\(\int (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^{5/2} \, dx\) [1059]

Optimal result

Integrand size = 33, antiderivative size = 67 \[ \int (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^{5/2} \, dx=\frac {i \operatorname {Hypergeometric2F1}\left (1,\frac {5}{2}+m,\frac {7}{2},\frac {1}{2} (1-i \tan (e+f x))\right ) (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^{5/2}}{5 f} \]

[Out]

1/5*I*hypergeom([1, 5/2+m],[7/2],1/2-1/2*I*tan(f*x+e))*(a+I*a*tan(f*x+e))^m*(c-I*c*tan(f*x+e))^(5/2)/f

Rubi [A] (verified)

Time = 0.14 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.31, number of steps used = 3, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.091, Rules used = {3604, 72, 71} \[ \int (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^{5/2} \, dx=\frac {i 2^m (c-i c \tan (e+f x))^{5/2} (1+i \tan (e+f x))^{-m} (a+i a \tan (e+f x))^m \operatorname {Hypergeometric2F1}\left (\frac {5}{2},1-m,\frac {7}{2},\frac {1}{2} (1-i \tan (e+f x))\right )}{5 f} \]

[In]

Int[(a + I*a*Tan[e + f*x])^m*(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

((I/5)*2^m*Hypergeometric2F1[5/2, 1 - m, 7/2, (1 - I*Tan[e + f*x])/2]*(a + I*a*Tan[e + f*x])^m*(c - I*c*Tan[e
+ f*x])^(5/2))/(f*(1 + I*Tan[e + f*x])^m)

Rule 71

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] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] && GtQ[b/(b*c - a*d), 0] && (RationalQ[m] ||  !(Ra
tionalQ[n] && GtQ[-d/(b*c - a*d), 0]))

Rule 72

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Dist[(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] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] &&  !IntegerQ[n] &&
(RationalQ[m] ||  !SimplerQ[n + 1, m + 1])

Rule 3604

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist
[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]

Rubi steps \begin{align*} \text {integral}& = \frac {(a c) \text {Subst}\left (\int (a+i a x)^{-1+m} (c-i c x)^{3/2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {\left (2^{-1+m} c (a+i a \tan (e+f x))^m \left (\frac {a+i a \tan (e+f x)}{a}\right )^{-m}\right ) \text {Subst}\left (\int \left (\frac {1}{2}+\frac {i x}{2}\right )^{-1+m} (c-i c x)^{3/2} \, dx,x,\tan (e+f x)\right )}{f} \\ & = \frac {i 2^m \operatorname {Hypergeometric2F1}\left (\frac {5}{2},1-m,\frac {7}{2},\frac {1}{2} (1-i \tan (e+f x))\right ) (1+i \tan (e+f x))^{-m} (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^{5/2}}{5 f} \\ \end{align*}

Mathematica [B] (verified)

Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(141\) vs. \(2(67)=134\).

Time = 20.07 (sec) , antiderivative size = 141, normalized size of antiderivative = 2.10 \[ \int (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^{5/2} \, dx=-\frac {i 2^{\frac {3}{2}+m} c \left (e^{i f x}\right )^m \left (\frac {c}{1+e^{2 i (e+f x)}}\right )^{3/2} \left (\frac {e^{i (e+f x)}}{1+e^{2 i (e+f x)}}\right )^m \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,1+m,-e^{2 i (e+f x)}\right ) \sec ^{-m}(e+f x) (\cos (f x)+i \sin (f x))^{-m} (a+i a \tan (e+f x))^m}{f m} \]

[In]

Integrate[(a + I*a*Tan[e + f*x])^m*(c - I*c*Tan[e + f*x])^(5/2),x]

[Out]

((-I)*2^(3/2 + m)*c*(E^(I*f*x))^m*(c/(1 + E^((2*I)*(e + f*x))))^(3/2)*(E^(I*(e + f*x))/(1 + E^((2*I)*(e + f*x)
)))^m*Hypergeometric2F1[-3/2, 1, 1 + m, -E^((2*I)*(e + f*x))]*(a + I*a*Tan[e + f*x])^m)/(f*m*Sec[e + f*x]^m*(C
os[f*x] + I*Sin[f*x])^m)

Maple [F(-1)]

Timed out.

\[\int \left (a +i a \tan \left (f x +e \right )\right )^{m} \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}d x\]

[In]

int((a+I*a*tan(f*x+e))^m*(c-I*c*tan(f*x+e))^(5/2),x)

[Out]

int((a+I*a*tan(f*x+e))^m*(c-I*c*tan(f*x+e))^(5/2),x)

Fricas [F]

\[ \int (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^{5/2} \, dx=\int { {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m} \,d x } \]

[In]

integrate((a+I*a*tan(f*x+e))^m*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="fricas")

[Out]

integral(4*sqrt(2)*c^2*(2*a*e^(2*I*f*x + 2*I*e)/(e^(2*I*f*x + 2*I*e) + 1))^m*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))
/(e^(4*I*f*x + 4*I*e) + 2*e^(2*I*f*x + 2*I*e) + 1), x)

Sympy [F(-1)]

Timed out. \[ \int (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^{5/2} \, dx=\text {Timed out} \]

[In]

integrate((a+I*a*tan(f*x+e))**m*(c-I*c*tan(f*x+e))**(5/2),x)

[Out]

Timed out

Maxima [F]

\[ \int (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^{5/2} \, dx=\int { {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m} \,d x } \]

[In]

integrate((a+I*a*tan(f*x+e))^m*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="maxima")

[Out]

integrate((-I*c*tan(f*x + e) + c)^(5/2)*(I*a*tan(f*x + e) + a)^m, x)

Giac [F]

\[ \int (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^{5/2} \, dx=\int { {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} {\left (i \, a \tan \left (f x + e\right ) + a\right )}^{m} \,d x } \]

[In]

integrate((a+I*a*tan(f*x+e))^m*(c-I*c*tan(f*x+e))^(5/2),x, algorithm="giac")

[Out]

integrate((-I*c*tan(f*x + e) + c)^(5/2)*(I*a*tan(f*x + e) + a)^m, x)

Mupad [F(-1)]

Timed out. \[ \int (a+i a \tan (e+f x))^m (c-i c \tan (e+f x))^{5/2} \, dx=\int {\left (a+a\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^m\,{\left (c-c\,\mathrm {tan}\left (e+f\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \]

[In]

int((a + a*tan(e + f*x)*1i)^m*(c - c*tan(e + f*x)*1i)^(5/2),x)

[Out]

int((a + a*tan(e + f*x)*1i)^m*(c - c*tan(e + f*x)*1i)^(5/2), x)