3.1.57 \(\int \cos ^7(c+d x) (a+i a \tan (c+d x))^4 \, dx\) [57]

Optimal. Leaf size=102 \[ \frac {3 a^4 \sin (c+d x)}{35 d}-\frac {a^4 \sin ^3(c+d x)}{35 d}-\frac {2 i a \cos ^7(c+d x) (a+i a \tan (c+d x))^3}{7 d}-\frac {2 i \cos ^5(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{35 d} \]

[Out]

3/35*a^4*sin(d*x+c)/d-1/35*a^4*sin(d*x+c)^3/d-2/7*I*a*cos(d*x+c)^7*(a+I*a*tan(d*x+c))^3/d-2/35*I*cos(d*x+c)^5*
(a^4+I*a^4*tan(d*x+c))/d

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Rubi [A]
time = 0.07, antiderivative size = 102, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 2, integrand size = 24, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.083, Rules used = {3577, 2713} \begin {gather*} -\frac {a^4 \sin ^3(c+d x)}{35 d}+\frac {3 a^4 \sin (c+d x)}{35 d}-\frac {2 i \cos ^5(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{35 d}-\frac {2 i a \cos ^7(c+d x) (a+i a \tan (c+d x))^3}{7 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^7*(a + I*a*Tan[c + d*x])^4,x]

[Out]

(3*a^4*Sin[c + d*x])/(35*d) - (a^4*Sin[c + d*x]^3)/(35*d) - (((2*I)/7)*a*Cos[c + d*x]^7*(a + I*a*Tan[c + d*x])
^3)/d - (((2*I)/35)*Cos[c + d*x]^5*(a^4 + I*a^4*Tan[c + d*x]))/d

Rule 2713

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> Dist[-d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rule 3577

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[2*b*(d
*Sec[e + f*x])^m*((a + b*Tan[e + f*x])^(n - 1)/(f*m)), x] - Dist[b^2*((m + 2*n - 2)/(d^2*m)), Int[(d*Sec[e + f
*x])^(m + 2)*(a + b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n,
1] && ((IGtQ[n/2, 0] && ILtQ[m - 1/2, 0]) || (EqQ[n, 2] && LtQ[m, 0]) || (LeQ[m, -1] && GtQ[m + n, 0]) || (ILt
Q[m, 0] && LtQ[m/2 + n - 1, 0] && IntegerQ[n]) || (EqQ[n, 3/2] && EqQ[m, -2^(-1)])) && IntegerQ[2*m]

Rubi steps

\begin {align*} \int \cos ^7(c+d x) (a+i a \tan (c+d x))^4 \, dx &=-\frac {2 i a \cos ^7(c+d x) (a+i a \tan (c+d x))^3}{7 d}+\frac {1}{7} a^2 \int \cos ^5(c+d x) (a+i a \tan (c+d x))^2 \, dx\\ &=-\frac {2 i a \cos ^7(c+d x) (a+i a \tan (c+d x))^3}{7 d}-\frac {2 i \cos ^5(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{35 d}+\frac {1}{35} \left (3 a^4\right ) \int \cos ^3(c+d x) \, dx\\ &=-\frac {2 i a \cos ^7(c+d x) (a+i a \tan (c+d x))^3}{7 d}-\frac {2 i \cos ^5(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{35 d}-\frac {\left (3 a^4\right ) \text {Subst}\left (\int \left (1-x^2\right ) \, dx,x,-\sin (c+d x)\right )}{35 d}\\ &=\frac {3 a^4 \sin (c+d x)}{35 d}-\frac {a^4 \sin ^3(c+d x)}{35 d}-\frac {2 i a \cos ^7(c+d x) (a+i a \tan (c+d x))^3}{7 d}-\frac {2 i \cos ^5(c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{35 d}\\ \end {align*}

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Mathematica [A]
time = 0.50, size = 73, normalized size = 0.72 \begin {gather*} \frac {a^4 (28 \cos (c+d x)+20 \cos (3 (c+d x))-i (7 \sin (c+d x)+15 \sin (3 (c+d x)))) (-i \cos (4 (c+d x))+\sin (4 (c+d x)))}{140 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^7*(a + I*a*Tan[c + d*x])^4,x]

[Out]

