\(\int \sec ^4(c+d x) (a+i a \tan (c+d x))^n \, dx\) [466]

Optimal result

Integrand size = 24, antiderivative size = 65 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^n \, dx=-\frac {2 i (a+i a \tan (c+d x))^{2+n}}{a^2 d (2+n)}+\frac {i (a+i a \tan (c+d x))^{3+n}}{a^3 d (3+n)} \]

[Out]

-2*I*(a+I*a*tan(d*x+c))^(2+n)/a^2/d/(2+n)+I*(a+I*a*tan(d*x+c))^(3+n)/a^3/d/(3+n)

Rubi [A] (verified)

Time = 0.07 (sec) , antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3568, 45} \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^n \, dx=\frac {i (a+i a \tan (c+d x))^{n+3}}{a^3 d (n+3)}-\frac {2 i (a+i a \tan (c+d x))^{n+2}}{a^2 d (n+2)} \]

[In]

Int[Sec[c + d*x]^4*(a + I*a*Tan[c + d*x])^n,x]

[Out]

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

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)^{1+n} \, dx,x,i a \tan (c+d x)\right )}{a^3 d} \\ & = -\frac {i \text {Subst}\left (\int \left (2 a (a+x)^{1+n}-(a+x)^{2+n}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^3 d} \\ & = -\frac {2 i (a+i a \tan (c+d x))^{2+n}}{a^2 d (2+n)}+\frac {i (a+i a \tan (c+d x))^{3+n}}{a^3 d (3+n)} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.28 (sec) , antiderivative size = 60, normalized size of antiderivative = 0.92 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^n \, dx=-\frac {i \left (\frac {2 a (a+i a \tan (c+d x))^{2+n}}{2+n}-\frac {(a+i a \tan (c+d x))^{3+n}}{3+n}\right )}{a^3 d} \]

[In]

Integrate[Sec[c + d*x]^4*(a + I*a*Tan[c + d*x])^n,x]

[Out]

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

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (61 ) = 122\).

Time = 1.01 (sec) , antiderivative size = 150, normalized size of antiderivative = 2.31

method	result	size
derivativedivides	\(\frac {\left (\tan ^{3}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (a +i a \tan \left (d x +c \right )\right )}}{d \left (3+n \right )}+\frac {\left (n +6\right ) \tan \left (d x +c \right ) {\mathrm e}^{n \ln \left (a +i a \tan \left (d x +c \right )\right )}}{d \left (2+n \right ) \left (3+n \right )}-\frac {i \left (4+n \right ) {\mathrm e}^{n \ln \left (a +i a \tan \left (d x +c \right )\right )}}{d \left (2+n \right ) \left (3+n \right )}-\frac {i n \left (\tan ^{2}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (a +i a \tan \left (d x +c \right )\right )}}{d \left (2+n \right ) \left (3+n \right )}\)	\(150\)
default	\(\frac {\left (\tan ^{3}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (a +i a \tan \left (d x +c \right )\right )}}{d \left (3+n \right )}+\frac {\left (n +6\right ) \tan \left (d x +c \right ) {\mathrm e}^{n \ln \left (a +i a \tan \left (d x +c \right )\right )}}{d \left (2+n \right ) \left (3+n \right )}-\frac {i \left (4+n \right ) {\mathrm e}^{n \ln \left (a +i a \tan \left (d x +c \right )\right )}}{d \left (2+n \right ) \left (3+n \right )}-\frac {i n \left (\tan ^{2}\left (d x +c \right )\right ) {\mathrm e}^{n \ln \left (a +i a \tan \left (d x +c \right )\right )}}{d \left (2+n \right ) \left (3+n \right )}\)	\(150\)
risch	\(\text {Expression too large to display}\)	\(1552\)

[In]

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

[Out]

1/d/(3+n)*tan(d*x+c)^3*exp(n*ln(a+I*a*tan(d*x+c)))+(n+6)/d/(2+n)/(3+n)*tan(d*x+c)*exp(n*ln(a+I*a*tan(d*x+c)))-
I*(4+n)/d/(2+n)/(3+n)*exp(n*ln(a+I*a*tan(d*x+c)))-I*n/d/(2+n)/(3+n)*tan(d*x+c)^2*exp(n*ln(a+I*a*tan(d*x+c)))

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 142 vs. \(2 (57) = 114\).

