Integrand size = 24, antiderivative size = 82 \[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x)) \, dx=\frac {i 2^{1+\frac {m}{2}} a \operatorname {Hypergeometric2F1}\left (-\frac {m}{2},\frac {m}{2},\frac {2+m}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^m (1+i \tan (c+d x))^{-m/2}}{d m} \] Output:
I*2^(1+1/2*m)*a*hypergeom([-1/2*m, 1/2*m],[1+1/2*m],1/2-1/2*I*tan(d*x+c))* (e*sec(d*x+c))^m/d/m/((1+I*tan(d*x+c))^(1/2*m))
Time = 0.14 (sec) , antiderivative size = 67, normalized size of antiderivative = 0.82 \[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x)) \, dx=\frac {a (e \sec (c+d x))^m \left (i+\cot (c+d x) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},\frac {m}{2},\frac {2+m}{2},\sec ^2(c+d x)\right ) \sqrt {-\tan ^2(c+d x)}\right )}{d m} \] Input:
Integrate[(e*Sec[c + d*x])^m*(a + I*a*Tan[c + d*x]),x]
Output:
(a*(e*Sec[c + d*x])^m*(I + Cot[c + d*x]*Hypergeometric2F1[1/2, m/2, (2 + m )/2, Sec[c + d*x]^2]*Sqrt[-Tan[c + d*x]^2]))/(d*m)
Time = 0.42 (sec) , antiderivative size = 114, normalized size of antiderivative = 1.39, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.250, Rules used = {3042, 3986, 3042, 4006, 80, 79}
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.
\(\displaystyle \int (a+i a \tan (c+d x)) (e \sec (c+d x))^m \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int (a+i a \tan (c+d x)) (e \sec (c+d x))^mdx\) |
\(\Big \downarrow \) 3986 |
\(\displaystyle (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \int (a-i a \tan (c+d x))^{m/2} (i \tan (c+d x) a+a)^{\frac {m+2}{2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \int (a-i a \tan (c+d x))^{m/2} (i \tan (c+d x) a+a)^{\frac {m+2}{2}}dx\) |
\(\Big \downarrow \) 4006 |
\(\displaystyle \frac {a^2 (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \int (a-i a \tan (c+d x))^{\frac {m-2}{2}} (i \tan (c+d x) a+a)^{m/2}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {a 2^{\frac {m}{2}-1} (1-i \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \int \left (\frac {1}{2}-\frac {1}{2} i \tan (c+d x)\right )^{\frac {m-2}{2}} (i \tan (c+d x) a+a)^{m/2}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle -\frac {i 2^{m/2} (1-i \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{\frac {m+2}{2}-\frac {m}{2}} (e \sec (c+d x))^m \operatorname {Hypergeometric2F1}\left (\frac {2-m}{2},\frac {m+2}{2},\frac {m+4}{2},\frac {1}{2} (i \tan (c+d x)+1)\right )}{d (m+2)}\) |
Input:
Int[(e*Sec[c + d*x])^m*(a + I*a*Tan[c + d*x]),x]
Output:
((-I)*2^(m/2)*Hypergeometric2F1[(2 - m)/2, (2 + m)/2, (4 + m)/2, (1 + I*Ta n[c + d*x])/2]*(e*Sec[c + d*x])^m*(a + I*a*Tan[c + d*x])^(-1/2*m + (2 + m) /2))/(d*(2 + m)*(1 - I*Tan[c + d*x])^(m/2))
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(( a + b*x)^(m + 1)/(b*(m + 1)*(b/(b*c - a*d))^n))*Hypergeometric2F1[-n, m + 1 , m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] || !(RationalQ[n] && GtQ[-d/(b*c - a*d), 0]))
Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(c + d*x)^FracPart[n]/((b/(b*c - a*d))^IntPart[n]*(b*((c + d*x)/(b*c - a*d))) ^FracPart[n]) Int[(a + b*x)^m*Simp[b*(c/(b*c - a*d)) + b*d*(x/(b*c - a*d) ), x]^n, x], x] /; FreeQ[{a, b, c, d, m, n}, x] && !IntegerQ[m] && !Integ erQ[n] && (RationalQ[m] || !SimplerQ[n + 1, m + 1])
Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*( x_)])^(n_.), x_Symbol] :> Simp[(d*Sec[e + f*x])^m/((a + b*Tan[e + f*x])^(m/ 2)*(a - b*Tan[e + f*x])^(m/2)) Int[(a + b*Tan[e + f*x])^(m/2 + n)*(a - b* Tan[e + f*x])^(m/2), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0]
Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + ( f_.)*(x_)])^(n_), x_Symbol] :> Simp[a*(c/f) Subst[Int[(a + b*x)^(m - 1)*( c + d*x)^(n - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n }, x] && EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0]
\[\int \left (e \sec \left (d x +c \right )\right )^{m} \left (a +i a \tan \left (d x +c \right )\right )d x\]
Input:
int((e*sec(d*x+c))^m*(a+I*a*tan(d*x+c)),x)
Output:
int((e*sec(d*x+c))^m*(a+I*a*tan(d*x+c)),x)
\[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x)) \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )} \left (e \sec \left (d x + c\right )\right )^{m} \,d x } \] Input:
integrate((e*sec(d*x+c))^m*(a+I*a*tan(d*x+c)),x, algorithm="fricas")
Output:
integral(2*a*(2*e*e^(I*d*x + I*c)/(e^(2*I*d*x + 2*I*c) + 1))^m*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1), x)
\[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x)) \, dx=i a \left (\int \left (- i \left (e \sec {\left (c + d x \right )}\right )^{m}\right )\, dx + \int \left (e \sec {\left (c + d x \right )}\right )^{m} \tan {\left (c + d x \right )}\, dx\right ) \] Input:
integrate((e*sec(d*x+c))**m*(a+I*a*tan(d*x+c)),x)
Output:
I*a*(Integral(-I*(e*sec(c + d*x))**m, x) + Integral((e*sec(c + d*x))**m*ta n(c + d*x), x))
\[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x)) \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )} \left (e \sec \left (d x + c\right )\right )^{m} \,d x } \] Input:
integrate((e*sec(d*x+c))^m*(a+I*a*tan(d*x+c)),x, algorithm="maxima")
Output:
integrate((I*a*tan(d*x + c) + a)*(e*sec(d*x + c))^m, x)
\[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x)) \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )} \left (e \sec \left (d x + c\right )\right )^{m} \,d x } \] Input:
integrate((e*sec(d*x+c))^m*(a+I*a*tan(d*x+c)),x, algorithm="giac")
Output:
integrate((I*a*tan(d*x + c) + a)*(e*sec(d*x + c))^m, x)
Timed out. \[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x)) \, dx=\int {\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^m\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right ) \,d x \] Input:
int((e/cos(c + d*x))^m*(a + a*tan(c + d*x)*1i),x)
Output:
int((e/cos(c + d*x))^m*(a + a*tan(c + d*x)*1i), x)
\[ \int (e \sec (c+d x))^m (a+i a \tan (c+d x)) \, dx=\frac {e^{m} a \left (\sec \left (d x +c \right )^{m} i +\left (\int \sec \left (d x +c \right )^{m}d x \right ) d m \right )}{d m} \] Input:
int((e*sec(d*x+c))^m*(a+I*a*tan(d*x+c)),x)
Output:
(e**m*a*(sec(c + d*x)**m*i + int(sec(c + d*x)**m,x)*d*m))/(d*m)