3.342 \(\int \frac{\sec ^5(c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=73 \[ \frac{8 i a^2 \sec ^5(c+d x)}{35 d (a+i a \tan (c+d x))^{5/2}}+\frac{2 i a \sec ^5(c+d x)}{7 d (a+i a \tan (c+d x))^{3/2}} \]

[Out]

(((8*I)/35)*a^2*Sec[c + d*x]^5)/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (((2*I)/7)*a*Sec[c + d*x]^5)/(d*(a + I*a*Ta
n[c + d*x])^(3/2))

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Rubi [A]  time = 0.118018, antiderivative size = 73, normalized size of antiderivative = 1., 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 = {3494, 3493} \[ \frac{8 i a^2 \sec ^5(c+d x)}{35 d (a+i a \tan (c+d x))^{5/2}}+\frac{2 i a \sec ^5(c+d x)}{7 d (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((8*I)/35)*a^2*Sec[c + d*x]^5)/(d*(a + I*a*Tan[c + d*x])^(5/2)) + (((2*I)/7)*a*Sec[c + d*x]^5)/(d*(a + I*a*Ta
n[c + d*x])^(3/2))

Rule 3494

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] + Dist[(a*(m + 2*n - 2))/(m + n - 1), Int[(
d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2, 0]
&& IGtQ[Simplify[m/2 + n - 1], 0] &&  !IntegerQ[n]

Rule 3493

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(2*b*
(d*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^(n - 1))/(f*m), x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2
, 0] && EqQ[Simplify[m/2 + n - 1], 0]

Rubi steps

\begin{align*} \int \frac{\sec ^5(c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx &=\frac{2 i a \sec ^5(c+d x)}{7 d (a+i a \tan (c+d x))^{3/2}}+\frac{1}{7} (4 a) \int \frac{\sec ^5(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\\ &=\frac{8 i a^2 \sec ^5(c+d x)}{35 d (a+i a \tan (c+d x))^{5/2}}+\frac{2 i a \sec ^5(c+d x)}{7 d (a+i a \tan (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.213615, size = 65, normalized size = 0.89 \[ -\frac{2 (5 \tan (c+d x)-9 i) \sec ^3(c+d x) (\cos (2 (c+d x))-i \sin (2 (c+d x)))}{35 d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sec[c + d*x]^3*(Cos[2*(c + d*x)] - I*Sin[2*(c + d*x)])*(-9*I + 5*Tan[c + d*x]))/(35*d*Sqrt[a + I*a*Tan[c +
d*x]])

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Maple [A]  time = 0.322, size = 100, normalized size = 1.4 \begin{align*}{\frac{32\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}+32\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) -4\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}+12\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) -10\,i}{35\,ad \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)^5/(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

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

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Maxima [B]  time = 1.64271, size = 459, normalized size = 6.29 \begin{align*} -\frac{2 \,{\left (-9 i \, \sqrt{a} - \frac{26 \, \sqrt{a} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{14 i \, \sqrt{a} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{14 \, \sqrt{a} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{14 \, \sqrt{a} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac{14 i \, \sqrt{a} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{26 \, \sqrt{a} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac{9 i \, \sqrt{a} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} \sqrt{\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1} \sqrt{\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1}}{35 \,{\left (a - \frac{4 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{6 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac{4 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} d \sqrt{-\frac{2 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac{\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-2/35*(-9*I*sqrt(a) - 26*sqrt(a)*sin(d*x + c)/(cos(d*x + c) + 1) + 14*I*sqrt(a)*sin(d*x + c)^2/(cos(d*x + c) +
1)^2 - 14*sqrt(a)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 14*sqrt(a)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 14*I
*sqrt(a)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 26*sqrt(a)*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + 9*I*sqrt(a)*si
n(d*x + c)^8/(cos(d*x + c) + 1)^8)*sqrt(sin(d*x + c)/(cos(d*x + c) + 1) + 1)*sqrt(sin(d*x + c)/(cos(d*x + c) +
1) - 1)/((a - 4*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 4*a*sin(d*x
+ c)^6/(cos(d*x + c) + 1)^6 + a*sin(d*x + c)^8/(cos(d*x + c) + 1)^8)*d*sqrt(-2*I*sin(d*x + c)/(cos(d*x + c) +
1) + sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1))

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Fricas [A]  time = 2.10777, size = 274, normalized size = 3.75 \begin{align*} \frac{\sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (112 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 32 i\right )} e^{\left (i \, d x + i \, c\right )}}{35 \,{\left (a d e^{\left (7 i \, d x + 7 i \, c\right )} + 3 \, a d e^{\left (5 i \, d x + 5 i \, c\right )} + 3 \, a d e^{\left (3 i \, d x + 3 i \, c\right )} + a d e^{\left (i \, d x + i \, c\right )}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(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))*(112*I*e^(2*I*d*x + 2*I*c) + 32*I)*e^(I*d*x + I*c)/(a*d*e^(7*I*
d*x + 7*I*c) + 3*a*d*e^(5*I*d*x + 5*I*c) + 3*a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{5}{\left (c + d x \right )}}{\sqrt{a \left (i \tan{\left (c + d x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Integral(sec(c + d*x)**5/sqrt(a*(I*tan(c + d*x) + 1)), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{5}}{\sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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