Time = 0.26 (sec) , antiderivative size = 142, normalized size of antiderivative = 2.18 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^n \, dx=-\frac {8 \, {\left ({\left (i \, n + 3 i\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + i \, e^{\left (6 i \, d x + 6 i \, c\right )}\right )} \left (\frac {2 \, a e^{\left (2 i \, d x + 2 i \, c\right )}}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}\right )^{n}}{d n^{2} + 5 \, d n + {\left (d n^{2} + 5 \, d n + 6 \, d\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, {\left (d n^{2} + 5 \, d n + 6 \, d\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, {\left (d n^{2} + 5 \, d n + 6 \, d\right )} e^{\left (2 i \, d x + 2 i \, c\right )} + 6 \, d} \]

[In]

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

[Out]

-8*((I*n + 3*I)*e^(4*I*d*x + 4*I*c) + I*e^(6*I*d*x + 6*I*c))*(2*a*e^(2*I*d*x + 2*I*c)/(e^(2*I*d*x + 2*I*c) + 1
))^n/(d*n^2 + 5*d*n + (d*n^2 + 5*d*n + 6*d)*e^(6*I*d*x + 6*I*c) + 3*(d*n^2 + 5*d*n + 6*d)*e^(4*I*d*x + 4*I*c)
+ 3*(d*n^2 + 5*d*n + 6*d)*e^(2*I*d*x + 2*I*c) + 6*d)

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F]

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

[In]

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

[Out]

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

Giac [F]

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

[In]

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

[Out]

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

Mupad [B] (verification not implemented)

Time = 2.82 (sec) , antiderivative size = 216, normalized size of antiderivative = 3.32 \[ \int \sec ^4(c+d x) (a+i a \tan (c+d x))^n \, dx=-\frac {4\,{\left (\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}\right )}^n\,\left (n\,3{}\mathrm {i}+\cos \left (2\,c+2\,d\,x\right )\,15{}\mathrm {i}+\cos \left (4\,c+4\,d\,x\right )\,6{}\mathrm {i}+\cos \left (6\,c+6\,d\,x\right )\,1{}\mathrm {i}-9\,\sin \left (2\,c+2\,d\,x\right )-6\,\sin \left (4\,c+4\,d\,x\right )-\sin \left (6\,c+6\,d\,x\right )+n\,\cos \left (2\,c+2\,d\,x\right )\,4{}\mathrm {i}+n\,\cos \left (4\,c+4\,d\,x\right )\,1{}\mathrm {i}-2\,n\,\sin \left (2\,c+2\,d\,x\right )-n\,\sin \left (4\,c+4\,d\,x\right )+10{}\mathrm {i}\right )}{d\,\left (n^2+5\,n+6\right )\,\left (15\,\cos \left (2\,c+2\,d\,x\right )+6\,\cos \left (4\,c+4\,d\,x\right )+\cos \left (6\,c+6\,d\,x\right )+10\right )} \]

[In]

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

[Out]

-(4*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^n*(n*3i + cos(2*c + 2*d*x)*15i +
cos(4*c + 4*d*x)*6i + cos(6*c + 6*d*x)*1i - 9*sin(2*c + 2*d*x) - 6*sin(4*c + 4*d*x) - sin(6*c + 6*d*x) + n*co
s(2*c + 2*d*x)*4i + n*cos(4*c + 4*d*x)*1i - 2*n*sin(2*c + 2*d*x) - n*sin(4*c + 4*d*x) + 10i))/(d*(5*n + n^2 +
6)*(15*cos(2*c + 2*d*x) + 6*cos(4*c + 4*d*x) + cos(6*c + 6*d*x) + 10))