Integrand size = 14, antiderivative size = 171 \[ \int e^{c-5 i \arctan (a+b x)} \, dx=\frac {2 i e^c (1-i a-i b x)^{5/2}}{3 b (1+i a+i b x)^{3/2}}-\frac {10 i e^c (1-i a-i b x)^{3/2}}{3 b \sqrt {1+i a+i b x}}-\frac {5 i e^c \sqrt {1-i a-i b x} \sqrt {1+i a+i b x}}{b}-\frac {10 i e^c \arcsin \left (\frac {\sqrt {1+i a+i b x}}{\sqrt {2}}\right )}{b} \] Output:
2/3*I*exp(c)*(1-I*a-I*b*x)^(5/2)/b/(1+I*a+I*b*x)^(3/2)-10/3*I*exp(c)*(1-I* a-I*b*x)^(3/2)/b/(1+I*a+I*b*x)^(1/2)-5*I*exp(c)*(1-I*a-I*b*x)^(1/2)*(1+I*a +I*b*x)^(1/2)/b-10*I*exp(c)*arcsin(1/2*(1+I*a+I*b*x)^(1/2)*2^(1/2))/b
Time = 0.17 (sec) , antiderivative size = 63, normalized size of antiderivative = 0.37 \[ \int e^{c-5 i \arctan (a+b x)} \, dx=\frac {e^c \left (\sqrt {1+(a+b x)^2} \left (-3 i-\frac {8 i}{(-i+a+b x)^2}-\frac {28}{-i+a+b x}\right )+15 \text {arcsinh}(a+b x)\right )}{3 b} \] Input:
Integrate[E^(c - (5*I)*ArcTan[a + b*x]),x]
Output:
(E^c*(Sqrt[1 + (a + b*x)^2]*(-3*I - (8*I)/(-I + a + b*x)^2 - 28/(-I + a + b*x)) + 15*ArcSinh[a + b*x]))/(3*b)
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.
\(\displaystyle \int e^{c-5 i \arctan (a+b x)} \, dx\) |
\(\Big \downarrow \) 7281 |
\(\displaystyle \frac {\int e^{c-5 i \arctan (a+b x)}d(a+b x)}{b}\) |
\(\Big \downarrow \) 7299 |
\(\displaystyle \frac {\int e^{c-5 i \arctan (a+b x)}d(a+b x)}{b}\) |
Input:
Int[E^(c - (5*I)*ArcTan[a + b*x]),x]
Output:
$Aborted
Int[u_, x_Symbol] :> With[{lst = FunctionOfLinear[u, x]}, Simp[1/lst[[3]] Subst[Int[lst[[1]], x], x, lst[[2]] + lst[[3]]*x], x] /; !FalseQ[lst]]
\[\int {\mathrm e}^{c -5 i \arctan \left (b x +a \right )}d x\]
Input:
int(exp(c-5*I*arctan(b*x+a)),x)
Output:
int(exp(c-5*I*arctan(b*x+a)),x)
Time = 0.12 (sec) , antiderivative size = 124, normalized size of antiderivative = 0.73 \[ \int e^{c-5 i \arctan (a+b x)} \, dx=-\frac {15 \, {\left (b x + a - i\right )} e^{c} \log \left (\sqrt {\frac {b x + a + i}{b x + a - i}} + 1\right ) - 15 \, {\left (b x + a - i\right )} e^{c} \log \left (\sqrt {\frac {b x + a + i}{b x + a - i}} - 1\right ) - {\left (3 i \, b^{2} x^{2} - 2 \, {\left (-3 i \, a - 17\right )} b x + 3 i \, a^{2} + 34 \, a - 23 i\right )} \sqrt {\frac {b x + a + i}{b x + a - i}} e^{c}}{3 \, {\left (b^{2} x + {\left (a - i\right )} b\right )}} \] Input:
integrate(exp(c-5*I*arctan(b*x+a)),x, algorithm="fricas")
Output:
-1/3*(15*(b*x + a - I)*e^c*log(sqrt((b*x + a + I)/(b*x + a - I)) + 1) - 15 *(b*x + a - I)*e^c*log(sqrt((b*x + a + I)/(b*x + a - I)) - 1) - (3*I*b^2*x ^2 - 2*(-3*I*a - 17)*b*x + 3*I*a^2 + 34*a - 23*I)*sqrt((b*x + a + I)/(b*x + a - I))*e^c)/(b^2*x + (a - I)*b)
\[ \int e^{c-5 i \arctan (a+b x)} \, dx=e^{c} \int e^{- 5 i \operatorname {atan}{\left (a + b x \right )}}\, dx \] Input:
integrate(exp(c-5*I*atan(b*x+a)),x)
Output:
exp(c)*Integral(exp(-5*I*atan(a + b*x)), x)
Time = 0.16 (sec) , antiderivative size = 145, normalized size of antiderivative = 0.85 \[ \int e^{c-5 i \arctan (a+b x)} \, dx=\frac {15 \, {\left (b^{2} x^{2} e^{c} + 2 \, {\left (a - i\right )} b x e^{c} + {\left (a^{2} - 2 i \, a - 1\right )} e^{c}\right )} \log \left (b x + a + \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}\right ) - {\left (3 i \, b^{2} x^{2} e^{c} - 2 \, {\left (-3 i \, a - 17\right )} b x e^{c} + {\left (3 i \, a^{2} + 34 \, a - 23 i\right )} e^{c}\right )} \sqrt {b^{2} x^{2} + 2 \, a b x + a^{2} + 1}}{3 \, {\left (b^{3} x^{2} + 2 \, {\left (a - i\right )} b^{2} x + {\left (a^{2} - 2 i \, a - 1\right )} b\right )}} \] Input:
integrate(exp(c-5*I*arctan(b*x+a)),x, algorithm="maxima")
Output:
1/3*(15*(b^2*x^2*e^c + 2*(a - I)*b*x*e^c + (a^2 - 2*I*a - 1)*e^c)*log(b*x + a + sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1)) - (3*I*b^2*x^2*e^c - 2*(-3*I*a - 17)*b*x*e^c + (3*I*a^2 + 34*a - 23*I)*e^c)*sqrt(b^2*x^2 + 2*a*b*x + a^2 + 1))/(b^3*x^2 + 2*(a - I)*b^2*x + (a^2 - 2*I*a - 1)*b)
Timed out. \[ \int e^{c-5 i \arctan (a+b x)} \, dx=\text {Timed out} \] Input:
integrate(exp(c-5*I*arctan(b*x+a)),x, algorithm="giac")
Output:
Timed out
Timed out. \[ \int e^{c-5 i \arctan (a+b x)} \, dx=\int {\mathrm {e}}^{c-\mathrm {atan}\left (a+b\,x\right )\,5{}\mathrm {i}} \,d x \] Input:
int(exp(c - atan(a + b*x)*5i),x)
Output:
int(exp(c - atan(a + b*x)*5i), x)
\[ \int e^{c-5 i \arctan (a+b x)} \, dx=e^{c} \left (\int \frac {1}{e^{5 \mathit {atan} \left (b x +a \right ) i}}d x \right ) \] Input:
int(exp(c-5*I*atan(b*x+a)),x)
Output:
e**c*int(1/e**(5*atan(a + b*x)*i),x)