3.4.48 \(\int \frac {\sec ^6(c+d x)}{(a+i a \tan (c+d x))^{3/2}} \, dx\) [348]

Optimal. Leaf size=88 \[ -\frac {8 i (a+i a \tan (c+d x))^{3/2}}{3 a^3 d}+\frac {8 i (a+i a \tan (c+d x))^{5/2}}{5 a^4 d}-\frac {2 i (a+i a \tan (c+d x))^{7/2}}{7 a^5 d} \]

[Out]

-8/3*I*(a+I*a*tan(d*x+c))^(3/2)/a^3/d+8/5*I*(a+I*a*tan(d*x+c))^(5/2)/a^4/d-2/7*I*(a+I*a*tan(d*x+c))^(7/2)/a^5/
d

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Rubi [A]
time = 0.07, antiderivative size = 88, 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))^{7/2}}{7 a^5 d}+\frac {8 i (a+i a \tan (c+d x))^{5/2}}{5 a^4 d}-\frac {8 i (a+i a \tan (c+d x))^{3/2}}{3 a^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6/(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

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

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Mathematica [A]
time = 0.40, size = 92, normalized size = 1.05 \begin {gather*} -\frac {2 \sec ^5(c+d x) (28+43 \cos (2 (c+d x))-27 i \sin (2 (c+d x))) (\cos (3 (c+d x))+i \sin (3 (c+d x)))}{105 a d (-i+\tan (c+d x)) \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])^(3/2),x]

[Out]

(-2*Sec[c + d*x]^5*(28 + 43*Cos[2*(c + d*x)] - (27*I)*Sin[2*(c + d*x)])*(Cos[3*(c + d*x)] + I*Sin[3*(c + d*x)]
))/(105*a*d*(-I + Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

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Maple [A]
time = 0.78, size = 90, normalized size = 1.02

method	result	size
default	\(-\frac {2 \left (32 i \left (\cos ^{3}\left (d x +c \right )\right )-32 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+39 i \cos \left (d x +c \right )+15 \sin \left (d x +c \right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}}{105 d \cos \left (d x +c \right )^{3} a^{2}}\)	\(90\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6/(a+I*a*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-2/105/d*(32*I*cos(d*x+c)^3-32*cos(d*x+c)^2*sin(d*x+c)+39*I*cos(d*x+c)+15*sin(d*x+c))*(a*(I*sin(d*x+c)+cos(d*x
+c))/cos(d*x+c))^(1/2)/cos(d*x+c)^3/a^2

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Maxima [A]
time = 0.28, size = 58, normalized size = 0.66 \begin {gather*} -\frac {2 i \, {\left (15 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} - 84 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a + 140 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{2}\right )}}{105 \, a^{5} 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))^(3/2),x, algorithm="maxima")

[Out]

-2/105*I*(15*(I*a*tan(d*x + c) + a)^(7/2) - 84*(I*a*tan(d*x + c) + a)^(5/2)*a + 140*(I*a*tan(d*x + c) + a)^(3/
2)*a^2)/(a^5*d)

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Fricas [A]
time = 0.38, size = 108, normalized size = 1.23 \begin {gather*} -\frac {16 \, \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (8 i \, e^{\left (7 i \, d x + 7 i \, c\right )} + 28 i \, e^{\left (5 i \, d x + 5 i \, c\right )} + 35 i \, e^{\left (3 i \, d x + 3 i \, c\right )}\right )}}{105 \, {\left (a^{2} d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, a^{2} d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, a^{2} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2} d\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))^(3/2),x, algorithm="fricas")

[Out]

-16/105*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(8*I*e^(7*I*d*x + 7*I*c) + 28*I*e^(5*I*d*x + 5*I*c) + 35*I*e
^(3*I*d*x + 3*I*c))/(a^2*d*e^(6*I*d*x + 6*I*c) + 3*a^2*d*e^(4*I*d*x + 4*I*c) + 3*a^2*d*e^(2*I*d*x + 2*I*c) + a
^2*d)

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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 {3}{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))**(3/2),x)

[Out]

Integral(sec(c + d*x)**6/(I*a*(tan(c + d*x) - I))**(3/2), 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))^(3/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^6/(I*a*tan(d*x + c) + a)^(3/2), x)

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Mupad [B]
time = 6.68, size = 242, normalized size = 2.75 \begin {gather*} -\frac {\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}}\,128{}\mathrm {i}}{105\,a^2\,d}-\frac {\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}}\,64{}\mathrm {i}}{105\,a^2\,d\,\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}-\frac {\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}}\,16{}\mathrm {i}}{35\,a^2\,d\,{\left ({\mathrm {e}}^{c\,2{}\mathrm {i}+d\,x\,2{}\mathrm {i}}+1\right )}^2}+\frac {\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}}\,16{}\mathrm {i}}{7\,a^2\,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(1/(cos(c + d*x)^6*(a + a*tan(c + d*x)*1i)^(3/2)),x)

[Out]

((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*16i)/(7*a^2*d*(exp(c*2i + d*x*2i) +
1)^3) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*64i)/(105*a^2*d*(exp(c*2i +
d*x*2i) + 1)) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*16i)/(35*a^2*d*(exp(
c*2i + d*x*2i) + 1)^2) - ((a - (a*(exp(c*2i + d*x*2i)*1i - 1i)*1i)/(exp(c*2i + d*x*2i) + 1))^(1/2)*128i)/(105*
a^2*d)

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