3.1.36 \(\int \sec ^8(c+d x) (a+i a \tan (c+d x))^3 \, dx\) [36]

Optimal. Leaf size=109 \[ -\frac {8 i (a+i a \tan (c+d x))^7}{7 a^4 d}+\frac {3 i (a+i a \tan (c+d x))^8}{2 a^5 d}-\frac {2 i (a+i a \tan (c+d x))^9}{3 a^6 d}+\frac {i (a+i a \tan (c+d x))^{10}}{10 a^7 d} \]

[Out]

-8/7*I*(a+I*a*tan(d*x+c))^7/a^4/d+3/2*I*(a+I*a*tan(d*x+c))^8/a^5/d-2/3*I*(a+I*a*tan(d*x+c))^9/a^6/d+1/10*I*(a+
I*a*tan(d*x+c))^10/a^7/d

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Rubi [A]
time = 0.05, antiderivative size = 109, 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 {i (a+i a \tan (c+d x))^{10}}{10 a^7 d}-\frac {2 i (a+i a \tan (c+d x))^9}{3 a^6 d}+\frac {3 i (a+i a \tan (c+d x))^8}{2 a^5 d}-\frac {8 i (a+i a \tan (c+d x))^7}{7 a^4 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^8*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(((-8*I)/7)*(a + I*a*Tan[c + d*x])^7)/(a^4*d) + (((3*I)/2)*(a + I*a*Tan[c + d*x])^8)/(a^5*d) - (((2*I)/3)*(a +
I*a*Tan[c + d*x])^9)/(a^6*d) + ((I/10)*(a + I*a*Tan[c + d*x])^10)/(a^7*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 \sec ^8(c+d x) (a+i a \tan (c+d x))^3 \, dx &=-\frac {i \text {Subst}\left (\int (a-x)^3 (a+x)^6 \, dx,x,i a \tan (c+d x)\right )}{a^7 d}\\ &=-\frac {i \text {Subst}\left (\int \left (8 a^3 (a+x)^6-12 a^2 (a+x)^7+6 a (a+x)^8-(a+x)^9\right ) \, dx,x,i a \tan (c+d x)\right )}{a^7 d}\\ &=-\frac {8 i (a+i a \tan (c+d x))^7}{7 a^4 d}+\frac {3 i (a+i a \tan (c+d x))^8}{2 a^5 d}-\frac {2 i (a+i a \tan (c+d x))^9}{3 a^6 d}+\frac {i (a+i a \tan (c+d x))^{10}}{10 a^7 d}\\ \end {align*}

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Mathematica [A]
time = 1.17, size = 117, normalized size = 1.07 \begin {gather*} \frac {a^3 \sec (c) \sec ^{10}(c+d x) (126 i \cos (c)+105 i \cos (c+2 d x)+105 i \cos (3 c+2 d x)-126 \sin (c)+105 \sin (c+2 d x)-105 \sin (3 c+2 d x)+120 \sin (3 c+4 d x)+45 \sin (5 c+6 d x)+10 \sin (7 c+8 d x)+\sin (9 c+10 d x))}{840 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^8*(a + I*a*Tan[c + d*x])^3,x]

[Out]

(a^3*Sec[c]*Sec[c + d*x]^10*((126*I)*Cos[c] + (105*I)*Cos[c + 2*d*x] + (105*I)*Cos[3*c + 2*d*x] - 126*Sin[c] +
105*Sin[c + 2*d*x] - 105*Sin[3*c + 2*d*x] + 120*Sin[3*c + 4*d*x] + 45*Sin[5*c + 6*d*x] + 10*Sin[7*c + 8*d*x]
+ Sin[9*c + 10*d*x]))/(840*d)

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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 219 vs. \(2 (93 ) = 186\).
time = 0.27, size = 220, normalized size = 2.02

