3.123 \(\int \frac{A+B \sinh (x)}{\sqrt{a+i a \sinh (x)}} \, dx\)

Optimal. Leaf size=66 \[ \frac{\sqrt{2} (-B+i A) \tanh ^{-1}\left (\frac{\sqrt{a} \cosh (x)}{\sqrt{2} \sqrt{a+i a \sinh (x)}}\right )}{\sqrt{a}}+\frac{2 B \cosh (x)}{\sqrt{a+i a \sinh (x)}} \]

[Out]

(Sqrt[2]*(I*A - B)*ArcTanh[(Sqrt[a]*Cosh[x])/(Sqrt[2]*Sqrt[a + I*a*Sinh[x]])])/Sqrt[a] + (2*B*Cosh[x])/Sqrt[a
+ I*a*Sinh[x]]

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Rubi [A]  time = 0.0659672, antiderivative size = 66, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.15, Rules used = {2751, 2649, 206} \[ \frac{\sqrt{2} (-B+i A) \tanh ^{-1}\left (\frac{\sqrt{a} \cosh (x)}{\sqrt{2} \sqrt{a+i a \sinh (x)}}\right )}{\sqrt{a}}+\frac{2 B \cosh (x)}{\sqrt{a+i a \sinh (x)}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Sinh[x])/Sqrt[a + I*a*Sinh[x]],x]

[Out]

(Sqrt[2]*(I*A - B)*ArcTanh[(Sqrt[a]*Cosh[x])/(Sqrt[2]*Sqrt[a + I*a*Sinh[x]])])/Sqrt[a] + (2*B*Cosh[x])/Sqrt[a
+ I*a*Sinh[x]]

Rule 2751

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

Rule 2649

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

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])

Rubi steps

\begin{align*} \int \frac{A+B \sinh (x)}{\sqrt{a+i a \sinh (x)}} \, dx &=\frac{2 B \cosh (x)}{\sqrt{a+i a \sinh (x)}}+(A+i B) \int \frac{1}{\sqrt{a+i a \sinh (x)}} \, dx\\ &=\frac{2 B \cosh (x)}{\sqrt{a+i a \sinh (x)}}+(2 (i A-B)) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cosh (x)}{\sqrt{a+i a \sinh (x)}}\right )\\ &=\frac{\sqrt{2} (i A-B) \tanh ^{-1}\left (\frac{\sqrt{a} \cosh (x)}{\sqrt{2} \sqrt{a+i a \sinh (x)}}\right )}{\sqrt{a}}+\frac{2 B \cosh (x)}{\sqrt{a+i a \sinh (x)}}\\ \end{align*}

Mathematica [A]  time = 0.11868, size = 85, normalized size = 1.29 \[ \frac{2 \left (\cosh \left (\frac{x}{2}\right )+i \sinh \left (\frac{x}{2}\right )\right ) \left ((1+i) \sqrt [4]{-1} (B-i A) \tan ^{-1}\left (\frac{\tanh \left (\frac{x}{4}\right )+i}{\sqrt{2}}\right )-i B \sinh \left (\frac{x}{2}\right )+B \cosh \left (\frac{x}{2}\right )\right )}{\sqrt{a+i a \sinh (x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Sinh[x])/Sqrt[a + I*a*Sinh[x]],x]

[Out]

(2*(Cosh[x/2] + I*Sinh[x/2])*((1 + I)*(-1)^(1/4)*((-I)*A + B)*ArcTan[(I + Tanh[x/4])/Sqrt[2]] + B*Cosh[x/2] -
I*B*Sinh[x/2]))/Sqrt[a + I*a*Sinh[x]]

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Maple [F]  time = 0.11, size = 0, normalized size = 0. \begin{align*} \int{(A+B\sinh \left ( x \right ) ){\frac{1}{\sqrt{a+ia\sinh \left ( x \right ) }}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*sinh(x))/(a+I*a*sinh(x))^(1/2),x)

[Out]

int((A+B*sinh(x))/(a+I*a*sinh(x))^(1/2),x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \sinh \left (x\right ) + A}{\sqrt{i \, a \sinh \left (x\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sinh(x))/(a+I*a*sinh(x))^(1/2),x, algorithm="maxima")

