3.118 \(\int \frac{A+B \sinh (x)}{(i+\sinh (x))^4} \, dx\)

Optimal. Leaf size=91 \[ \frac{2 (3 A+4 i B) \cosh (x)}{105 (\sinh (x)+i)}+\frac{2 (-4 B+3 i A) \cosh (x)}{105 (\sinh (x)+i)^2}-\frac{(3 A+4 i B) \cosh (x)}{35 (\sinh (x)+i)^3}-\frac{(B+i A) \cosh (x)}{7 (\sinh (x)+i)^4} \]

[Out]

-((I*A + B)*Cosh[x])/(7*(I + Sinh[x])^4) - ((3*A + (4*I)*B)*Cosh[x])/(35*(I + Sinh[x])^3) + (2*((3*I)*A - 4*B)
*Cosh[x])/(105*(I + Sinh[x])^2) + (2*(3*A + (4*I)*B)*Cosh[x])/(105*(I + Sinh[x]))

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Rubi [A]  time = 0.0676414, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 15, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {2750, 2650, 2648} \[ \frac{2 (3 A+4 i B) \cosh (x)}{105 (\sinh (x)+i)}+\frac{2 (-4 B+3 i A) \cosh (x)}{105 (\sinh (x)+i)^2}-\frac{(3 A+4 i B) \cosh (x)}{35 (\sinh (x)+i)^3}-\frac{(B+i A) \cosh (x)}{7 (\sinh (x)+i)^4} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sinh[x])/(I + Sinh[x])^4,x]

[Out]

-((I*A + B)*Cosh[x])/(7*(I + Sinh[x])^4) - ((3*A + (4*I)*B)*Cosh[x])/(35*(I + Sinh[x])^3) + (2*((3*I)*A - 4*B)
*Cosh[x])/(105*(I + Sinh[x])^2) + (2*(3*A + (4*I)*B)*Cosh[x])/(105*(I + Sinh[x]))

Rule 2750

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((b
*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m)/(a*f*(2*m + 1)), x] + Dist[(a*d*m + b*c*(m + 1))/(a*b*(2*m + 1)
), Int[(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 -
b^2, 0] && LtQ[m, -2^(-1)]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rubi steps

\begin{align*} \int \frac{A+B \sinh (x)}{(i+\sinh (x))^4} \, dx &=-\frac{(i A+B) \cosh (x)}{7 (i+\sinh (x))^4}+\frac{1}{7} (-3 i A+4 B) \int \frac{1}{(i+\sinh (x))^3} \, dx\\ &=-\frac{(i A+B) \cosh (x)}{7 (i+\sinh (x))^4}-\frac{(3 A+4 i B) \cosh (x)}{35 (i+\sinh (x))^3}-\frac{1}{35} (2 (3 A+4 i B)) \int \frac{1}{(i+\sinh (x))^2} \, dx\\ &=-\frac{(i A+B) \cosh (x)}{7 (i+\sinh (x))^4}-\frac{(3 A+4 i B) \cosh (x)}{35 (i+\sinh (x))^3}+\frac{2 (3 i A-4 B) \cosh (x)}{105 (i+\sinh (x))^2}+\frac{1}{105} (2 (3 i A-4 B)) \int \frac{1}{i+\sinh (x)} \, dx\\ &=-\frac{(i A+B) \cosh (x)}{7 (i+\sinh (x))^4}-\frac{(3 A+4 i B) \cosh (x)}{35 (i+\sinh (x))^3}+\frac{2 (3 i A-4 B) \cosh (x)}{105 (i+\sinh (x))^2}+\frac{2 (3 A+4 i B) \cosh (x)}{105 (i+\sinh (x))}\\ \end{align*}

Mathematica [A]  time = 0.0490861, size = 67, normalized size = 0.74 \[ \frac{\cosh (x) \left ((6 A+8 i B) \sinh ^3(x)+8 i (3 A+4 i B) \sinh ^2(x)-13 (3 A+4 i B) \sinh (x)-36 i A+13 B\right )}{105 (\sinh (x)+i)^4} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sinh[x])/(I + Sinh[x])^4,x]

[Out]

(Cosh[x]*((-36*I)*A + 13*B - 13*(3*A + (4*I)*B)*Sinh[x] + (8*I)*(3*A + (4*I)*B)*Sinh[x]^2 + (6*A + (8*I)*B)*Si
nh[x]^3))/(105*(I + Sinh[x])^4)

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Maple [A]  time = 0.03, size = 128, normalized size = 1.4 \begin{align*} -{(6\,iA+2\,B) \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-2}}-{\frac{16\,A-16\,iB}{7} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-7}}+2\,{\frac{A}{\tanh \left ( x/2 \right ) +i}}-{\frac{-32\,iA-24\,B}{2} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-4}}-{\frac{-72\,A+64\,iB}{5} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-5}}-{\frac{36\,A-20\,iB}{3} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-3}}-{\frac{24\,iA+24\,B}{3} \left ( \tanh \left ({\frac{x}{2}} \right ) +i \right ) ^{-6}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sinh(x))/(I+sinh(x))^4,x)

[Out]