method	result	size
risch	\(\frac {128 i a^{3} \left (210 \,{\mathrm e}^{12 i \left (d x +c \right )}+252 \,{\mathrm e}^{10 i \left (d x +c \right )}+210 \,{\mathrm e}^{8 i \left (d x +c \right )}+120 \,{\mathrm e}^{6 i \left (d x +c \right )}+45 \,{\mathrm e}^{4 i \left (d x +c \right )}+10 \,{\mathrm e}^{2 i \left (d x +c \right )}+1\right )}{105 d \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )^{10}}\)	\(91\)
derivativedivides	\(\frac {-i a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{10 \cos \left (d x +c \right )^{10}}+\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{40 \cos \left (d x +c \right )^{8}}+\frac {\sin ^{4}\left (d x +c \right )}{20 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{40 \cos \left (d x +c \right )^{4}}\right )-3 a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{9 \cos \left (d x +c \right )^{9}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{21 \cos \left (d x +c \right )^{7}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{5}}+\frac {16 \left (\sin ^{3}\left (d x +c \right )\right )}{315 \cos \left (d x +c \right )^{3}}\right )+\frac {3 i a^{3}}{8 \cos \left (d x +c \right )^{8}}-a^{3} \left (-\frac {16}{35}-\frac {\left (\sec ^{6}\left (d x +c \right )\right )}{7}-\frac {6 \left (\sec ^{4}\left (d x +c \right )\right )}{35}-\frac {8 \left (\sec ^{2}\left (d x +c \right )\right )}{35}\right ) \tan \left (d x +c \right )}{d}\)	\(220\)
default	\(\frac {-i a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{10 \cos \left (d x +c \right )^{10}}+\frac {3 \left (\sin ^{4}\left (d x +c \right )\right )}{40 \cos \left (d x +c \right )^{8}}+\frac {\sin ^{4}\left (d x +c \right )}{20 \cos \left (d x +c \right )^{6}}+\frac {\sin ^{4}\left (d x +c \right )}{40 \cos \left (d x +c \right )^{4}}\right )-3 a^{3} \left (\frac {\sin ^{3}\left (d x +c \right )}{9 \cos \left (d x +c \right )^{9}}+\frac {2 \left (\sin ^{3}\left (d x +c \right )\right )}{21 \cos \left (d x +c \right )^{7}}+\frac {8 \left (\sin ^{3}\left (d x +c \right )\right )}{105 \cos \left (d x +c \right )^{5}}+\frac {16 \left (\sin ^{3}\left (d x +c \right )\right )}{315 \cos \left (d x +c \right )^{3}}\right )+\frac {3 i a^{3}}{8 \cos \left (d x +c \right )^{8}}-a^{3} \left (-\frac {16}{35}-\frac {\left (\sec ^{6}\left (d x +c \right )\right )}{7}-\frac {6 \left (\sec ^{4}\left (d x +c \right )\right )}{35}-\frac {8 \left (\sec ^{2}\left (d x +c \right )\right )}{35}\right ) \tan \left (d x +c \right )}{d}\)	\(220\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

1/d*(-I*a^3*(1/10*sin(d*x+c)^4/cos(d*x+c)^10+3/40*sin(d*x+c)^4/cos(d*x+c)^8+1/20*sin(d*x+c)^4/cos(d*x+c)^6+1/4
0*sin(d*x+c)^4/cos(d*x+c)^4)-3*a^3*(1/9*sin(d*x+c)^3/cos(d*x+c)^9+2/21*sin(d*x+c)^3/cos(d*x+c)^7+8/105*sin(d*x
+c)^3/cos(d*x+c)^5+16/315*sin(d*x+c)^3/cos(d*x+c)^3)+3/8*I*a^3/cos(d*x+c)^8-a^3*(-16/35-1/7*sec(d*x+c)^6-6/35*
sec(d*x+c)^4-8/35*sec(d*x+c)^2)*tan(d*x+c))

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Maxima [A]
time = 0.28, size = 108, normalized size = 0.99 \begin {gather*} -\frac {21 i \, a^{3} \tan \left (d x + c\right )^{10} + 70 \, a^{3} \tan \left (d x + c\right )^{9} + 240 \, a^{3} \tan \left (d x + c\right )^{7} - 210 i \, a^{3} \tan \left (d x + c\right )^{6} + 252 \, a^{3} \tan \left (d x + c\right )^{5} - 420 i \, a^{3} \tan \left (d x + c\right )^{4} - 315 i \, a^{3} \tan \left (d x + c\right )^{2} - 210 \, a^{3} \tan \left (d x + c\right )}{210 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^3,x, algorithm="maxima")

[Out]

-1/210*(21*I*a^3*tan(d*x + c)^10 + 70*a^3*tan(d*x + c)^9 + 240*a^3*tan(d*x + c)^7 - 210*I*a^3*tan(d*x + c)^6 +
252*a^3*tan(d*x + c)^5 - 420*I*a^3*tan(d*x + c)^4 - 315*I*a^3*tan(d*x + c)^2 - 210*a^3*tan(d*x + c))/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. 215 vs. \(2 (85) = 170\).
time = 0.37, size = 215, normalized size = 1.97 \begin {gather*} -\frac {128 \, {\left (-210 i \, a^{3} e^{\left (12 i \, d x + 12 i \, c\right )} - 252 i \, a^{3} e^{\left (10 i \, d x + 10 i \, c\right )} - 210 i \, a^{3} e^{\left (8 i \, d x + 8 i \, c\right )} - 120 i \, a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} - 45 i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} - 10 i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - i \, a^{3}\right )}}{105 \, {\left (d e^{\left (20 i \, d x + 20 i \, c\right )} + 10 \, d e^{\left (18 i \, d x + 18 i \, c\right )} + 45 \, d e^{\left (16 i \, d x + 16 i \, c\right )} + 120 \, d e^{\left (14 i \, d x + 14 i \, c\right )} + 210 \, d e^{\left (12 i \, d x + 12 i \, c\right )} + 252 \, d e^{\left (10 i \, d x + 10 i \, c\right )} + 210 \, d e^{\left (8 i \, d x + 8 i \, c\right )} + 120 \, d e^{\left (6 i \, d x + 6 i \, c\right )} + 45 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 10 \, 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)^8*(a+I*a*tan(d*x+c))^3,x, algorithm="fricas")

