3.4.1 \(\int \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\) [301]

Optimal. Leaf size=110 \[ \frac {64 i a^3 \sec ^3(c+d x)}{105 d (a+i a \tan (c+d x))^{3/2}}+\frac {16 i a^2 \sec ^3(c+d x)}{35 d \sqrt {a+i a \tan (c+d x)}}+\frac {2 i a \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d} \]

[Out]

16/35*I*a^2*sec(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(1/2)+2/7*I*a*sec(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2)/d+64/105*I*a
^3*sec(d*x+c)^3/d/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]
time = 0.13, antiderivative size = 110, normalized size of antiderivative = 1.00, 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 = {3575, 3574} \begin {gather*} \frac {64 i a^3 \sec ^3(c+d x)}{105 d (a+i a \tan (c+d x))^{3/2}}+\frac {16 i a^2 \sec ^3(c+d x)}{35 d \sqrt {a+i a \tan (c+d x)}}+\frac {2 i a \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((64*I)/105)*a^3*Sec[c + d*x]^3)/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (((16*I)/35)*a^2*Sec[c + d*x]^3)/(d*Sqrt[
a + I*a*Tan[c + d*x]]) + (((2*I)/7)*a*Sec[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d

Rule 3574

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]

Rule 3575

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]

Rubi steps

\begin {align*} \int \sec ^3(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx &=\frac {2 i a \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}+\frac {1}{7} (8 a) \int \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\\ &=\frac {16 i a^2 \sec ^3(c+d x)}{35 d \sqrt {a+i a \tan (c+d x)}}+\frac {2 i a \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}+\frac {1}{35} \left (32 a^2\right ) \int \frac {\sec ^3(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\\ &=\frac {64 i a^3 \sec ^3(c+d x)}{105 d (a+i a \tan (c+d x))^{3/2}}+\frac {16 i a^2 \sec ^3(c+d x)}{35 d \sqrt {a+i a \tan (c+d x)}}+\frac {2 i a \sec ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{7 d}\\ \end {align*}

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Mathematica [A]
time = 0.58, size = 91, normalized size = 0.83 \begin {gather*} \frac {2 a \sec ^3(c+d x) (\cos (d x)-i \sin (d x)) (28+43 \cos (2 (c+d x))+27 i \sin (2 (c+d x))) (i \cos (2 c+d x)+\sin (2 c+d x)) \sqrt {a+i a \tan (c+d x)}}{105 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*Sec[c + d*x]^3*(Cos[d*x] - I*Sin[d*x])*(28 + 43*Cos[2*(c + d*x)] + (27*I)*Sin[2*(c + d*x)])*(I*Cos[2*c +
d*x] + Sin[2*c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(105*d)

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Maple [A]
time = 0.73, size = 98, normalized size = 0.89

method	result	size
default	\(\frac {2 \left (64 i \left (\cos ^{4}\left (d x +c \right )\right )+64 \sin \left (d x +c \right ) \left (\cos ^{3}\left (d x +c \right )\right )-8 i \left (\cos ^{2}\left (d x +c \right )\right )+24 \sin \left (d x +c \right ) \cos \left (d x +c \right )+15 i\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a}{105 d \cos \left (d x +c \right )^{3}}\)	\(98\)

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,method=_RETURNVERBOSE)

[Out]

2/105/d*(64*I*cos(d*x+c)^4+64*sin(d*x+c)*cos(d*x+c)^3-8*I*cos(d*x+c)^2+24*sin(d*x+c)*cos(d*x+c)+15*I)*(a*(I*si
n(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^3*a

