Integrand size = 17, antiderivative size = 122 \[ \int \frac {1}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\frac {3 i \text {arctanh}\left (\frac {\sqrt {a} \cosh (c+d x)}{\sqrt {2} \sqrt {a+i a \sinh (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {i \cosh (c+d x)}{4 d (a+i a \sinh (c+d x))^{5/2}}+\frac {3 i \cosh (c+d x)}{16 a d (a+i a \sinh (c+d x))^{3/2}} \]
1/4*I*cosh(d*x+c)/d/(a+I*a*sinh(d*x+c))^(5/2)+3/16*I*cosh(d*x+c)/a/d/(a+I* a*sinh(d*x+c))^(3/2)+3/32*I*arctanh(1/2*cosh(d*x+c)*a^(1/2)*2^(1/2)/(a+I*a *sinh(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)
Time = 0.26 (sec) , antiderivative size = 210, normalized size of antiderivative = 1.72 \[ \int \frac {1}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\frac {\left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right ) \left (4 i \cosh \left (\frac {1}{2} (c+d x)\right )+(3-3 i) \sqrt [4]{-1} \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt [4]{-1} \left (1-i \tanh \left (\frac {1}{4} (c+d x)\right )\right )\right ) \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^4+4 \sinh \left (\frac {1}{2} (c+d x)\right )+6 \left (\cosh \left (\frac {1}{2} (c+d x)\right )+i \sinh \left (\frac {1}{2} (c+d x)\right )\right )^2 \sinh \left (\frac {1}{2} (c+d x)\right )+3 \left (-i \cosh \left (\frac {1}{2} (c+d x)\right )+\sinh \left (\frac {1}{2} (c+d x)\right )\right )^3\right )}{16 d (a+i a \sinh (c+d x))^{5/2}} \]
((Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])*((4*I)*Cosh[(c + d*x)/2] + (3 - 3*I)*(-1)^(1/4)*ArcTan[(1/2 + I/2)*(-1)^(1/4)*(1 - I*Tanh[(c + d*x)/4])]* (Cosh[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^4 + 4*Sinh[(c + d*x)/2] + 6*(Cos h[(c + d*x)/2] + I*Sinh[(c + d*x)/2])^2*Sinh[(c + d*x)/2] + 3*((-I)*Cosh[( c + d*x)/2] + Sinh[(c + d*x)/2])^3))/(16*d*(a + I*a*Sinh[c + d*x])^(5/2))
Time = 0.38 (sec) , antiderivative size = 127, normalized size of antiderivative = 1.04, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.412, Rules used = {3042, 3129, 3042, 3129, 3042, 3128, 219}
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.
\(\displaystyle \int \frac {1}{(a+i a \sinh (c+d x))^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {1}{(a+a \sin (i c+i d x))^{5/2}}dx\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle \frac {3 \int \frac {1}{(i \sinh (c+d x) a+a)^{3/2}}dx}{8 a}+\frac {i \cosh (c+d x)}{4 d (a+i a \sinh (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \int \frac {1}{(\sin (i c+i d x) a+a)^{3/2}}dx}{8 a}+\frac {i \cosh (c+d x)}{4 d (a+i a \sinh (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3129 |
\(\displaystyle \frac {3 \left (\frac {\int \frac {1}{\sqrt {i \sinh (c+d x) a+a}}dx}{4 a}+\frac {i \cosh (c+d x)}{2 d (a+i a \sinh (c+d x))^{3/2}}\right )}{8 a}+\frac {i \cosh (c+d x)}{4 d (a+i a \sinh (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {3 \left (\frac {\int \frac {1}{\sqrt {\sin (i c+i d x) a+a}}dx}{4 a}+\frac {i \cosh (c+d x)}{2 d (a+i a \sinh (c+d x))^{3/2}}\right )}{8 a}+\frac {i \cosh (c+d x)}{4 d (a+i a \sinh (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 3128 |
\(\displaystyle \frac {3 \left (\frac {i \int \frac {1}{2 a-\frac {a^2 \cosh ^2(c+d x)}{i \sinh (c+d x) a+a}}d\frac {a \cosh (c+d x)}{\sqrt {i \sinh (c+d x) a+a}}}{2 a d}+\frac {i \cosh (c+d x)}{2 d (a+i a \sinh (c+d x))^{3/2}}\right )}{8 a}+\frac {i \cosh (c+d x)}{4 d (a+i a \sinh (c+d x))^{5/2}}\) |
\(\Big \downarrow \) 219 |
\(\displaystyle \frac {3 \left (\frac {i \text {arctanh}\left (\frac {\sqrt {a} \cosh (c+d x)}{\sqrt {2} \sqrt {a+i a \sinh (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}+\frac {i \cosh (c+d x)}{2 d (a+i a \sinh (c+d x))^{3/2}}\right )}{8 a}+\frac {i \cosh (c+d x)}{4 d (a+i a \sinh (c+d x))^{5/2}}\) |
((I/4)*Cosh[c + d*x])/(d*(a + I*a*Sinh[c + d*x])^(5/2)) + (3*(((I/2)*ArcTa nh[(Sqrt[a]*Cosh[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Sinh[c + d*x]])])/(Sqrt[2 ]*a^(3/2)*d) + ((I/2)*Cosh[c + d*x])/(d*(a + I*a*Sinh[c + d*x])^(3/2))))/( 8*a)
3.1.71.3.1 Defintions of rubi rules used
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))* ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && NegQ[a/b] && (Gt Q[a, 0] || LtQ[b, 0])
Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[-2/d Subst[Int[1/(2*a - x^2), x], x, b*(Cos[c + d*x]/Sqrt[a + b*Sin[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[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/(a*d*(2*n + 1))), x] + Simp[(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]
\[\int \frac {1}{\left (a +i a \sinh \left (d x +c \right )\right )^{\frac {5}{2}}}d x\]
Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 348 vs. \(2 (91) = 182\).
