e)^5 + 4*cosh(f*x + e)^3 + cosh(f*x + e))*e^(f*x + e))*ar
ctan(cosh(f*x + e) + sinh(f*x + e)) + (2*cosh(f*x + e)^10 + 15*cosh(f*x + e)^8 + 5*cosh(f*x + e)^6 - 5*cosh(f*
x + e)^4 - 15*cosh(f*x + e)^2 - 2)*e^(f*x + e))*sqrt(a*e^(4*f*x + 4*e) + 2*a*e^(2*f*x + 2*e) + a)*e^(-f*x - e)
/(f*cosh(f*x + e)^9 + (f*e^(2*f*x + 2*e) + f)*sinh(f*x + e)^9 + 9*(f*cosh(f*x + e)*e^(2*f*x + 2*e) + f*cosh(f*
x + e))*sinh(f*x + e)^8 + 4*f*cosh(f*x + e)^7 + 4*(9*f*cosh(f*x + e)^2 + (9*f*cosh(f*x + e)^2 + f)*e^(2*f*x +
2*e) + f)*sinh(f*x + e)^7 + 28*(3*f*cosh(f*x + e)^3 + f*cosh(f*x + e) + (3*f*cosh(f*x + e)^3 + f*cosh(f*x + e)
)*e^(2*f*x + 2*e))*sinh(f*x + e)^6 + 6*f*cosh(f*x + e)^5 + 6*(21*f*cosh(f*x + e)^4 + 14*f*cosh(f*x + e)^2 + (2
1*f*cosh(f*x + e)^4 + 14*f*cosh(f*x + e)^2 + f)*e^(2*f*x + 2*e) + f)*sinh(f*x + e)^5 + 2*(63*f*cosh(f*x + e)^5
+ 70*f*cosh(f*x + e)^3 + 15*f*cosh(f*x + e) + (63*f*cosh(f*x + e)^5 + 70*f*cosh(f*x + e)^3 + 15*f*cosh(f*x +
e))*e^(2*f*x + 2*e))*sinh(f*x + e)^4 + 4*f*cosh(f*x + e)^3 + 4*(21*f*cosh(f*x + e)^6 + 35*f*cosh(f*x + e)^4 +
15*f*cosh(f*x + e)^2 + (21*f*cosh(f*x + e)^6 + 35*f*cosh(f*x + e)^4 + 15*f*cosh(f*x + e)^2 + f)*e^(2*f*x + 2*e
) + f)*sinh(f*x + e)^3 + 12*(3*f*cosh(f*x + e)^7 + 7*f*cosh(f*x + e)^5 + 5*f*cosh(f*x + e)^3 + f*cosh(f*x + e)
+ (3*f*cosh(f*x + e)^7 + 7*f*cosh(f*x + e)^5 + 5*f*cosh(f*x + e)^3 + f*cosh(f*x + e))*e^(2*f*x + 2*e))*sinh(f
*x + e)^2 + f*cosh(f*x + e) + (f*cosh(f*x + e)^9 + 4*f*cosh(f*x + e)^7 + 6*f*cosh(f*x + e)^5 + 4*f*cosh(f*x +
e)^3 + f*cosh(f*x + e))*e^(2*f*x + 2*e) + (9*f*cosh(f*x + e)^8 + 28*f*cosh(f*x + e)^6 + 30*f*cosh(f*x + e)^4 +
12*f*cosh(f*x + e)^2 + (9*f*cosh(f*x + e)^8 + 28*f*cosh(f*x + e)^6 + 30*f*cosh(f*x + e)^4 + 12*f*cosh(f*x + e
)^2 + f)*e^(2*f*x + 2*e) + f)*sinh(f*x + e))
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sinh(f*x+e)**2)**(1/2)*tanh(f*x+e)**6,x)
[Out]
Timed out
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Giac [A] time = 1.40308, size = 134, normalized size = 1.12 \begin{align*} \frac{\sqrt{a}{\left (\frac{9 \, e^{\left (7 \, f x + 7 \, e\right )} + e^{\left (5 \, f x + 5 \, e\right )} - e^{\left (3 \, f x + 3 \, e\right )} - 9 \, e^{\left (f x + e\right )}}{{\left (e^{\left (2 \, f x + 2 \, e\right )} + 1\right )}^{4}} - 15 \, \arctan \left (e^{\left (f x + e\right )}\right ) + 2 \, e^{\left (f x + e\right )} - 2 \, e^{\left (-f x - e\right )}\right )}}{4 \, f} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sinh(f*x+e)^2)^(1/2)*tanh(f*x+e)^6,x, algorithm="giac")
[Out]
1/4*sqrt(a)*((9*e^(7*f*x + 7*e) + e^(5*f*x + 5*e) - e^(3*f*x + 3*e) - 9*e^(f*x + e))/(e^(2*f*x + 2*e) + 1)^4 -
15*arctan(e^(f*x + e)) + 2*e^(f*x + e) - 2*e^(-f*x - e))/f