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

Optimal. Leaf size=59 \[ \frac{2 i (a+i a \tan (c+d x))^{7/2}}{7 a^3 d}-\frac{4 i (a+i a \tan (c+d x))^{5/2}}{5 a^2 d} \]

[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)

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Rubi [A]  time = 0.0622915, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.077, Rules used = {3487, 43} \[ \frac{2 i (a+i a \tan (c+d x))^{7/2}}{7 a^3 d}-\frac{4 i (a+i a \tan (c+d x))^{5/2}}{5 a^2 d} \]

Antiderivative was successfully verified.

[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 3487

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]

Rule 43

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])

Rubi steps

\begin{align*} \int \sec ^4(c+d x) \sqrt{a+i a \tan (c+d x)} \, dx &=-\frac{i \operatorname{Subst}\left (\int (a-x) (a+x)^{3/2} \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=-\frac{i \operatorname{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]  time = 0.267442, size = 58, normalized size = 0.98 \[ \frac{2 \sqrt{a+i a \tan (c+d x)} \left (8 (\tan (c+d x)-i)+(5 \tan (c+d x)-i) \sec ^2(c+d x)\right )}{35 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Sqrt[a + I*a*Tan[c + d*x]]*(8*(-I + Tan[c + d*x]) + Sec[c + d*x]^2*(-I + 5*Tan[c + d*x])))/(35*d)

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Maple [A]  time = 0.34, size = 87, normalized size = 1.5 \begin{align*} -{\frac{16\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}-16\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +2\,i\cos \left ( dx+c \right ) -10\,\sin \left ( dx+c \right ) }{35\,d \left ( \cos \left ( dx+c \right ) \right ) ^{3}}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/35/d*(8*I*cos(d*x+c)^3-8*cos(d*x+c)^2*sin(d*x+c)+I*cos(d*x+c)-5*sin(d*x+c))*(a*(I*sin(d*x+c)+cos(d*x+c))/co
s(d*x+c))^(1/2)/cos(d*x+c)^3

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Maxima [A]  time = 0.947808, size = 54, normalized size = 0.92 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)

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Fricas [B]  time = 2.32816, size = 270, normalized size = 4.58 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-32 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 112 i \, e^{\left (4 i \, d x + 4 i \, c\right )}\right )} e^{\left (i \, d x + i \, c\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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-32*I*e^(6*I*d*x + 6*I*c) - 112*I*e^(4*I*d*x + 4*I*c))*e^(I*d*
x + 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)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a \left (i \tan{\left (c + d x \right )} + 1\right )} \sec ^{4}{\left (c + d x \right )}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sqrt(a*(I*tan(c + d*x) + 1))*sec(c + d*x)**4, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{i \, a \tan \left (d x + c\right ) + a} \sec \left (d x + c\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[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)