3.4.35 \(\int \frac {\sec ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [335]

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

[Out]

-4/3*I*(a+I*a*tan(d*x+c))^(3/2)/a^2/d+2/5*I*(a+I*a*tan(d*x+c))^(5/2)/a^3/d

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Rubi [A]
time = 0.05, antiderivative size = 59, 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 = {3568, 45} \begin {gather*} \frac {2 i (a+i a \tan (c+d x))^{5/2}}{5 a^3 d}-\frac {4 i (a+i a \tan (c+d x))^{3/2}}{3 a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((-4*I)/3)*(a + I*a*Tan[c + d*x])^(3/2))/(a^2*d) + (((2*I)/5)*(a + I*a*Tan[c + d*x])^(5/2))/(a^3*d)

Rule 45

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])

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 \frac {\sec ^4(c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx &=-\frac {i \text {Subst}\left (\int (a-x) \sqrt {a+x} \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=-\frac {i \text {Subst}\left (\int \left (2 a \sqrt {a+x}-(a+x)^{3/2}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^3 d}\\ &=-\frac {4 i (a+i a \tan (c+d x))^{3/2}}{3 a^2 d}+\frac {2 i (a+i a \tan (c+d x))^{5/2}}{5 a^3 d}\\ \end {align*}

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Mathematica [A]
time = 0.28, size = 65, normalized size = 1.10 \begin {gather*} -\frac {2 \sec ^2(c+d x) (\cos (2 (c+d x))+i \sin (2 (c+d x))) (7 i+3 \tan (c+d x))}{15 d \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sec[c + d*x]^2*(Cos[2*(c + d*x)] + I*Sin[2*(c + d*x)])*(7*I + 3*Tan[c + d*x]))/(15*d*Sqrt[a + I*a*Tan[c +
d*x]])

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Maple [A]
time = 0.82, size = 73, normalized size = 1.24

method	result	size
default	\(\frac {2 \left (-4 i \left (\cos ^{2}\left (d x +c \right )\right )+4 \sin \left (d x +c \right ) \cos \left (d x +c \right )-3 i\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{15 d \cos \left (d x +c \right )^{2} a}\)	\(73\)

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,method=_RETURNVERBOSE)

[Out]

2/15/d*(-4*I*cos(d*x+c)^2+4*sin(d*x+c)*cos(d*x+c)-3*I)*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/cos(d*x+
c)^2/a

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Maxima [A]
time = 0.29, size = 79, normalized size = 1.34 \begin {gather*} -\frac {2 i \, {\left (15 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} - \frac {3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} - 10 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a + 15 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{2}}{a^{2}}\right )}}{15 \, a d} \end {gather*}

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]

-2/15*I*(15*sqrt(I*a*tan(d*x + c) + a) - (3*(I*a*tan(d*x + c) + a)^(5/2) - 10*(I*a*tan(d*x + c) + a)^(3/2)*a +
15*sqrt(I*a*tan(d*x + c) + a)*a^2)/a^2)/(a*d)

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Fricas [A]
time = 0.39, size = 76, normalized size = 1.29 \begin {gather*} -\frac {8 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (2 i \, e^{\left (5 i \, d x + 5 i \, c\right )} + 5 i \, e^{\left (3 i \, d x + 3 i \, c\right )}\right )}}{15 \, {\left (a d e^{\left (4 i \, d x + 4 i \, c\right )} + 2 \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + a d\right )}} \end {gather*}

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]

-8/15*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(2*I*e^(5*I*d*x + 5*I*c) + 5*I*e^(3*I*d*x + 3*I*c))/(a*d*e^(4*
I*d*x + 4*I*c) + 2*a*d*e^(2*I*d*x + 2*I*c) + a*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{4}{\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \end {gather*}

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

[Out]

integrate(sec(d*x + c)^4/sqrt(I*a*tan(d*x + c) + a), x)

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Mupad [B]
time = 1.31, size = 155, normalized size = 2.63 \begin {gather*} -\frac {8\,\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 )\,27{}\mathrm {i}+\cos \left (4\,c+4\,d\,x\right )\,9{}\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 )+19{}\mathrm {i}\right )}{15\,a\,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(1/(cos(c + d*x)^4*(a + a*tan(c + d*x)*1i)^(1/2)),x)

[Out]

-(8*((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)*27i + co
s(4*c + 4*d*x)*9i + 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) + 19i))/(
15*a*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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