Time = 0.30 (sec) , antiderivative size = 348, normalized size of antiderivative = 2.85 \[ \int \frac {1}{(a+i a \sinh (c+d x))^{5/2}} \, dx=-\frac {3 \, \sqrt {\frac {1}{2}} {\left (-i \, a^{3} d e^{\left (4 \, d x + 4 \, c\right )} - 4 \, a^{3} d e^{\left (3 \, d x + 3 \, c\right )} + 6 i \, a^{3} d e^{\left (2 \, d x + 2 \, c\right )} + 4 \, a^{3} d e^{\left (d x + c\right )} - i \, a^{3} d\right )} \sqrt {\frac {1}{a^{5} d^{2}}} \log \left (\sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} + \sqrt {\frac {1}{2} i \, a e^{\left (-d x - c\right )}}\right ) + 3 \, \sqrt {\frac {1}{2}} {\left (i \, a^{3} d e^{\left (4 \, d x + 4 \, c\right )} + 4 \, a^{3} d e^{\left (3 \, d x + 3 \, c\right )} - 6 i \, a^{3} d e^{\left (2 \, d x + 2 \, c\right )} - 4 \, a^{3} d e^{\left (d x + c\right )} + i \, a^{3} d\right )} \sqrt {\frac {1}{a^{5} d^{2}}} \log \left (-\sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {1}{a^{5} d^{2}}} + \sqrt {\frac {1}{2} i \, a e^{\left (-d x - c\right )}}\right ) - 2 \, \sqrt {\frac {1}{2} i \, a e^{\left (-d x - c\right )}} {\left (-3 i \, e^{\left (4 \, d x + 4 \, c\right )} - 11 \, e^{\left (3 \, d x + 3 \, c\right )} - 11 i \, e^{\left (2 \, d x + 2 \, c\right )} - 3 \, e^{\left (d x + c\right )}\right )}}{16 \, {\left (a^{3} d e^{\left (4 \, d x + 4 \, c\right )} - 4 i \, a^{3} d e^{\left (3 \, d x + 3 \, c\right )} - 6 \, a^{3} d e^{\left (2 \, d x + 2 \, c\right )} + 4 i \, a^{3} d e^{\left (d x + c\right )} + a^{3} d\right )}} \]
-1/16*(3*sqrt(1/2)*(-I*a^3*d*e^(4*d*x + 4*c) - 4*a^3*d*e^(3*d*x + 3*c) + 6 *I*a^3*d*e^(2*d*x + 2*c) + 4*a^3*d*e^(d*x + c) - I*a^3*d)*sqrt(1/(a^5*d^2) )*log(sqrt(1/2)*a^3*d*sqrt(1/(a^5*d^2)) + sqrt(1/2*I*a*e^(-d*x - c))) + 3* sqrt(1/2)*(I*a^3*d*e^(4*d*x + 4*c) + 4*a^3*d*e^(3*d*x + 3*c) - 6*I*a^3*d*e ^(2*d*x + 2*c) - 4*a^3*d*e^(d*x + c) + I*a^3*d)*sqrt(1/(a^5*d^2))*log(-sqr t(1/2)*a^3*d*sqrt(1/(a^5*d^2)) + sqrt(1/2*I*a*e^(-d*x - c))) - 2*sqrt(1/2* I*a*e^(-d*x - c))*(-3*I*e^(4*d*x + 4*c) - 11*e^(3*d*x + 3*c) - 11*I*e^(2*d *x + 2*c) - 3*e^(d*x + c)))/(a^3*d*e^(4*d*x + 4*c) - 4*I*a^3*d*e^(3*d*x + 3*c) - 6*a^3*d*e^(2*d*x + 2*c) + 4*I*a^3*d*e^(d*x + c) + a^3*d)
Timed out. \[ \int \frac {1}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\text {Timed out} \]
\[ \int \frac {1}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
\[ \int \frac {1}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\int { \frac {1}{{\left (i \, a \sinh \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]
Timed out. \[ \int \frac {1}{(a+i a \sinh (c+d x))^{5/2}} \, dx=\int \frac {1}{{\left (a+a\,\mathrm {sinh}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]