Integrand size = 20, antiderivative size = 48 \[ \int \sqrt {a+i a \sinh (x)} (A+B \sinh (x)) \, dx=\frac {2 a (3 i A+B) \cosh (x)}{3 \sqrt {a+i a \sinh (x)}}+\frac {2}{3} B \cosh (x) \sqrt {a+i a \sinh (x)} \] Output:
2/3*a*(3*I*A+B)*cosh(x)/(a+I*a*sinh(x))^(1/2)+2/3*B*cosh(x)*(a+I*a*sinh(x) )^(1/2)
Time = 0.14 (sec) , antiderivative size = 66, normalized size of antiderivative = 1.38 \[ \int \sqrt {a+i a \sinh (x)} (A+B \sinh (x)) \, dx=\frac {2 \left (i \cosh \left (\frac {x}{2}\right )+\sinh \left (\frac {x}{2}\right )\right ) \sqrt {a+i a \sinh (x)} (3 A-2 i B+B \sinh (x))}{3 \left (\cosh \left (\frac {x}{2}\right )+i \sinh \left (\frac {x}{2}\right )\right )} \] Input:
Integrate[Sqrt[a + I*a*Sinh[x]]*(A + B*Sinh[x]),x]
Output:
(2*(I*Cosh[x/2] + Sinh[x/2])*Sqrt[a + I*a*Sinh[x]]*(3*A - (2*I)*B + B*Sinh [x]))/(3*(Cosh[x/2] + I*Sinh[x/2]))
Time = 0.29 (sec) , antiderivative size = 52, normalized size of antiderivative = 1.08, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3042, 3230, 3042, 3125}
Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.
\(\displaystyle \int \sqrt {a+i a \sinh (x)} (A+B \sinh (x)) \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \sqrt {a+a \sin (i x)} (A-i B \sin (i x))dx\) |
\(\Big \downarrow \) 3230 |
\(\displaystyle \frac {1}{3} (3 A-i B) \int \sqrt {i \sinh (x) a+a}dx+\frac {2}{3} B \cosh (x) \sqrt {a+i a \sinh (x)}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {1}{3} (3 A-i B) \int \sqrt {\sin (i x) a+a}dx+\frac {2}{3} B \cosh (x) \sqrt {a+i a \sinh (x)}\) |
\(\Big \downarrow \) 3125 |
\(\displaystyle \frac {2 i a (3 A-i B) \cosh (x)}{3 \sqrt {a+i a \sinh (x)}}+\frac {2}{3} B \cosh (x) \sqrt {a+i a \sinh (x)}\) |
Input:
Int[Sqrt[a + I*a*Sinh[x]]*(A + B*Sinh[x]),x]
Output:
(((2*I)/3)*a*(3*A - I*B)*Cosh[x])/Sqrt[a + I*a*Sinh[x]] + (2*B*Cosh[x]*Sqr t[a + I*a*Sinh[x]])/3
Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2*b*(Cos [c + d*x]/(d*Sqrt[a + b*Sin[c + d*x]])), x] /; FreeQ[{a, b, c, d}, x] && Eq Q[a^2 - b^2, 0]
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] + Simp[(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)]
\[\int \sqrt {a +i a \sinh \left (x \right )}\, \left (A +B \sinh \left (x \right )\right )d x\]
Input:
int((a+I*a*sinh(x))^(1/2)*(A+B*sinh(x)),x)
Output:
int((a+I*a*sinh(x))^(1/2)*(A+B*sinh(x)),x)
Time = 0.08 (sec) , antiderivative size = 49, normalized size of antiderivative = 1.02 \[ \int \sqrt {a+i a \sinh (x)} (A+B \sinh (x)) \, dx=\frac {1}{3} \, {\left (B e^{\left (3 \, x\right )} + 3 \, {\left (2 \, A - i \, B\right )} e^{\left (2 \, x\right )} - 3 \, {\left (-2 i \, A - B\right )} e^{x} - i \, B\right )} \sqrt {\frac {1}{2} i \, a e^{\left (-x\right )}} e^{\left (-x\right )} \] Input:
integrate((a+I*a*sinh(x))^(1/2)*(A+B*sinh(x)),x, algorithm="fricas")
Output:
1/3*(B*e^(3*x) + 3*(2*A - I*B)*e^(2*x) - 3*(-2*I*A - B)*e^x - I*B)*sqrt(1/ 2*I*a*e^(-x))*e^(-x)
\[ \int \sqrt {a+i a \sinh (x)} (A+B \sinh (x)) \, dx=\int \sqrt {i a \left (\sinh {\left (x \right )} - i\right )} \left (A + B \sinh {\left (x \right )}\right )\, dx \] Input:
integrate((a+I*a*sinh(x))**(1/2)*(A+B*sinh(x)),x)
Output:
Integral(sqrt(I*a*(sinh(x) - I))*(A + B*sinh(x)), x)
\[ \int \sqrt {a+i a \sinh (x)} (A+B \sinh (x)) \, dx=\int { {\left (B \sinh \left (x\right ) + A\right )} \sqrt {i \, a \sinh \left (x\right ) + a} \,d x } \] Input:
integrate((a+I*a*sinh(x))^(1/2)*(A+B*sinh(x)),x, algorithm="maxima")
Output:
integrate((B*sinh(x) + A)*sqrt(I*a*sinh(x) + a), x)
\[ \int \sqrt {a+i a \sinh (x)} (A+B \sinh (x)) \, dx=\int { {\left (B \sinh \left (x\right ) + A\right )} \sqrt {i \, a \sinh \left (x\right ) + a} \,d x } \] Input:
integrate((a+I*a*sinh(x))^(1/2)*(A+B*sinh(x)),x, algorithm="giac")
Output:
integrate((B*sinh(x) + A)*sqrt(I*a*sinh(x) + a), x)
Timed out. \[ \int \sqrt {a+i a \sinh (x)} (A+B \sinh (x)) \, dx=\int \left (A+B\,\mathrm {sinh}\left (x\right )\right )\,\sqrt {a+a\,\mathrm {sinh}\left (x\right )\,1{}\mathrm {i}} \,d x \] Input:
int((A + B*sinh(x))*(a + a*sinh(x)*1i)^(1/2),x)
Output:
int((A + B*sinh(x))*(a + a*sinh(x)*1i)^(1/2), x)
\[ \int \sqrt {a+i a \sinh (x)} (A+B \sinh (x)) \, dx=\sqrt {a}\, \left (\left (\int \sqrt {\sinh \left (x \right ) i +1}d x \right ) a +\left (\int \sqrt {\sinh \left (x \right ) i +1}\, \sinh \left (x \right )d x \right ) b \right ) \] Input:
int((a+I*a*sinh(x))^(1/2)*(A+B*sinh(x)),x)
Output:
sqrt(a)*(int(sqrt(sinh(x)*i + 1),x)*a + int(sqrt(sinh(x)*i + 1)*sinh(x),x) *b)