Integrand size = 26, antiderivative size = 86 \[ \int \frac {(e \sec (c+d x))^m}{(a+i a \tan (c+d x))^3} \, dx=\frac {i 2^{-3+\frac {m}{2}} \operatorname {Hypergeometric2F1}\left (4-\frac {m}{2},\frac {m}{2},\frac {2+m}{2},\frac {1}{2} (1-i \tan (c+d x))\right ) (e \sec (c+d x))^m (1+i \tan (c+d x))^{-m/2}}{a^3 d m} \] Output:
I*2^(-3+1/2*m)*hypergeom([1/2*m, 4-1/2*m],[1+1/2*m],1/2-1/2*I*tan(d*x+c))* (e*sec(d*x+c))^m/a^3/d/m/((1+I*tan(d*x+c))^(1/2*m))
Time = 1.58 (sec) , antiderivative size = 153, normalized size of antiderivative = 1.78 \[ \int \frac {(e \sec (c+d x))^m}{(a+i a \tan (c+d x))^3} \, dx=\frac {2^{-3+m} e^{-3 i (c+2 d x)} \left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^m \left (1+e^{2 i (c+d x)}\right )^m \operatorname {Hypergeometric2F1}\left (\frac {1}{2} (-6+m),-3+m,\frac {1}{2} (-4+m),-e^{2 i (c+d x)}\right ) \sec ^{3-m}(c+d x) (e \sec (c+d x))^m (\cos (d x)+i \sin (d x))^3}{a^3 d (-6+m) (-i+\tan (c+d x))^3} \] Input:
Integrate[(e*Sec[c + d*x])^m/(a + I*a*Tan[c + d*x])^3,x]
Output:
(2^(-3 + m)*(E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^m*(1 + E^((2*I)*(c + d*x)))^m*Hypergeometric2F1[(-6 + m)/2, -3 + m, (-4 + m)/2, -E^((2*I)*(c + d*x))]*Sec[c + d*x]^(3 - m)*(e*Sec[c + d*x])^m*(Cos[d*x] + I*Sin[d*x])^ 3)/(a^3*d*E^((3*I)*(c + 2*d*x))*(-6 + m)*(-I + Tan[c + d*x])^3)
Time = 0.47 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.02, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.231, 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.
\(\displaystyle \int \frac {(e \sec (c+d x))^m}{(a+i a \tan (c+d x))^3} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {(e \sec (c+d x))^m}{(a+i a \tan (c+d x))^3}dx\) |
\(\Big \downarrow \) 3986 |
\(\displaystyle (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \int (a-i a \tan (c+d x))^{m/2} (i \tan (c+d x) a+a)^{\frac {m-6}{2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \int (a-i a \tan (c+d x))^{m/2} (i \tan (c+d x) a+a)^{\frac {m-6}{2}}dx\) |
\(\Big \downarrow \) 4006 |
\(\displaystyle \frac {a^2 (a-i a \tan (c+d x))^{-m/2} (a+i a \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \int (a-i a \tan (c+d x))^{\frac {m-2}{2}} (i \tan (c+d x) a+a)^{\frac {m-8}{2}}d\tan (c+d x)}{d}\) |
\(\Big \downarrow \) 80 |
\(\displaystyle \frac {2^{\frac {m}{2}-4} (1+i \tan (c+d x))^{-m/2} (a-i a \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \int \left (\frac {1}{2} i \tan (c+d x)+\frac {1}{2}\right )^{\frac {m-8}{2}} (a-i a \tan (c+d x))^{\frac {m-2}{2}}d\tan (c+d x)}{a^2 d}\) |
\(\Big \downarrow \) 79 |
\(\displaystyle \frac {i 2^{\frac {m}{2}-3} (1+i \tan (c+d x))^{-m/2} (e \sec (c+d x))^m \operatorname {Hypergeometric2F1}\left (\frac {8-m}{2},\frac {m}{2},\frac {m+2}{2},\frac {1}{2} (1-i \tan (c+d x))\right )}{a^3 d m}\) |
