3.2.32 \(\int \frac {\sec ^{10}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx\) [132]

Optimal. Leaf size=55 \[ \frac {2 i (a-i a \tan (c+d x))^5}{5 a^8 d}-\frac {i (a-i a \tan (c+d x))^6}{6 a^9 d} \]

[Out]

2/5*I*(a-I*a*tan(d*x+c))^5/a^8/d-1/6*I*(a-I*a*tan(d*x+c))^6/a^9/d

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Rubi [A]
time = 0.04, antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3568, 45} \begin {gather*} \frac {2 i (a-i a \tan (c+d x))^5}{5 a^8 d}-\frac {i (a-i a \tan (c+d x))^6}{6 a^9 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(((2*I)/5)*(a - I*a*Tan[c + d*x])^5)/(a^8*d) - ((I/6)*(a - I*a*Tan[c + d*x])^6)/(a^9*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 ^{10}(c+d x)}{(a+i a \tan (c+d x))^3} \, dx &=-\frac {i \text {Subst}\left (\int (a-x)^4 (a+x) \, dx,x,i a \tan (c+d x)\right )}{a^9 d}\\ &=-\frac {i \text {Subst}\left (\int \left (2 a (a-x)^4-(a-x)^5\right ) \, dx,x,i a \tan (c+d x)\right )}{a^9 d}\\ &=\frac {2 i (a-i a \tan (c+d x))^5}{5 a^8 d}-\frac {i (a-i a \tan (c+d x))^6}{6 a^9 d}\\ \end {align*}

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Mathematica [A]
time = 0.66, size = 97, normalized size = 1.76 \begin {gather*} \frac {\sec (c) \sec ^6(c+d x) (-20 i \cos (c)-15 i \cos (c+2 d x)-15 i \cos (3 c+2 d x)-20 \sin (c)+15 \sin (c+2 d x)-15 \sin (3 c+2 d x)+12 \sin (3 c+4 d x)+2 \sin (5 c+6 d x))}{60 a^3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^10/(a + I*a*Tan[c + d*x])^3,x]

[Out]

(Sec[c]*Sec[c + d*x]^6*((-20*I)*Cos[c] - (15*I)*Cos[c + 2*d*x] - (15*I)*Cos[3*c + 2*d*x] - 20*Sin[c] + 15*Sin[
c + 2*d*x] - 15*Sin[3*c + 2*d*x] + 12*Sin[3*c + 4*d*x] + 2*Sin[5*c + 6*d*x]))/(60*a^3*d)

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Maple [A]
time = 0.29, size = 72, normalized size = 1.31

method	result	size
risch	\(\frac {32 i \left (6 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{15 d \,a^{3} \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{6}}\)	\(36\)
derivativedivides	\(-\frac {i \left (i \tan \left (d x +c \right )-\frac {\left (\tan ^{6}\left (d x +c \right )\right )}{6}-\frac {3 i \left (\tan ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{2}-\frac {2 i \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {3 \left (\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d \,a^{3}}\)	\(72\)
default	\(-\frac {i \left (i \tan \left (d x +c \right )-\frac {\left (\tan ^{6}\left (d x +c \right )\right )}{6}-\frac {3 i \left (\tan ^{5}\left (d x +c \right )\right )}{5}+\frac {\left (\tan ^{4}\left (d x +c \right )\right )}{2}-\frac {2 i \left (\tan ^{3}\left (d x +c \right )\right )}{3}+\frac {3 \left (\tan ^{2}\left (d x +c \right )\right )}{2}\right )}{d \,a^{3}}\)	\(72\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

-I/d/a^3*(I*tan(d*x+c)-1/6*tan(d*x+c)^6-3/5*I*tan(d*x+c)^5+1/2*tan(d*x+c)^4-2/3*I*tan(d*x+c)^3+3/2*tan(d*x+c)^
2)

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Maxima [A]
time = 0.29, size = 67, normalized size = 1.22 \begin {gather*} \frac {5 i \, \tan \left (d x + c\right )^{6} - 18 \, \tan \left (d x + c\right )^{5} - 15 i \, \tan \left (d x + c\right )^{4} - 20 \, \tan \left (d x + c\right )^{3} - 45 i \, \tan \left (d x + c\right )^{2} + 30 \, \tan \left (d x + c\right )}{30 \, a^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