[Out]

-128/105*(-210*I*a^3*e^(12*I*d*x + 12*I*c) - 252*I*a^3*e^(10*I*d*x + 10*I*c) - 210*I*a^3*e^(8*I*d*x + 8*I*c) -
120*I*a^3*e^(6*I*d*x + 6*I*c) - 45*I*a^3*e^(4*I*d*x + 4*I*c) - 10*I*a^3*e^(2*I*d*x + 2*I*c) - I*a^3)/(d*e^(20
*I*d*x + 20*I*c) + 10*d*e^(18*I*d*x + 18*I*c) + 45*d*e^(16*I*d*x + 16*I*c) + 120*d*e^(14*I*d*x + 14*I*c) + 210
*d*e^(12*I*d*x + 12*I*c) + 252*d*e^(10*I*d*x + 10*I*c) + 210*d*e^(8*I*d*x + 8*I*c) + 120*d*e^(6*I*d*x + 6*I*c)
+ 45*d*e^(4*I*d*x + 4*I*c) + 10*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*} - i a^{3} \left (\int i \sec ^{8}{\left (c + d x \right )}\, dx + \int \left (- 3 \tan {\left (c + d x \right )} \sec ^{8}{\left (c + d x \right )}\right )\, dx + \int \tan ^{3}{\left (c + d x \right )} \sec ^{8}{\left (c + d x \right )}\, dx + \int \left (- 3 i \tan ^{2}{\left (c + d x \right )} \sec ^{8}{\left (c + d x \right )}\right )\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**8*(a+I*a*tan(d*x+c))**3,x)

[Out]

-I*a**3*(Integral(I*sec(c + d*x)**8, x) + Integral(-3*tan(c + d*x)*sec(c + d*x)**8, x) + Integral(tan(c + d*x)
**3*sec(c + d*x)**8, x) + Integral(-3*I*tan(c + d*x)**2*sec(c + d*x)**8, x))

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Giac [A]
time = 0.65, size = 108, normalized size = 0.99 \begin {gather*} -\frac {21 i \, a^{3} \tan \left (d x + c\right )^{10} + 70 \, a^{3} \tan \left (d x + c\right )^{9} + 240 \, a^{3} \tan \left (d x + c\right )^{7} - 210 i \, a^{3} \tan \left (d x + c\right )^{6} + 252 \, a^{3} \tan \left (d x + c\right )^{5} - 420 i \, a^{3} \tan \left (d x + c\right )^{4} - 315 i \, a^{3} \tan \left (d x + c\right )^{2} - 210 \, a^{3} \tan \left (d x + c\right )}{210 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^8*(a+I*a*tan(d*x+c))^3,x, algorithm="giac")

[Out]

-1/210*(21*I*a^3*tan(d*x + c)^10 + 70*a^3*tan(d*x + c)^9 + 240*a^3*tan(d*x + c)^7 - 210*I*a^3*tan(d*x + c)^6 +
252*a^3*tan(d*x + c)^5 - 420*I*a^3*tan(d*x + c)^4 - 315*I*a^3*tan(d*x + c)^2 - 210*a^3*tan(d*x + c))/d

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Mupad [B]
time = 3.31, size = 151, normalized size = 1.39 \begin {gather*} -\frac {a^3\,\sin \left (c+d\,x\right )\,\left (-210\,{\cos \left (c+d\,x\right )}^9-{\cos \left (c+d\,x\right )}^8\,\sin \left (c+d\,x\right )\,315{}\mathrm {i}-{\cos \left (c+d\,x\right )}^6\,{\sin \left (c+d\,x\right )}^3\,420{}\mathrm {i}+252\,{\cos \left (c+d\,x\right )}^5\,{\sin \left (c+d\,x\right )}^4-{\cos \left (c+d\,x\right )}^4\,{\sin \left (c+d\,x\right )}^5\,210{}\mathrm {i}+240\,{\cos \left (c+d\,x\right )}^3\,{\sin \left (c+d\,x\right )}^6+70\,\cos \left (c+d\,x\right )\,{\sin \left (c+d\,x\right )}^8+{\sin \left (c+d\,x\right )}^9\,21{}\mathrm {i}\right )}{210\,d\,{\cos \left (c+d\,x\right )}^{10}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*tan(c + d*x)*1i)^3/cos(c + d*x)^8,x)

[Out]

-(a^3*sin(c + d*x)*(70*cos(c + d*x)*sin(c + d*x)^8 - cos(c + d*x)^8*sin(c + d*x)*315i - 210*cos(c + d*x)^9 + s
in(c + d*x)^9*21i + 240*cos(c + d*x)^3*sin(c + d*x)^6 - cos(c + d*x)^4*sin(c + d*x)^5*210i + 252*cos(c + d*x)^
5*sin(c + d*x)^4 - cos(c + d*x)^6*sin(c + d*x)^3*420i))/(210*d*cos(c + d*x)^10)

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