3.31 \(\int \frac{\sinh ^4(c+d x)}{a+b \sinh ^2(c+d x)} \, dx\)

Optimal. Leaf size=79 \[ \frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh (c+d x)}{\sqrt{a}}\right )}{b^2 d \sqrt{a-b}}-\frac{x (2 a+b)}{2 b^2}+\frac{\sinh (c+d x) \cosh (c+d x)}{2 b d} \]

[Out]

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

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Rubi [A]  time = 0.121325, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {3187, 470, 522, 206, 208} \[ \frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh (c+d x)}{\sqrt{a}}\right )}{b^2 d \sqrt{a-b}}-\frac{x (2 a+b)}{2 b^2}+\frac{\sinh (c+d x) \cosh (c+d x)}{2 b d} \]

Antiderivative was successfully verified.

[In]

Int[Sinh[c + d*x]^4/(a + b*Sinh[c + d*x]^2),x]

[Out]

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

Rule 3187

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Tan[e + f*x], x]}, Dist[ff^(m + 1)/f, Subst[Int[(x^m*(a + (a + b)*ff^2*x^2)^p)/(1 + ff^2*x^2)^(m/2 + p
+ 1), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f}, x] && IntegerQ[m/2] && IntegerQ[p]

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, n, p, q, x]

Rule 522

Int[((e_) + (f_.)*(x_)^(n_))/(((a_) + (b_.)*(x_)^(n_))*((c_) + (d_.)*(x_)^(n_))), x_Symbol] :> Dist[(b*e - a*f
)/(b*c - a*d), Int[1/(a + b*x^n), x], x] - Dist[(d*e - c*f)/(b*c - a*d), Int[1/(c + d*x^n), x], x] /; FreeQ[{a
, b, c, d, e, f, n}, x]

Rule 206

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

Rule 208

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

Rubi steps

\begin{align*} \int \frac{\sinh ^4(c+d x)}{a+b \sinh ^2(c+d x)} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^4}{\left (1-x^2\right )^2 \left (a-(a-b) x^2\right )} \, dx,x,\tanh (c+d x)\right )}{d}\\ &=\frac{\cosh (c+d x) \sinh (c+d x)}{2 b d}-\frac{\operatorname{Subst}\left (\int \frac{a+(a+b) x^2}{\left (1-x^2\right ) \left (a+(-a+b) x^2\right )} \, dx,x,\tanh (c+d x)\right )}{2 b d}\\ &=\frac{\cosh (c+d x) \sinh (c+d x)}{2 b d}+\frac{a^2 \operatorname{Subst}\left (\int \frac{1}{a+(-a+b) x^2} \, dx,x,\tanh (c+d x)\right )}{b^2 d}-\frac{(2 a+b) \operatorname{Subst}\left (\int \frac{1}{1-x^2} \, dx,x,\tanh (c+d x)\right )}{2 b^2 d}\\ &=-\frac{(2 a+b) x}{2 b^2}+\frac{a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh (c+d x)}{\sqrt{a}}\right )}{\sqrt{a-b} b^2 d}+\frac{\cosh (c+d x) \sinh (c+d x)}{2 b d}\\ \end{align*}

Mathematica [A]  time = 0.24996, size = 71, normalized size = 0.9 \[ \frac{\frac{4 a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a-b} \tanh (c+d x)}{\sqrt{a}}\right )}{\sqrt{a-b}}-2 (2 a+b) (c+d x)+b \sinh (2 (c+d x))}{4 b^2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sinh[c + d*x]^4/(a + b*Sinh[c + d*x]^2),x]

[Out]

(-2*(2*a + b)*(c + d*x) + (4*a^(3/2)*ArcTanh[(Sqrt[a - b]*Tanh[c + d*x])/Sqrt[a]])/Sqrt[a - b] + b*Sinh[2*(c +
d*x)])/(4*b^2*d)

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Maple [B]  time = 0.043, size = 454, normalized size = 5.8 \begin{align*} -{\frac{1}{2\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}+{\frac{1}{2\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}}-{\frac{a}{d{b}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }-{\frac{1}{2\,bd}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) }+{\frac{1}{2\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{1}{2\,bd} \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}+{\frac{a}{d{b}^{2}}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{1}{2\,bd}\ln \left ( \tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) }+{\frac{{a}^{2}}{d{b}^{2}}{\it Artanh} \left ({a\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{-b \left ( a-b \right ) }+a-2\,b \right ) a}}}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{-b \left ( a-b \right ) }+a-2\,b \right ) a}}}}-{\frac{{a}^{2}}{bd}{\it Artanh} \left ({a\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{-b \left ( a-b \right ) }+a-2\,b \right ) a}}}} \right ){\frac{1}{\sqrt{-b \left ( a-b \right ) }}}{\frac{1}{\sqrt{ \left ( 2\,\sqrt{-b \left ( a-b \right ) }+a-2\,b \right ) a}}}}-{\frac{{a}^{2}}{d{b}^{2}}\arctan \left ({a\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{-b \left ( a-b \right ) }-a+2\,b \right ) a}}}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{-b \left ( a-b \right ) }-a+2\,b \right ) a}}}}-{\frac{{a}^{2}}{bd}\arctan \left ({a\tanh \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ){\frac{1}{\sqrt{ \left ( 2\,\sqrt{-b \left ( a-b \right ) }-a+2\,b \right ) a}}}} \right ){\frac{1}{\sqrt{-b \left ( a-b \right ) }}}{\frac{1}{\sqrt{ \left ( 2\,\sqrt{-b \left ( a-b \right ) }-a+2\,b \right ) a}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sinh(d*x+c)^4/(a+b*sinh(d*x+c)^2),x)

