3.3.96 \(\int \sec ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\) [296]

Optimal. Leaf size=29 \[ -\frac {2 i (a+i a \tan (c+d x))^{5/2}}{5 a d} \]

[Out]

-2/5*I*(a+I*a*tan(d*x+c))^(5/2)/a/d

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Rubi [A]
time = 0.04, antiderivative size = 29, normalized size of antiderivative = 1.00, 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 = {3568, 32} \begin {gather*} -\frac {2 i (a+i a \tan (c+d x))^{5/2}}{5 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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*} \int \sec ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx &=-\frac {i \text {Subst}\left (\int (a+x)^{3/2} \, dx,x,i a \tan (c+d x)\right )}{a d}\\ &=-\frac {2 i (a+i a \tan (c+d x))^{5/2}}{5 a d}\\ \end {align*}

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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice the leaf count of optimal. \(69\) vs. \(2(29)=58\).
time = 0.40, size = 69, normalized size = 2.38 \begin {gather*} \frac {2 a \sec ^2(c+d x) (\cos (d x)-i \sin (d x)) (-i \cos (2 c+3 d x)+\sin (2 c+3 d x)) \sqrt {a+i a \tan (c+d x)}}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]
time = 0.21, size = 24, normalized size = 0.83

method	result	size
derivativedivides	\(-\frac {2 i \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5 d a}\)	\(24\)
default	\(-\frac {2 i \left (a +i a \tan \left (d x +c \right )\right )^{\frac {5}{2}}}{5 d a}\)	\(24\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*I*(a+I*a*tan(d*x+c))^(5/2)/d/a

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Maxima [A]
time = 0.28, size = 21, normalized size = 0.72 \begin {gather*} -\frac {2 i \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{5 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/5*I*(I*a*tan(d*x + c) + a)^(5/2)/(a*d)

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 59 vs. \(2 (21) = 42\).
time = 0.41, size = 59, normalized size = 2.03 \begin {gather*} -\frac {8 i \, \sqrt {2} a \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (5 i \, d x + 5 i \, c\right )}}{5 \, {\left (d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, 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)^2*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-8/5*I*sqrt(2)*a*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(5*I*d*x + 5*I*c)/(d*e^(4*I*d*x + 4*I*c) + 2*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 ^{2}{\left (c + d x \right )}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*(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)**2, 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)^2*(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)^2, x)

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Mupad [B]
time = 1.36, size = 153, normalized size = 5.28 \begin {gather*} -\frac {4\,a\,\sqrt {\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}}\,\left (\cos \left (2\,c+2\,d\,x\right )\,7{}\mathrm {i}+\cos \left (4\,c+4\,d\,x\right )\,4{}\mathrm {i}+\cos \left (6\,c+6\,d\,x\right )\,1{}\mathrm {i}-5\,\sin \left (2\,c+2\,d\,x\right )-4\,\sin \left (4\,c+4\,d\,x\right )-\sin \left (6\,c+6\,d\,x\right )+4{}\mathrm {i}\right )}{5\,d\,\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 )} \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)^2,x)

[Out]

-(4*a*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2)*(cos(2*c + 2*d*x)*7i + c
os(4*c + 4*d*x)*4i + cos(6*c + 6*d*x)*1i - 5*sin(2*c + 2*d*x) - 4*sin(4*c + 4*d*x) - sin(6*c + 6*d*x) + 4i))/(
5*d*(15*cos(2*c + 2*d*x) + 6*cos(4*c + 4*d*x) + cos(6*c + 6*d*x) + 10))

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