(a^4*(28*Cos[c + d*x] + 20*Cos[3*(c + d*x)] - I*(7*Sin[c + d*x] + 15*Sin[3*(c + d*x)]))*((-I)*Cos[4*(c + d*x)]
+ Sin[4*(c + d*x)]))/(140*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. 202 vs. \(2 (90 ) = 180\).
time = 0.25, size = 203, normalized size = 1.99

method	result	size
risch	\(-\frac {i a^{4} {\mathrm e}^{7 i \left (d x +c \right )}}{56 d}-\frac {3 i a^{4} {\mathrm e}^{5 i \left (d x +c \right )}}{40 d}-\frac {i a^{4} {\mathrm e}^{3 i \left (d x +c \right )}}{8 d}-\frac {i a^{4} {\mathrm e}^{i \left (d x +c \right )}}{8 d}\)	\(74\)
derivativedivides	\(\frac {a^{4} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{7}-\frac {3 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{35}+\frac {\left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )}{35}\right )-4 i a^{4} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{7}-\frac {2 \left (\cos ^{5}\left (d x +c \right )\right )}{35}\right )-6 a^{4} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{7}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{35}\right )-\frac {4 i a^{4} \left (\cos ^{7}\left (d x +c \right )\right )}{7}+\frac {a^{4} \left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{7}}{d}\)	\(203\)
default	\(\frac {a^{4} \left (-\frac {\left (\sin ^{3}\left (d x +c \right )\right ) \left (\cos ^{4}\left (d x +c \right )\right )}{7}-\frac {3 \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )}{35}+\frac {\left (\cos ^{2}\left (d x +c \right )+2\right ) \sin \left (d x +c \right )}{35}\right )-4 i a^{4} \left (-\frac {\left (\sin ^{2}\left (d x +c \right )\right ) \left (\cos ^{5}\left (d x +c \right )\right )}{7}-\frac {2 \left (\cos ^{5}\left (d x +c \right )\right )}{35}\right )-6 a^{4} \left (-\frac {\sin \left (d x +c \right ) \left (\cos ^{6}\left (d x +c \right )\right )}{7}+\frac {\left (\frac {8}{3}+\cos ^{4}\left (d x +c \right )+\frac {4 \left (\cos ^{2}\left (d x +c \right )\right )}{3}\right ) \sin \left (d x +c \right )}{35}\right )-\frac {4 i a^{4} \left (\cos ^{7}\left (d x +c \right )\right )}{7}+\frac {a^{4} \left (\frac {16}{5}+\cos ^{6}\left (d x +c \right )+\frac {6 \left (\cos ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\cos ^{2}\left (d x +c \right )\right )}{5}\right ) \sin \left (d x +c \right )}{7}}{d}\)	\(203\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^7*(a+I*a*tan(d*x+c))^4,x,method=_RETURNVERBOSE)

[Out]

1/d*(a^4*(-1/7*sin(d*x+c)^3*cos(d*x+c)^4-3/35*sin(d*x+c)*cos(d*x+c)^4+1/35*(cos(d*x+c)^2+2)*sin(d*x+c))-4*I*a^
4*(-1/7*sin(d*x+c)^2*cos(d*x+c)^5-2/35*cos(d*x+c)^5)-6*a^4*(-1/7*sin(d*x+c)*cos(d*x+c)^6+1/35*(8/3+cos(d*x+c)^
4+4/3*cos(d*x+c)^2)*sin(d*x+c))-4/7*I*a^4*cos(d*x+c)^7+1/7*a^4*(16/5+cos(d*x+c)^6+6/5*cos(d*x+c)^4+8/5*cos(d*x
+c)^2)*sin(d*x+c))

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Maxima [A]
time = 0.28, size = 149, normalized size = 1.46 \begin {gather*} -\frac {20 i \, a^{4} \cos \left (d x + c\right )^{7} + 4 i \, {\left (5 \, \cos \left (d x + c\right )^{7} - 7 \, \cos \left (d x + c\right )^{5}\right )} a^{4} + 2 \, {\left (15 \, \sin \left (d x + c\right )^{7} - 42 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3}\right )} a^{4} + {\left (5 \, \sin \left (d x + c\right )^{7} - 7 \, \sin \left (d x + c\right )^{5}\right )} a^{4} + {\left (5 \, \sin \left (d x + c\right )^{7} - 21 \, \sin \left (d x + c\right )^{5} + 35 \, \sin \left (d x + c\right )^{3} - 35 \, \sin \left (d x + c\right )\right )} a^{4}}{35 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")