Input:
Int[(e*Sec[c + d*x])^m/(a + I*a*Tan[c + d*x])^3,x]
Output:
(I*2^(-3 + m/2)*Hypergeometric2F1[(8 - m)/2, m/2, (2 + m)/2, (1 - I*Tan[c + d*x])/2]*(e*Sec[c + d*x])^m)/(a^3*d*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]
\[\int \frac {\left (e \sec \left (d x +c \right )\right )^{m}}{\left (a +i a \tan \left (d x +c \right )\right )^{3}}d x\]
Input:
int((e*sec(d*x+c))^m/(a+I*a*tan(d*x+c))^3,x)
Output:
int((e*sec(d*x+c))^m/(a+I*a*tan(d*x+c))^3,x)
\[ \int \frac {(e \sec (c+d x))^m}{(a+i a \tan (c+d x))^3} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{m}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:
integrate((e*sec(d*x+c))^m/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")
Output:
integral(1/8*(2*e*e^(I*d*x + I*c)/(e^(2*I*d*x + 2*I*c) + 1))^m*(e^(6*I*d*x + 6*I*c) + 3*e^(4*I*d*x + 4*I*c) + 3*e^(2*I*d*x + 2*I*c) + 1)*e^(-6*I*d*x - 6*I*c)/a^3, x)
\[ \int \frac {(e \sec (c+d x))^m}{(a+i a \tan (c+d x))^3} \, dx=\frac {i \int \frac {\left (e \sec {\left (c + d x \right )}\right )^{m}}{\tan ^{3}{\left (c + d x \right )} - 3 i \tan ^{2}{\left (c + d x \right )} - 3 \tan {\left (c + d x \right )} + i}\, dx}{a^{3}} \] Input:
integrate((e*sec(d*x+c))**m/(a+I*a*tan(d*x+c))**3,x)
Output:
I*Integral((e*sec(c + d*x))**m/(tan(c + d*x)**3 - 3*I*tan(c + d*x)**2 - 3* tan(c + d*x) + I), x)/a**3
Exception generated. \[ \int \frac {(e \sec (c+d x))^m}{(a+i a \tan (c+d x))^3} \, dx=\text {Exception raised: RuntimeError} \] Input:
integrate((e*sec(d*x+c))^m/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")
Output:
Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is un defined.
\[ \int \frac {(e \sec (c+d x))^m}{(a+i a \tan (c+d x))^3} \, dx=\int { \frac {\left (e \sec \left (d x + c\right )\right )^{m}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3}} \,d x } \] Input:
integrate((e*sec(d*x+c))^m/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")
Output:
integrate((e*sec(d*x + c))^m/(I*a*tan(d*x + c) + a)^3, x)
Timed out. \[ \int \frac {(e \sec (c+d x))^m}{(a+i a \tan (c+d x))^3} \, dx=\int \frac {{\left (\frac {e}{\cos \left (c+d\,x\right )}\right )}^m}{{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3} \,d x \] Input:
int((e/cos(c + d*x))^m/(a + a*tan(c + d*x)*1i)^3,x)
Output:
int((e/cos(c + d*x))^m/(a + a*tan(c + d*x)*1i)^3, x)
\[ \int \frac {(e \sec (c+d x))^m}{(a+i a \tan (c+d x))^3} \, dx=-\frac {e^{m} \left (\int \frac {\sec \left (d x +c \right )^{m}}{\tan \left (d x +c \right )^{3} i +3 \tan \left (d x +c \right )^{2}-3 \tan \left (d x +c \right ) i -1}d x \right )}{a^{3}} \] Input:
int((e*sec(d*x+c))^m/(a+I*a*tan(d*x+c))^3,x)
Output:
( - e**m*int(sec(c + d*x)**m/(tan(c + d*x)**3*i + 3*tan(c + d*x)**2 - 3*ta n(c + d*x)*i - 1),x))/a**3