1/30*(5*I*tan(d*x + c)^6 - 18*tan(d*x + c)^5 - 15*I*tan(d*x + c)^4 - 20*tan(d*x + c)^3 - 45*I*tan(d*x + c)^2 +
30*tan(d*x + c))/(a^3*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. 112 vs. \(2 (43) = 86\).
time = 0.38, size = 112, normalized size = 2.04 \begin {gather*} -\frac {32 \, {\left (-6 i \, e^{\left (2 i \, d x + 2 i \, c\right )} - i\right )}}{15 \, {\left (a^{3} d e^{\left (12 i \, d x + 12 i \, c\right )} + 6 \, a^{3} d e^{\left (10 i \, d x + 10 i \, c\right )} + 15 \, a^{3} d e^{\left (8 i \, d x + 8 i \, c\right )} + 20 \, a^{3} d e^{\left (6 i \, d x + 6 i \, c\right )} + 15 \, a^{3} d e^{\left (4 i \, d x + 4 i \, c\right )} + 6 \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + a^{3} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-32/15*(-6*I*e^(2*I*d*x + 2*I*c) - I)/(a^3*d*e^(12*I*d*x + 12*I*c) + 6*a^3*d*e^(10*I*d*x + 10*I*c) + 15*a^3*d*
e^(8*I*d*x + 8*I*c) + 20*a^3*d*e^(6*I*d*x + 6*I*c) + 15*a^3*d*e^(4*I*d*x + 4*I*c) + 6*a^3*d*e^(2*I*d*x + 2*I*c
) + a^3*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {i \int \frac {\sec ^{10}{\left (c + d x \right )}}{\tan ^{3}{\left (c + d x \right )} - 3 i \tan ^{2}{\left (c + d x \right )} - 3 \tan {\left (c + d x \right )} + i}\, dx}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**10/(a+I*a*tan(d*x+c))**3,x)

[Out]

I*Integral(sec(c + d*x)**10/(tan(c + d*x)**3 - 3*I*tan(c + d*x)**2 - 3*tan(c + d*x) + I), x)/a**3

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Giac [A]
time = 1.01, size = 67, normalized size = 1.22 \begin {gather*} -\frac {-5 i \, \tan \left (d x + c\right )^{6} + 18 \, \tan \left (d x + c\right )^{5} + 15 i \, \tan \left (d x + c\right )^{4} + 20 \, \tan \left (d x + c\right )^{3} + 45 i \, \tan \left (d x + c\right )^{2} - 30 \, \tan \left (d x + c\right )}{30 \, a^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10/(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/30*(-5*I*tan(d*x + c)^6 + 18*tan(d*x + c)^5 + 15*I*tan(d*x + c)^4 + 20*tan(d*x + c)^3 + 45*I*tan(d*x + c)^2
- 30*tan(d*x + c))/(a^3*d)

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Mupad [B]
time = 3.34, size = 114, normalized size = 2.07 \begin {gather*} -\frac {\sin \left (c+d\,x\right )\,\left (-30\,{\cos \left (c+d\,x\right )}^5+{\cos \left (c+d\,x\right )}^4\,\sin \left (c+d\,x\right )\,45{}\mathrm {i}+20\,{\cos \left (c+d\,x\right )}^3\,{\sin \left (c+d\,x\right )}^2+{\cos \left (c+d\,x\right )}^2\,{\sin \left (c+d\,x\right )}^3\,15{}\mathrm {i}+18\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^4-{\sin \left (c+d\,x\right )}^5\,5{}\mathrm {i}\right )}{30\,a^3\,d\,{\cos \left (c+d\,x\right )}^6} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^10*(a + a*tan(c + d*x)*1i)^3),x)

[Out]

-(sin(c + d*x)*(18*cos(c + d*x)*sin(c + d*x)^4 + cos(c + d*x)^4*sin(c + d*x)*45i - 30*cos(c + d*x)^5 - sin(c +
d*x)^5*5i + cos(c + d*x)^2*sin(c + d*x)^3*15i + 20*cos(c + d*x)^3*sin(c + d*x)^2))/(30*a^3*d*cos(c + d*x)^6)

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