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Maxima [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 580 vs. \(2 (86) = 172\).
time = 0.83, size = 580, normalized size = 5.27 \begin {gather*} \frac {16 \, {\left (35 i \, \sqrt {2} a \cos \left (4 \, d x + 4 \, c\right ) + 28 i \, \sqrt {2} a \cos \left (2 \, d x + 2 \, c\right ) - 35 \, \sqrt {2} a \sin \left (4 \, d x + 4 \, c\right ) - 28 \, \sqrt {2} a \sin \left (2 \, d x + 2 \, c\right ) + 8 i \, \sqrt {2} a\right )} {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )}^{\frac {1}{4}} \sqrt {a}}{105 \, {\left ({\left (2 \, \cos \left (2 \, d x + 2 \, c\right )^{3} + {\left (2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 i \, \sin \left (2 \, d x + 2 \, c\right )^{3} + {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (4 \, d x + 4 \, c\right ) + 5 \, \cos \left (2 \, d x + 2 \, c\right )^{2} + {\left (i \, \cos \left (2 \, d x + 2 \, c\right )^{2} + i \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 i \, \cos \left (2 \, d x + 2 \, c\right ) + i\right )} \sin \left (4 \, d x + 4 \, c\right ) + 2 \, {\left (i \, \cos \left (2 \, d x + 2 \, c\right )^{2} + 2 i \, \cos \left (2 \, d x + 2 \, c\right ) + i\right )} \sin \left (2 \, d x + 2 \, c\right ) + 4 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \cos \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right ) + {\left (2 i \, \cos \left (2 \, d x + 2 \, c\right )^{3} + {\left (2 i \, \cos \left (2 \, d x + 2 \, c\right ) + i\right )} \sin \left (2 \, d x + 2 \, c\right )^{2} - 2 \, \sin \left (2 \, d x + 2 \, c\right )^{3} + {\left (i \, \cos \left (2 \, d x + 2 \, c\right )^{2} + i \, \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 i \, \cos \left (2 \, d x + 2 \, c\right ) + i\right )} \cos \left (4 \, d x + 4 \, c\right ) + 5 i \, \cos \left (2 \, d x + 2 \, c\right )^{2} - {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + \sin \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (4 \, d x + 4 \, c\right ) - 2 \, {\left (\cos \left (2 \, d x + 2 \, c\right )^{2} + 2 \, \cos \left (2 \, d x + 2 \, c\right ) + 1\right )} \sin \left (2 \, d x + 2 \, c\right ) + 4 i \, \cos \left (2 \, d x + 2 \, c\right ) + i\right )} \sin \left (\frac {3}{2} \, \arctan \left (\sin \left (2 \, d x + 2 \, c\right ), \cos \left (2 \, d x + 2 \, c\right ) + 1\right )\right )\right )} d} \end {gather*}

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]

16/105*(35*I*sqrt(2)*a*cos(4*d*x + 4*c) + 28*I*sqrt(2)*a*cos(2*d*x + 2*c) - 35*sqrt(2)*a*sin(4*d*x + 4*c) - 28
*sqrt(2)*a*sin(2*d*x + 2*c) + 8*I*sqrt(2)*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)*sqrt(a)/(((2*cos(2*d*x + 2*c)^3 + (2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c)^2 + 2*I*sin(2*d*x + 2*c)^3
+ (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(4*d*x + 4*c) + 5*cos(2*d*x + 2*c)^2
+ (I*cos(2*d*x + 2*c)^2 + I*sin(2*d*x + 2*c)^2 + 2*I*cos(2*d*x + 2*c) + I)*sin(4*d*x + 4*c) + 2*(I*cos(2*d*x +
2*c)^2 + 2*I*cos(2*d*x + 2*c) + I)*sin(2*d*x + 2*c) + 4*cos(2*d*x + 2*c) + 1)*cos(3/2*arctan2(sin(2*d*x + 2*c
), cos(2*d*x + 2*c) + 1)) + (2*I*cos(2*d*x + 2*c)^3 + (2*I*cos(2*d*x + 2*c) + I)*sin(2*d*x + 2*c)^2 - 2*sin(2*
d*x + 2*c)^3 + (I*cos(2*d*x + 2*c)^2 + I*sin(2*d*x + 2*c)^2 + 2*I*cos(2*d*x + 2*c) + I)*cos(4*d*x + 4*c) + 5*I
*cos(2*d*x + 2*c)^2 - (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(4*d*x + 4*c) - 2*
(cos(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(2*d*x + 2*c) + 4*I*cos(2*d*x + 2*c) + I)*sin(3/2*arctan2(sin
(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))*d)

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Fricas [A]
time = 0.48, size = 89, normalized size = 0.81 \begin {gather*} -\frac {16 \, \sqrt {2} {\left (-35 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 28 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} - 8 i \, a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{105 \, {\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 {gather*}

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]

-16/105*sqrt(2)*(-35*I*a*e^(4*I*d*x + 4*I*c) - 28*I*a*e^(2*I*d*x + 2*I*c) - 8*I*a)*sqrt(a/(e^(2*I*d*x + 2*I*c)
+ 1))/(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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}} \sec ^{3}{\left (c + d x \right )}\, dx \end {gather*}

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

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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

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Mupad [B]
time = 6.02, size = 103, normalized size = 0.94 \begin {gather*} \frac {16\,a\,{\mathrm {e}}^{-c\,1{}\mathrm {i}-d\,x\,1{}\mathrm {i}}\,\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}}\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}\,28{}\mathrm {i}+{\mathrm {e}}^{c\,4{}\mathrm {i}+d\,x\,4{}\mathrm {i}}\,35{}\mathrm {i}+8{}\mathrm {i}\right )}{105\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^(3/2)/cos(c + d*x)^3,x)

[Out]

(16*a*exp(- c*1i - d*x*1i)*(a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*(exp(c*2i
+ d*x*2i)*28i + exp(c*4i + d*x*4i)*35i + 8i))/(105*d*(exp(c*2i + d*x*2i) + 1)^3)

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