∫ sec2(c + dx)(a + ia tan(c + dx))5∕2 dx [309]

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Rubi [A]
time = 0.05, antiderivative size = 29, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3568, 32} \begin {gather*} -\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a d} \end {gather*}

Int[Sec[c + d*x]^2*(a + I*a*Tan[c + d*x])^(5/2),x]

(((-2*I)/7)*(a + I*a*Tan[c + d*x])^(7/2))/(a*d)

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

\begin {align*} \int \sec ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx &=-\frac {i \text {Subst}\left (\int (a+x)^{5/2} \, dx,x,i a \tan (c+d x)\right )}{a d}\\ &=-\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a d}\\ \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. \(73\) vs. \(2(29)=58\).
time = 0.49, size = 73, normalized size = 2.52 \begin {gather*} \frac {2 a^2 \sec ^3(c+d x) (-i \cos (3 c+5 d x)+\sin (3 c+5 d x)) \sqrt {a+i a \tan (c+d x)}}{7 d (\cos (d x)+i \sin (d x))^2} \end {gather*}

Integrate[Sec[c + d*x]^2*(a + I*a*Tan[c + d*x])^(5/2),x]

(2*a^2*Sec[c + d*x]^3*((-I)*Cos[3*c + 5*d*x] + Sin[3*c + 5*d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(7*d*(Cos[d*x] +

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method	result	size

derivativedivides	\(-\frac {2 i \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7 d a}\)	\(24\)
default	\(-\frac {2 i \left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7 d a}\)	\(24\)

int(sec(d*x+c)^2*(a+I*a*tan(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

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Maxima [A]
time = 0.28, size = 21, normalized size = 0.72 \begin {gather*} -\frac {2 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}}}{7 \, a d} \end {gather*}

integrate(sec(d*x+c)^2*(a+I*a*tan(d*x+c))^(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. 73 vs. \(2 (21) = 42\).
time = 0.39, size = 73, normalized size = 2.52 \begin {gather*} -\frac {16 i \, \sqrt {2} a^{2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (7 i \, d x + 7 i \, c\right )}}{7 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end {gather*}

integrate(sec(d*x+c)^2*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

-16/7*I*sqrt(2)*a^2*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(7*I*d*x + 7*I*c)/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {5}{2}} \sec ^{2}{\left (c + d x \right )}\, dx \end {gather*}

integrate(sec(d*x+c)**2*(a+I*a*tan(d*x+c))**(5/2),x)

Integral((I*a*(tan(c + d*x) - I))**(5/2)*sec(c + d*x)**2, x)

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

integrate(sec(d*x+c)^2*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

integrate((I*a*tan(d*x + c) + a)^(5/2)*sec(d*x + c)^2, x)

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Mupad [B]
time = 6.35, size = 242, normalized size = 8.34 \begin {gather*} -\frac {a^2\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,16{}\mathrm {i}}{7\,d}+\frac {a^2\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,48{}\mathrm {i}}{7\,d\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}-\frac {a^2\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,48{}\mathrm {i}}{7\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}+\frac {a^2\,\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,16{}\mathrm {i}}{7\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3} \end {gather*}

int((a + a*tan(c + d*x)*1i)^(5/2)/cos(c + d*x)^2,x)

(a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*48i)/(7*d*(exp(c*2i + d*x*2i) +

1)) - (a^2*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*16i)/(7*d) - (a^2*(a - (a*

(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*48i)/(7*d*(exp(c*2i + d*x*2i) + 1)^2) + (a^2*

(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*16i)/(7*d*(exp(c*2i + d*x*2i) + 1)^3)

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3.4.9 \(\int \sec ^2(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) [309]