3.358 \(\int \frac{\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=86 \[ \frac{2 i \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}-\frac{2 i \sec (c+d x)}{a d \sqrt{a+i a \tan (c+d x)}} \]

[Out]

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

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Rubi [A]  time = 0.101564, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.115, Rules used = {3491, 3489, 206} \[ \frac{2 i \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}-\frac{2 i \sec (c+d x)}{a d \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

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

Rule 3491

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

Rule 3489

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(-2*a)/(b*f), Subst[
Int[1/(2 - a*x^2), x], x, Sec[e + f*x]/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 + b^
2, 0]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \frac{\sec ^3(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx &=-\frac{2 i \sec (c+d x)}{a d \sqrt{a+i a \tan (c+d x)}}+\frac{2 \int \frac{\sec (c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx}{a}\\ &=-\frac{2 i \sec (c+d x)}{a d \sqrt{a+i a \tan (c+d x)}}+\frac{(4 i) \operatorname{Subst}\left (\int \frac{1}{2-a x^2} \, dx,x,\frac{\sec (c+d x)}{\sqrt{a+i a \tan (c+d x)}}\right )}{a d}\\ &=\frac{2 i \sqrt{2} \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{a^{3/2} d}-\frac{2 i \sec (c+d x)}{a d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.72189, size = 101, normalized size = 1.17 \[ \frac{8 e^{3 i (c+d x)} \left (-1+\sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )\right )}{a d \left (1+e^{2 i (c+d x)}\right )^2 (\tan (c+d x)-i) \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(8*E^((3*I)*(c + d*x))*(-1 + Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]]))/(a*d*(1 +
E^((2*I)*(c + d*x)))^2*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [B]  time = 0.238, size = 158, normalized size = 1.8 \begin{align*} -2\,{\frac{1}{{a}^{2}d \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) -1 \right ) }\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}} \left ( \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\arctan \left ( 1/2\,{\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i+\sin \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) \sqrt{2}\sin \left ( dx+c \right ) -i\cos \left ( dx+c \right ) +i-\sin \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

-2/d/a^2*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*((-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/
2)*(I*cos(d*x+c)-I+sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)*sin(d*x+c)-I*cos(d*x+c
)+I-sin(d*x+c))/(I*sin(d*x+c)+cos(d*x+c)-1)

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Maxima [B]  time = 2.0252, size = 1099, normalized size = 12.78 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

-1/2*((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*(2*sqrt(2)*arctan2((cos(2*d*x +
2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c
) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x +
2*c), cos(2*d*x + 2*c) + 1)) + 1) - 2*sqrt(2)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x +
2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2
*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 1) - I*sqrt(2
)*log(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(1/2*arctan2(sin(2*d*x + 2*c),
cos(2*d*x + 2*c) + 1))^2 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(1/2*arc
tan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 + 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2
*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) + I*sqrt(2)*log(sqrt(cos(2*d*x +
2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))
^2 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(1/2*arctan2(sin(2*d*x + 2*c),
cos(2*d*x + 2*c) + 1))^2 - 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*
arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1))*sqrt(a) - (-4*I*sqrt(2)*cos(1/2*arctan2(sin(2*d*x + 2*c
), cos(2*d*x + 2*c) + 1)) - 4*sqrt(2)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*sqrt(a))/((cos
(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*a^2*d)

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Fricas [B]  time = 2.10683, size = 690, normalized size = 8.02 \begin{align*} \frac{{\left (i \, \sqrt{2} a^{2} d \sqrt{\frac{1}{a^{3} d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left ({\left (\sqrt{2} a^{2} d \sqrt{\frac{1}{a^{3} d^{2}}} e^{\left (i \, d x + i \, c\right )} + \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - i \, \sqrt{2} a^{2} d \sqrt{\frac{1}{a^{3} d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (-{\left (\sqrt{2} a^{2} d \sqrt{\frac{1}{a^{3} d^{2}}} e^{\left (i \, d x + i \, c\right )} - \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 2 i \, \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{a^{2} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

(I*sqrt(2)*a^2*d*sqrt(1/(a^3*d^2))*e^(I*d*x + I*c)*log((sqrt(2)*a^2*d*sqrt(1/(a^3*d^2))*e^(I*d*x + I*c) + sqrt
(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - I*sqrt(2)
*a^2*d*sqrt(1/(a^3*d^2))*e^(I*d*x + I*c)*log(-(sqrt(2)*a^2*d*sqrt(1/(a^3*d^2))*e^(I*d*x + I*c) - sqrt(2)*sqrt(
a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 2*I*sqrt(2)*sqrt(a
/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(a^2*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec ^{3}{\left (c + d x \right )}}{\left (a \left (i \tan{\left (c + d x \right )} + 1\right )\right )^{\frac{3}{2}}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3/(a+I*a*tan(d*x+c))**(3/2),x)

[Out]

Integral(sec(c + d*x)**3/(a*(I*tan(c + d*x) + 1))**(3/2), x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (d x + c\right )^{3}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^3/(I*a*tan(d*x + c) + a)^(3/2), x)