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]

(I*A-B)*arctanh(1/2*cosh(x)*a^(1/2)*2^(1/2)/(a+I*a*sinh(x))^(1/2))*2^(1/2)/a^(1/2)+2*B*cosh(x)/(a+I*a*sinh(x))
^(1/2)

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Rubi [A]  time = 0.07, antiderivative size = 66, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 20, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.150, 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 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 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 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)]

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*}

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Mathematica [A]  time = 0.12, 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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fricas [B]  time = 0.48, size = 186, normalized size = 2.82 \[ \frac {a \sqrt {-\frac {8 \, A^{2} + 16 i \, A B - 8 \, B^{2}}{a}} \log \left (-\frac {4 \, \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}} {\left (i \, A - B\right )} + a \sqrt {-\frac {8 \, A^{2} + 16 i \, A B - 8 \, B^{2}}{a}}}{-4 i \, A + 4 \, B}\right ) - a \sqrt {-\frac {8 \, A^{2} + 16 i \, A B - 8 \, B^{2}}{a}} \log \left (-\frac {4 \, \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}} {\left (i \, A - B\right )} - a \sqrt {-\frac {8 \, A^{2} + 16 i \, A B - 8 \, B^{2}}{a}}}{-4 i \, A + 4 \, B}\right ) - 4 \, \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}} {\left (i \, B e^{x} - B\right )}}{2 \, a} \]

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*(a*sqrt(-(8*A^2 + 16*I*A*B - 8*B^2)/a)*log(-(4*sqrt(1/2*I*a*e^(-x))*(I*A - B) + a*sqrt(-(8*A^2 + 16*I*A*B
- 8*B^2)/a))/(-4*I*A + 4*B)) - a*sqrt(-(8*A^2 + 16*I*A*B - 8*B^2)/a)*log(-(4*sqrt(1/2*I*a*e^(-x))*(I*A - B) -
a*sqrt(-(8*A^2 + 16*I*A*B - 8*B^2)/a))/(-4*I*A + 4*B)) - 4*sqrt(1/2*I*a*e^(-x))*(I*B*e^x - B))/a

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \sinh \relax (x) + A}{\sqrt {i \, a \sinh \relax (x) + a}}\,{d x} \]

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)

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maple [F]  time = 0.15, size = 0, normalized size = 0.00 \[ \int \frac {A +B \sinh \relax (x )}{\sqrt {a +i a \sinh \relax (x )}}\, dx \]

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 = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \sinh \relax (x) + A}{\sqrt {i \, a \sinh \relax (x) + a}}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.02 \[ \int \frac {A+B\,\mathrm {sinh}\relax (x)}{\sqrt {a+a\,\mathrm {sinh}\relax (x)\,1{}\mathrm {i}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {A + B \sinh {\relax (x )}}{\sqrt {i a \left (\sinh {\relax (x )} - i\right )}}\, dx \]

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(I*a*(sinh(x) - I)), x)

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