[Out]

integrate((B*sinh(x) + A)/sqrt(I*a*sinh(x) + a), x)

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Fricas [B]  time = 1.80588, size = 738, normalized size = 11.18 \begin{align*} \frac{\sqrt{\frac{1}{2}} \sqrt{i \, a e^{\left (2 \, x\right )} + 2 \, a e^{x} - i \, a}{\left (-4 i \, B e^{x} + 4 \, B\right )} e^{\left (-\frac{1}{2} \, x\right )} - 2 \, \sqrt{2}{\left (a e^{x} - i \, a\right )} \sqrt{-\frac{A^{2} + 2 i \, A B - B^{2}}{a}} \log \left (\frac{\sqrt{\frac{1}{2}} \sqrt{i \, a e^{\left (2 \, x\right )} + 2 \, a e^{x} - i \, a}{\left (-4 i \, A + 4 \, B\right )} e^{\left (-\frac{1}{2} \, x\right )} + 2 \, \sqrt{2}{\left (a e^{x} - i \, a\right )} \sqrt{-\frac{A^{2} + 2 i \, A B - B^{2}}{a}}}{{\left (-4 i \, A + 4 \, B\right )} e^{x} - 4 \, A - 4 i \, B}\right ) + 2 \, \sqrt{2}{\left (a e^{x} - i \, a\right )} \sqrt{-\frac{A^{2} + 2 i \, A B - B^{2}}{a}} \log \left (\frac{\sqrt{\frac{1}{2}} \sqrt{i \, a e^{\left (2 \, x\right )} + 2 \, a e^{x} - i \, a}{\left (-4 i \, A + 4 \, B\right )} e^{\left (-\frac{1}{2} \, x\right )} - 2 \, \sqrt{2}{\left (a e^{x} - i \, a\right )} \sqrt{-\frac{A^{2} + 2 i \, A B - B^{2}}{a}}}{{\left (-4 i \, A + 4 \, B\right )} e^{x} - 4 \, A - 4 i \, B}\right )}{2 \,{\left (a e^{x} - i \, a\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sinh(x))/(a+I*a*sinh(x))^(1/2),x, algorithm="fricas")

[Out]

1/2*(sqrt(1/2)*sqrt(I*a*e^(2*x) + 2*a*e^x - I*a)*(-4*I*B*e^x + 4*B)*e^(-1/2*x) - 2*sqrt(2)*(a*e^x - I*a)*sqrt(
-(A^2 + 2*I*A*B - B^2)/a)*log((sqrt(1/2)*sqrt(I*a*e^(2*x) + 2*a*e^x - I*a)*(-4*I*A + 4*B)*e^(-1/2*x) + 2*sqrt(
2)*(a*e^x - I*a)*sqrt(-(A^2 + 2*I*A*B - B^2)/a))/((-4*I*A + 4*B)*e^x - 4*A - 4*I*B)) + 2*sqrt(2)*(a*e^x - I*a)
*sqrt(-(A^2 + 2*I*A*B - B^2)/a)*log((sqrt(1/2)*sqrt(I*a*e^(2*x) + 2*a*e^x - I*a)*(-4*I*A + 4*B)*e^(-1/2*x) - 2
*sqrt(2)*(a*e^x - I*a)*sqrt(-(A^2 + 2*I*A*B - B^2)/a))/((-4*I*A + 4*B)*e^x - 4*A - 4*I*B)))/(a*e^x - I*a)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{A + B \sinh{\left (x \right )}}{\sqrt{a \left (i \sinh{\left (x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sinh(x))/(a+I*a*sinh(x))**(1/2),x)

[Out]

Integral((A + B*sinh(x))/sqrt(a*(I*sinh(x) + 1)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{B \sinh \left (x\right ) + A}{\sqrt{i \, a \sinh \left (x\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*sinh(x))/(a+I*a*sinh(x))^(1/2),x, algorithm="giac")

[Out]

integrate((B*sinh(x) + A)/sqrt(I*a*sinh(x) + a), x)