[Out]

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

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^4/(a+b*sinh(d*x+c)^2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.12917, size = 2191, normalized size = 27.73 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^4/(a+b*sinh(d*x+c)^2),x, algorithm="fricas")

[Out]

[-1/8*(4*(2*a + b)*d*x*cosh(d*x + c)^2 - b*cosh(d*x + c)^4 - 4*b*cosh(d*x + c)*sinh(d*x + c)^3 - b*sinh(d*x +
c)^4 + 2*(2*(2*a + b)*d*x - 3*b*cosh(d*x + c)^2)*sinh(d*x + c)^2 - 4*(a*cosh(d*x + c)^2 + 2*a*cosh(d*x + c)*si
nh(d*x + c) + a*sinh(d*x + c)^2)*sqrt(a/(a - b))*log((b^2*cosh(d*x + c)^4 + 4*b^2*cosh(d*x + c)*sinh(d*x + c)^
3 + b^2*sinh(d*x + c)^4 + 2*(2*a*b - b^2)*cosh(d*x + c)^2 + 2*(3*b^2*cosh(d*x + c)^2 + 2*a*b - b^2)*sinh(d*x +
c)^2 + 8*a^2 - 8*a*b + b^2 + 4*(b^2*cosh(d*x + c)^3 + (2*a*b - b^2)*cosh(d*x + c))*sinh(d*x + c) - 4*((a*b -
b^2)*cosh(d*x + c)^2 + 2*(a*b - b^2)*cosh(d*x + c)*sinh(d*x + c) + (a*b - b^2)*sinh(d*x + c)^2 + 2*a^2 - 3*a*b
+ b^2)*sqrt(a/(a - b)))/(b*cosh(d*x + c)^4 + 4*b*cosh(d*x + c)*sinh(d*x + c)^3 + b*sinh(d*x + c)^4 + 2*(2*a -
b)*cosh(d*x + c)^2 + 2*(3*b*cosh(d*x + c)^2 + 2*a - b)*sinh(d*x + c)^2 + 4*(b*cosh(d*x + c)^3 + (2*a - b)*cos
h(d*x + c))*sinh(d*x + c) + b)) + 4*(2*(2*a + b)*d*x*cosh(d*x + c) - b*cosh(d*x + c)^3)*sinh(d*x + c) + b)/(b^
2*d*cosh(d*x + c)^2 + 2*b^2*d*cosh(d*x + c)*sinh(d*x + c) + b^2*d*sinh(d*x + c)^2), -1/8*(4*(2*a + b)*d*x*cosh
(d*x + c)^2 - b*cosh(d*x + c)^4 - 4*b*cosh(d*x + c)*sinh(d*x + c)^3 - b*sinh(d*x + c)^4 + 2*(2*(2*a + b)*d*x -
3*b*cosh(d*x + c)^2)*sinh(d*x + c)^2 - 8*(a*cosh(d*x + c)^2 + 2*a*cosh(d*x + c)*sinh(d*x + c) + a*sinh(d*x +
c)^2)*sqrt(-a/(a - b))*arctan(1/2*(b*cosh(d*x + c)^2 + 2*b*cosh(d*x + c)*sinh(d*x + c) + b*sinh(d*x + c)^2 + 2
*a - b)*sqrt(-a/(a - b))/a) + 4*(2*(2*a + b)*d*x*cosh(d*x + c) - b*cosh(d*x + c)^3)*sinh(d*x + c) + b)/(b^2*d*
cosh(d*x + c)^2 + 2*b^2*d*cosh(d*x + c)*sinh(d*x + c) + b^2*d*sinh(d*x + c)^2)]

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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(sinh(d*x+c)**4/(a+b*sinh(d*x+c)**2),x)

[Out]

Timed out

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Giac [B]  time = 1.23783, size = 181, normalized size = 2.29 \begin{align*} \frac{a^{2} \arctan \left (\frac{b e^{\left (2 \, d x + 2 \, c\right )} + 2 \, a - b}{2 \, \sqrt{-a^{2} + a b}}\right )}{\sqrt{-a^{2} + a b} b^{2} d} - \frac{{\left (d x + c\right )}{\left (2 \, a + b\right )}}{2 \, b^{2} d} + \frac{e^{\left (2 \, d x + 2 \, c\right )}}{8 \, b d} + \frac{{\left (4 \, a e^{\left (2 \, d x + 2 \, c\right )} + 2 \, b e^{\left (2 \, d x + 2 \, c\right )} - b\right )} e^{\left (-2 \, d x - 2 \, c\right )}}{8 \, b^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sinh(d*x+c)^4/(a+b*sinh(d*x+c)^2),x, algorithm="giac")

[Out]

a^2*arctan(1/2*(b*e^(2*d*x + 2*c) + 2*a - b)/sqrt(-a^2 + a*b))/(sqrt(-a^2 + a*b)*b^2*d) - 1/2*(d*x + c)*(2*a +
b)/(b^2*d) + 1/8*e^(2*d*x + 2*c)/(b*d) + 1/8*(4*a*e^(2*d*x + 2*c) + 2*b*e^(2*d*x + 2*c) - b)*e^(-2*d*x - 2*c)
/(b^2*d)