3.4.80 \(\int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx\) [380]

Optimal. Leaf size=84 \[ \frac {8 i}{a^3 d \sqrt {a+i a \tan (c+d x)}}+\frac {8 i \sqrt {a+i a \tan (c+d x)}}{a^4 d}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{3 a^5 d} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 84, 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))^{3/2}}{3 a^5 d}+\frac {8 i \sqrt {a+i a \tan (c+d x)}}{a^4 d}+\frac {8 i}{a^3 d \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(8*I)/(a^3*d*Sqrt[a + I*a*Tan[c + d*x]]) + ((8*I)*Sqrt[a + I*a*Tan[c + d*x]])/(a^4*d) - (((2*I)/3)*(a + I*a*Ta
n[c + d*x])^(3/2))/(a^5*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 ^6(c+d x)}{(a+i a \tan (c+d x))^{7/2}} \, dx &=-\frac {i \text {Subst}\left (\int \frac {(a-x)^2}{(a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{a^5 d}\\ &=-\frac {i \text {Subst}\left (\int \left (\frac {4 a^2}{(a+x)^{3/2}}-\frac {4 a}{\sqrt {a+x}}+\sqrt {a+x}\right ) \, dx,x,i a \tan (c+d x)\right )}{a^5 d}\\ &=\frac {8 i}{a^3 d \sqrt {a+i a \tan (c+d x)}}+\frac {8 i \sqrt {a+i a \tan (c+d x)}}{a^4 d}-\frac {2 i (a+i a \tan (c+d x))^{3/2}}{3 a^5 d}\\ \end {align*}

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Mathematica [A]
time = 0.53, size = 61, normalized size = 0.73 \begin {gather*} \frac {2 i \sec ^2(c+d x) (12+11 \cos (2 (c+d x))+5 i \sin (2 (c+d x)))}{3 a^3 d \sqrt {a+i a \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6/(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(((2*I)/3)*Sec[c + d*x]^2*(12 + 11*Cos[2*(c + d*x)] + (5*I)*Sin[2*(c + d*x)]))/(a^3*d*Sqrt[a + I*a*Tan[c + d*x
]])

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Maple [A]
time = 0.88, size = 88, normalized size = 1.05

method	result	size
default	\(\frac {2 \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (12 i \left (\cos ^{3}\left (d x +c \right )\right )+12 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+11 i \cos \left (d x +c \right )+\sin \left (d x +c \right )\right )}{3 d \cos \left (d x +c \right ) a^{4}}\)	\(88\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

2/3/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(12*I*cos(d*x+c)^3+12*cos(d*x+c)^2*sin(d*x+c)+11*I*cos(d*
x+c)+sin(d*x+c))/cos(d*x+c)/a^4

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Maxima [A]
time = 0.29, size = 62, normalized size = 0.74 \begin {gather*} \frac {2 i \, {\left (\frac {12}{\sqrt {i \, a \tan \left (d x + c\right ) + a} a^{2}} - \frac {{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} - 12 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a}{a^{4}}\right )}}{3 \, a d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

2/3*I*(12/(sqrt(I*a*tan(d*x + c) + a)*a^2) - ((I*a*tan(d*x + c) + a)^(3/2) - 12*sqrt(I*a*tan(d*x + c) + a)*a)/
a^4)/(a*d)

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Fricas [A]
time = 0.39, size = 77, normalized size = 0.92 \begin {gather*} -\frac {4 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-8 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 12 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i\right )}}{3 \, {\left (a^{4} d e^{\left (3 i \, d x + 3 i \, c\right )} + a^{4} d e^{\left (i \, d x + i \, c\right )}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

-4/3*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-8*I*e^(4*I*d*x + 4*I*c) - 12*I*e^(2*I*d*x + 2*I*c) - 3*I)/(a^
4*d*e^(3*I*d*x + 3*I*c) + a^4*d*e^(I*d*x + I*c))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{6}{\left (c + d x \right )}}{\left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {7}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**6/(a+I*a*tan(d*x+c))**(7/2),x)

[Out]

Integral(sec(c + d*x)**6/(I*a*(tan(c + d*x) - I))**(7/2), 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)^6/(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^6/(I*a*tan(d*x + c) + a)^(7/2), x)

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Mupad [B]
time = 0.74, size = 110, normalized size = 1.31 \begin {gather*} \frac {2\,\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 )\,23{}\mathrm {i}+\cos \left (4\,c+4\,d\,x\right )\,3{}\mathrm {i}+7\,\sin \left (2\,c+2\,d\,x\right )+3\,\sin \left (4\,c+4\,d\,x\right )+20{}\mathrm {i}\right )}{3\,a^4\,d\,\left (\cos \left (2\,c+2\,d\,x\right )+1\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^6*(a + a*tan(c + d*x)*1i)^(7/2)),x)

[Out]

(2*((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)*23i + cos
(4*c + 4*d*x)*3i + 7*sin(2*c + 2*d*x) + 3*sin(4*c + 4*d*x) + 20i))/(3*a^4*d*(cos(2*c + 2*d*x) + 1))

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