-(6*I*A+2*B)/(tanh(1/2*x)+I)^2-2/7*(8*A-8*I*B)/(tanh(1/2*x)+I)^7+2*A/(tanh(1/2*x)+I)-1/2*(-32*I*A-24*B)/(tanh(
1/2*x)+I)^4-2/5*(-36*A+32*I*B)/(tanh(1/2*x)+I)^5-2/3*(18*A-10*I*B)/(tanh(1/2*x)+I)^3-1/3*(24*I*A+24*B)/(tanh(1
/2*x)+I)^6

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Maxima [B]  time = 1.30924, size = 632, normalized size = 6.95 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sinh(x))/(I+sinh(x))^4,x, algorithm="maxima")

[Out]

1/2*B*(224*I*e^(-x)/(735*e^(-x) + 2205*I*e^(-2*x) - 3675*e^(-3*x) - 3675*I*e^(-4*x) + 2205*e^(-5*x) + 735*I*e^
(-6*x) - 105*e^(-7*x) - 105*I) - 672*e^(-2*x)/(735*e^(-x) + 2205*I*e^(-2*x) - 3675*e^(-3*x) - 3675*I*e^(-4*x)
+ 2205*e^(-5*x) + 735*I*e^(-6*x) - 105*e^(-7*x) - 105*I) - 560*I*e^(-3*x)/(735*e^(-x) + 2205*I*e^(-2*x) - 3675
*e^(-3*x) - 3675*I*e^(-4*x) + 2205*e^(-5*x) + 735*I*e^(-6*x) - 105*e^(-7*x) - 105*I) + 560*e^(-4*x)/(735*e^(-x
) + 2205*I*e^(-2*x) - 3675*e^(-3*x) - 3675*I*e^(-4*x) + 2205*e^(-5*x) + 735*I*e^(-6*x) - 105*e^(-7*x) - 105*I)
+ 32/(735*e^(-x) + 2205*I*e^(-2*x) - 3675*e^(-3*x) - 3675*I*e^(-4*x) + 2205*e^(-5*x) + 735*I*e^(-6*x) - 105*e
^(-7*x) - 105*I)) + A*(28*e^(-x)/(245*e^(-x) + 735*I*e^(-2*x) - 1225*e^(-3*x) - 1225*I*e^(-4*x) + 735*e^(-5*x)
+ 245*I*e^(-6*x) - 35*e^(-7*x) - 35*I) + 84*I*e^(-2*x)/(245*e^(-x) + 735*I*e^(-2*x) - 1225*e^(-3*x) - 1225*I*
e^(-4*x) + 735*e^(-5*x) + 245*I*e^(-6*x) - 35*e^(-7*x) - 35*I) - 140*e^(-3*x)/(245*e^(-x) + 735*I*e^(-2*x) - 1
225*e^(-3*x) - 1225*I*e^(-4*x) + 735*e^(-5*x) + 245*I*e^(-6*x) - 35*e^(-7*x) - 35*I) - 4*I/(245*e^(-x) + 735*I
*e^(-2*x) - 1225*e^(-3*x) - 1225*I*e^(-4*x) + 735*e^(-5*x) + 245*I*e^(-6*x) - 35*e^(-7*x) - 35*I))

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Fricas [A]  time = 1.83645, size = 305, normalized size = 3.35 \begin{align*} -\frac{280 \, B e^{\left (4 \, x\right )} + 140 \,{\left (3 \, A + 2 i \, B\right )} e^{\left (3 \, x\right )} -{\left (-252 i \, A + 336 \, B\right )} e^{\left (2 \, x\right )} - 28 \,{\left (3 \, A + 4 i \, B\right )} e^{x} - 12 i \, A + 16 \, B}{105 \, e^{\left (7 \, x\right )} + 735 i \, e^{\left (6 \, x\right )} - 2205 \, e^{\left (5 \, x\right )} - 3675 i \, e^{\left (4 \, x\right )} + 3675 \, e^{\left (3 \, x\right )} + 2205 i \, e^{\left (2 \, x\right )} - 735 \, e^{x} - 105 i} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sinh(x))/(I+sinh(x))^4,x, algorithm="fricas")

[Out]

-(280*B*e^(4*x) + 140*(3*A + 2*I*B)*e^(3*x) - (-252*I*A + 336*B)*e^(2*x) - 28*(3*A + 4*I*B)*e^x - 12*I*A + 16*
B)/(105*e^(7*x) + 735*I*e^(6*x) - 2205*e^(5*x) - 3675*I*e^(4*x) + 3675*e^(3*x) + 2205*I*e^(2*x) - 735*e^x - 10
5*I)

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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+B*sinh(x))/(I+sinh(x))**4,x)

[Out]

Timed out

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Giac [A]  time = 1.14118, size = 81, normalized size = 0.89 \begin{align*} -\frac{280 \, B e^{\left (4 \, x\right )} + 420 \, A e^{\left (3 \, x\right )} + 280 i \, B e^{\left (3 \, x\right )} + 252 i \, A e^{\left (2 \, x\right )} - 336 \, B e^{\left (2 \, x\right )} - 84 \, A e^{x} - 112 i \, B e^{x} - 12 i \, A + 16 \, B}{105 \,{\left (e^{x} + i\right )}^{7}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sinh(x))/(I+sinh(x))^4,x, algorithm="giac")

[Out]

-1/105*(280*B*e^(4*x) + 420*A*e^(3*x) + 280*I*B*e^(3*x) + 252*I*A*e^(2*x) - 336*B*e^(2*x) - 84*A*e^x - 112*I*B
*e^x - 12*I*A + 16*B)/(e^x + I)^7