Integrand size = 17, antiderivative size = 69 \[ \int (a+i a \sinh (c+d x))^{3/2} \, dx=\frac {8 i a^2 \cosh (c+d x)}{3 d \sqrt {a+i a \sinh (c+d x)}}+\frac {2 i a \cosh (c+d x) \sqrt {a+i a \sinh (c+d x)}}{3 d} \] Output:
8/3*I*a^2*cosh(d*x+c)/d/(a+I*a*sinh(d*x+c))^(1/2)+2/3*I*a*cosh(d*x+c)*(a+I *a*sinh(d*x+c))^(1/2)/d
Time = 5.31 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.64 \[ \int (a+i a \sinh (c+d x))^{3/2} \, dx=-\frac {a (-i+\sinh (c+d x)) \sqrt {a+i a \sinh (c+d x)} \left (9 \cosh \left (\frac {1}{2} (c+d x)\right )+\cosh \left (\frac {3}{2} (c+d x)\right )-9 i \sinh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {3}{2} (c+d x)\right )\right )}{3 d \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^3} \] Input:
Integrate[(a + I*a*Sinh[c + d*x])^(3/2),x]
Output:
-1/3*(a*(-I + Sinh[c + d*x])*Sqrt[a + I*a*Sinh[c + d*x]]*(9*Cosh[(c + d*x) /2] + Cosh[(3*(c + d*x))/2] - (9*I)*Sinh[(c + d*x)/2] + I*Sinh[(3*(c + d*x ))/2]))/(d*(Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^3)
Time = 0.28 (sec) , antiderivative size = 69, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.235, Rules used = {3042, 3126, 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.
\(\displaystyle \int (a+i a \sinh (c+d x))^{3/2} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+a \sin (i c+i d x))^{3/2}dx\) |
\(\Big \downarrow \) 3126 |
\(\displaystyle \frac {4}{3} a \int \sqrt {i \sinh (c+d x) a+a}dx+\frac {2 i a \cosh (c+d x) \sqrt {a+i a \sinh (c+d x)}}{3 d}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {4}{3} a \int \sqrt {\sin (i c+i d x) a+a}dx+\frac {2 i a \cosh (c+d x) \sqrt {a+i a \sinh (c+d x)}}{3 d}\) |
\(\Big \downarrow \) 3125 |
\(\displaystyle \frac {8 i a^2 \cosh (c+d x)}{3 d \sqrt {a+i a \sinh (c+d x)}}+\frac {2 i a \cosh (c+d x) \sqrt {a+i a \sinh (c+d x)}}{3 d}\) |
Input:
Int[(a + I*a*Sinh[c + d*x])^(3/2),x]
Output:
(((8*I)/3)*a^2*Cosh[c + d*x])/(d*Sqrt[a + I*a*Sinh[c + d*x]]) + (((2*I)/3) *a*Cosh[c + d*x]*Sqrt[a + I*a*Sinh[c + d*x]])/d
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[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b)*Cos [c + d*x]*((a + b*Sin[c + d*x])^(n - 1)/(d*n)), x] + Simp[a*((2*n - 1)/n) Int[(a + b*Sin[c + d*x])^(n - 1), x], x] /; FreeQ[{a, b, c, d}, x] && EqQ[ a^2 - b^2, 0] && IGtQ[n - 1/2, 0]
Time = 0.27 (sec) , antiderivative size = 74, normalized size of antiderivative = 1.07
method | result | size |
default | \(\frac {4 a^{2} \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \sinh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right ) \left (\cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )^{2}+2\right ) \sqrt {2}}{3 \sqrt {a \cosh \left (\frac {c}{2}+\frac {i \pi }{4}+\frac {d x}{2}\right )^{2}}\, d}\) | \(74\) |
