\(\int \sec ^4(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\) [281]

Optimal result

Integrand size = 26, antiderivative size = 59 \[ \int \sec ^4(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {4 i (a+i a \tan (c+d x))^{5/2}}{5 a^2 d}+\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a^3 d} \]

[Out]

-4/5*I*(a+I*a*tan(d*x+c))^(5/2)/a^2/d+2/7*I*(a+I*a*tan(d*x+c))^(7/2)/a^3/d

Rubi [A] (verified)

Time = 0.08 (sec) , antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3568, 45} \[ \int \sec ^4(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a^3 d}-\frac {4 i (a+i a \tan (c+d x))^{5/2}}{5 a^2 d} \]

[In]

Int[Sec[c + d*x]^4*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(((-4*I)/5)*(a + I*a*Tan[c + d*x])^(5/2))/(a^2*d) + (((2*I)/7)*(a + I*a*Tan[c + d*x])^(7/2))/(a^3*d)

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 3568

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
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rubi steps \begin{align*} \text {integral}& = -\frac {i \text {Subst}\left (\int (a-x) (a+x)^{3/2} \, dx,x,i a \tan (c+d x)\right )}{a^3 d} \\ & = -\frac {i \text {Subst}\left (\int \left (2 a (a+x)^{3/2}-(a+x)^{5/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^3 d} \\ & = -\frac {4 i (a+i a \tan (c+d x))^{5/2}}{5 a^2 d}+\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a^3 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.14 (sec) , antiderivative size = 48, normalized size of antiderivative = 0.81 \[ \int \sec ^4(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {2 (-i+\tan (c+d x))^2 (9 i+5 \tan (c+d x)) \sqrt {a+i a \tan (c+d x)}}{35 d} \]

[In]

Integrate[Sec[c + d*x]^4*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(2*(-I + Tan[c + d*x])^2*(9*I + 5*Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(35*d)

Maple [A] (verified)

Time = 1.04 (sec) , antiderivative size = 44, normalized size of antiderivative = 0.75

method	result	size
derivativedivides	\(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}\right )}{d \,a^{3}}\)	\(44\)
default	\(\frac {2 i \left (\frac {\left (a +i a \tan \left (d x +c \right )\right )^{\frac {7}{2}}}{7}-\frac {2 a \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5}\right )}{d \,a^{3}}\)	\(44\)

[In]

int(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

2*I/d/a^3*(1/7*(a+I*a*tan(d*x+c))^(7/2)-2/5*a*(a+I*a*tan(d*x+c))^(5/2))

Fricas [A] (verification not implemented)

none

Time = 0.25 (sec) , antiderivative size = 84, normalized size of antiderivative = 1.42 \[ \int \sec ^4(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {16 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (2 i \, e^{\left (7 i \, d x + 7 i \, c\right )} + 7 i \, e^{\left (5 i \, d x + 5 i \, c\right )}\right )}}{35 \, {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

-16/35*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(2*I*e^(7*I*d*x + 7*I*c) + 7*I*e^(5*I*d*x + 5*I*c))/(d*e^(6*I
*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)

Sympy [F]

\[ \int \sec ^4(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\int \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \sec ^{4}{\left (c + d x \right )}\, dx \]

[In]

integrate(sec(d*x+c)**4*(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Integral(sqrt(I*a*(tan(c + d*x) - I))*sec(c + d*x)**4, x)

Maxima [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 40, normalized size of antiderivative = 0.68 \[ \int \sec ^4(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\frac {2 i \, {\left (5 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} - 14 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a\right )}}{35 \, a^{3} d} \]

[In]

integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

2/35*I*(5*(I*a*tan(d*x + c) + a)^(7/2) - 14*(I*a*tan(d*x + c) + a)^(5/2)*a)/(a^3*d)

Giac [F]

\[ \int \sec ^4(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=\int { \sqrt {i \, a \tan \left (d x + c\right ) + a} \sec \left (d x + c\right )^{4} \,d x } \]

[In]

integrate(sec(d*x+c)^4*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(I*a*tan(d*x + c) + a)*sec(d*x + c)^4, x)

Mupad [B] (verification not implemented)

Time = 7.37 (sec) , antiderivative size = 230, normalized size of antiderivative = 3.90 \[ \int \sec ^4(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx=-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,32{}\mathrm {i}}{35\,d}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,16{}\mathrm {i}}{35\,d\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}+\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,128{}\mathrm {i}}{35\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}-\frac {\sqrt {a-\frac {a\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,1{}\mathrm {i}-\mathrm {i}\right )\,1{}\mathrm {i}}{{\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1}}\,16{}\mathrm {i}}{7\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3} \]

[In]

int((a + a*tan(c + d*x)*1i)^(1/2)/cos(c + d*x)^4,x)

[Out]

((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*128i)/(35*d*(exp(c*2i + d*x*2i) + 1)
^2) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*16i)/(35*d*(exp(c*2i + d*x*2i)
+ 1)) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*32i)/(35*d) - ((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)