∫ (a + ia tan(e + fx))(c − ic tan(e + fx))5∕2 dx [970]

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Rubi [A]
time = 0.08, antiderivative size = 27, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {3603, 3568, 32} \begin {gather*} \frac {2 i a (c-i c \tan (e+f x))^{5/2}}{5 f} \end {gather*}

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

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b

*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x

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

st[a^m*c^m, Int[Sec[e + f*x]^(2*m)*(c + d*Tan[e + f*x])^(n - m), x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&

EqQ[b*c + a*d, 0] && EqQ[a^2 + b^2, 0] && IntegerQ[m] && !(IGtQ[n, 0] && (LtQ[m, 0] || GtQ[m, n]))

\begin {align*} \int (a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2} \, dx &=(a c) \int \sec ^2(e+f x) (c-i c \tan (e+f x))^{3/2} \, dx\\ &=\frac {(i a) \text {Subst}\left (\int (c+x)^{3/2} \, dx,x,-i c \tan (e+f x)\right )}{f}\\ &=\frac {2 i a (c-i c \tan (e+f x))^{5/2}}{5 f}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(70\) vs. \(2(27)=54\).
time = 0.68, size = 70, normalized size = 2.59 \begin {gather*} \frac {2 a c^2 \sec ^2(e+f x) (\cos (f x)-i \sin (f x)) (i \cos (2 e+f x)+\sin (2 e+f x)) \sqrt {c-i c \tan (e+f x)}}{5 f} \end {gather*}

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

(2*a*c^2*Sec[e + f*x]^2*(Cos[f*x] - I*Sin[f*x])*(I*Cos[2*e + f*x] + Sin[2*e + f*x])*Sqrt[c - I*c*Tan[e + f*x]]

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method	result	size

derivativedivides	\(\frac {2 i a \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5 f}\)	\(22\)
default	\(\frac {2 i a \left (c -i c \tan \left (f x +e \right )\right )^{\frac {5}{2}}}{5 f}\)	\(22\)

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

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Maxima [A]
time = 0.31, size = 20, normalized size = 0.74 \begin {gather*} \frac {2 i \, {\left (-i \, c \tan \left (f x + e\right ) + c\right )}^{\frac {5}{2}} a}{5 \, f} \end {gather*}

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

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 56 vs. \(2 (20) = 40\).
time = 0.94, size = 56, normalized size = 2.07 \begin {gather*} \frac {8 i \, \sqrt {2} a c^{2} \sqrt {\frac {c}{e^{\left (2 i \, f x + 2 i \, e\right )} + 1}}}{5 \, {\left (f e^{\left (4 i \, f x + 4 i \, e\right )} + 2 \, f e^{\left (2 i \, f x + 2 i \, e\right )} + f\right )}} \end {gather*}

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

8/5*I*sqrt(2)*a*c^2*sqrt(c/(e^(2*I*f*x + 2*I*e) + 1))/(f*e^(4*I*f*x + 4*I*e) + 2*f*e^(2*I*f*x + 2*I*e) + f)

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Sympy [A]
time = 4.43, size = 44, normalized size = 1.63 \begin {gather*} \begin {cases} \frac {2 i a \left (- i c \tan {\left (e + f x \right )} + c\right )^{\frac {5}{2}}}{5 f} & \text {for}\: f \neq 0 \\x \left (i a \tan {\left (e \right )} + a\right ) \left (- i c \tan {\left (e \right )} + c\right )^{\frac {5}{2}} & \text {otherwise} \end {cases} \end {gather*}

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

Piecewise((2*I*a*(-I*c*tan(e + f*x) + c)**(5/2)/(5*f), Ne(f, 0)), (x*(I*a*tan(e) + a)*(-I*c*tan(e) + c)**(5/2)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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

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Mupad [B]
time = 0.31, size = 120, normalized size = 4.44 \begin {gather*} \frac {a\,c^2\,\sqrt {\frac {c\,\left (\cos \left (2\,e+2\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,e+2\,f\,x\right )+1}}\,\left (2\,\cos \left (2\,e+2\,f\,x\right )+\cos \left (4\,e+4\,f\,x\right )+1-\sin \left (2\,e+2\,f\,x\right )\,2{}\mathrm {i}-\sin \left (4\,e+4\,f\,x\right )\,1{}\mathrm {i}\right )\,4{}\mathrm {i}}{5\,f\,\left (4\,\cos \left (2\,e+2\,f\,x\right )+\cos \left (4\,e+4\,f\,x\right )+3\right )} \end {gather*}

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

(a*c^2*((c*(cos(2*e + 2*f*x) - sin(2*e + 2*f*x)*1i + 1))/(cos(2*e + 2*f*x) + 1))^(1/2)*(2*cos(2*e + 2*f*x) + c

os(4*e + 4*f*x) - sin(2*e + 2*f*x)*2i - sin(4*e + 4*f*x)*1i + 1)*4i)/(5*f*(4*cos(2*e + 2*f*x) + cos(4*e + 4*f*

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3.10.70 \(\int (a+i a \tan (e+f x)) (c-i c \tan (e+f x))^{5/2} \, dx\) [970]