Input:
int((a+I*a*sinh(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
Output:
4/3*a^2*cosh(1/2*c+1/4*I*Pi+1/2*d*x)*sinh(1/2*c+1/4*I*Pi+1/2*d*x)*(cosh(1/ 2*c+1/4*I*Pi+1/2*d*x)^2+2)*2^(1/2)/(a*cosh(1/2*c+1/4*I*Pi+1/2*d*x)^2)^(1/2 )/d
Time = 0.08 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.91 \[ \int (a+i a \sinh (c+d x))^{3/2} \, dx=\frac {{\left (i \, a e^{\left (3 \, d x + 3 \, c\right )} + 9 \, a e^{\left (2 \, d x + 2 \, c\right )} + 9 i \, a e^{\left (d x + c\right )} + a\right )} \sqrt {\frac {1}{2} i \, a e^{\left (-d x - c\right )}} e^{\left (-d x - c\right )}}{3 \, d} \] Input:
integrate((a+I*a*sinh(d*x+c))^(3/2),x, algorithm="fricas")
Output:
1/3*(I*a*e^(3*d*x + 3*c) + 9*a*e^(2*d*x + 2*c) + 9*I*a*e^(d*x + c) + a)*sq rt(1/2*I*a*e^(-d*x - c))*e^(-d*x - c)/d
\[ \int (a+i a \sinh (c+d x))^{3/2} \, dx=\int \left (i a \sinh {\left (c + d x \right )} + a\right )^{\frac {3}{2}}\, dx \] Input:
integrate((a+I*a*sinh(d*x+c))**(3/2),x)
Output:
Integral((I*a*sinh(c + d*x) + a)**(3/2), x)
\[ \int (a+i a \sinh (c+d x))^{3/2} \, dx=\int { {\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \,d x } \] Input:
integrate((a+I*a*sinh(d*x+c))^(3/2),x, algorithm="maxima")
Output:
integrate((I*a*sinh(d*x + c) + a)^(3/2), x)
Time = 0.21 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.86 \[ \int (a+i a \sinh (c+d x))^{3/2} \, dx=-\frac {{\left (-\left (i - 1\right ) \, a^{\frac {3}{2}} e^{\left (3 \, d x + 3 \, c\right )} - \left (9 i + 9\right ) \, a^{\frac {3}{2}} e^{\left (2 \, d x + 2 \, c\right )} - \left (9 i - 9\right ) \, a^{\frac {3}{2}} e^{\left (d x + c\right )} - \left (i + 1\right ) \, a^{\frac {3}{2}}\right )} e^{\left (-\frac {3}{2} \, d x - \frac {3}{2} \, c\right )}}{6 \, d} \] Input:
integrate((a+I*a*sinh(d*x+c))^(3/2),x, algorithm="giac")
Output:
-1/6*(-(I - 1)*a^(3/2)*e^(3*d*x + 3*c) - (9*I + 9)*a^(3/2)*e^(2*d*x + 2*c) - (9*I - 9)*a^(3/2)*e^(d*x + c) - (I + 1)*a^(3/2))*e^(-3/2*d*x - 3/2*c)/d
Timed out. \[ \int (a+i a \sinh (c+d x))^{3/2} \, dx=\int {\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \] Input:
int((a + a*sinh(c + d*x)*1i)^(3/2),x)
Output:
int((a + a*sinh(c + d*x)*1i)^(3/2), x)
\[ \int (a+i a \sinh (c+d x))^{3/2} \, dx=\sqrt {a}\, a \left (\int \sqrt {\sinh \left (d x +c \right ) i +1}d x +\left (\int \sqrt {\sinh \left (d x +c \right ) i +1}\, \sinh \left (d x +c \right )d x \right ) i \right ) \] Input:
int((a+I*a*sinh(d*x+c))^(3/2),x)
Output:
sqrt(a)*a*(int(sqrt(sinh(c + d*x)*i + 1),x) + int(sqrt(sinh(c + d*x)*i + 1 )*sinh(c + d*x),x)*i)