[Out]

-1/35*(20*I*a^4*cos(d*x + c)^7 + 4*I*(5*cos(d*x + c)^7 - 7*cos(d*x + c)^5)*a^4 + 2*(15*sin(d*x + c)^7 - 42*sin
(d*x + c)^5 + 35*sin(d*x + c)^3)*a^4 + (5*sin(d*x + c)^7 - 7*sin(d*x + c)^5)*a^4 + (5*sin(d*x + c)^7 - 21*sin(
d*x + c)^5 + 35*sin(d*x + c)^3 - 35*sin(d*x + c))*a^4)/d

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Fricas [A]
time = 0.36, size = 62, normalized size = 0.61 \begin {gather*} \frac {-5 i \, a^{4} e^{\left (7 i \, d x + 7 i \, c\right )} - 21 i \, a^{4} e^{\left (5 i \, d x + 5 i \, c\right )} - 35 i \, a^{4} e^{\left (3 i \, d x + 3 i \, c\right )} - 35 i \, a^{4} e^{\left (i \, d x + i \, c\right )}}{280 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")

[Out]

1/280*(-5*I*a^4*e^(7*I*d*x + 7*I*c) - 21*I*a^4*e^(5*I*d*x + 5*I*c) - 35*I*a^4*e^(3*I*d*x + 3*I*c) - 35*I*a^4*e
^(I*d*x + I*c))/d

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Sympy [A]
time = 0.36, size = 156, normalized size = 1.53 \begin {gather*} \begin {cases} \frac {- 2560 i a^{4} d^{3} e^{7 i c} e^{7 i d x} - 10752 i a^{4} d^{3} e^{5 i c} e^{5 i d x} - 17920 i a^{4} d^{3} e^{3 i c} e^{3 i d x} - 17920 i a^{4} d^{3} e^{i c} e^{i d x}}{143360 d^{4}} & \text {for}\: d^{4} \neq 0 \\x \left (\frac {a^{4} e^{7 i c}}{8} + \frac {3 a^{4} e^{5 i c}}{8} + \frac {3 a^{4} e^{3 i c}}{8} + \frac {a^{4} e^{i c}}{8}\right ) & \text {otherwise} \end {cases} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**7*(a+I*a*tan(d*x+c))**4,x)

[Out]

Piecewise(((-2560*I*a**4*d**3*exp(7*I*c)*exp(7*I*d*x) - 10752*I*a**4*d**3*exp(5*I*c)*exp(5*I*d*x) - 17920*I*a*
*4*d**3*exp(3*I*c)*exp(3*I*d*x) - 17920*I*a**4*d**3*exp(I*c)*exp(I*d*x))/(143360*d**4), Ne(d**4, 0)), (x*(a**4
*exp(7*I*c)/8 + 3*a**4*exp(5*I*c)/8 + 3*a**4*exp(3*I*c)/8 + a**4*exp(I*c)/8), True))

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Giac [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1327 vs. \(2 (86) = 172\).
time = 0.91, size = 1327, normalized size = 13.01 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+I*a*tan(d*x+c))^4,x, algorithm="giac")

[Out]

1/143360*(89950*a^4*e^(12*I*d*x + 6*I*c)*log(I*e^(I*d*x + I*c) + 1) + 539700*a^4*e^(10*I*d*x + 4*I*c)*log(I*e^
(I*d*x + I*c) + 1) + 1349250*a^4*e^(8*I*d*x + 2*I*c)*log(I*e^(I*d*x + I*c) + 1) + 1349250*a^4*e^(4*I*d*x - 2*I
*c)*log(I*e^(I*d*x + I*c) + 1) + 539700*a^4*e^(2*I*d*x - 4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 1799000*a^4*e^(6*
I*d*x)*log(I*e^(I*d*x + I*c) + 1) + 89950*a^4*e^(-6*I*c)*log(I*e^(I*d*x + I*c) + 1) + 86065*a^4*e^(12*I*d*x +
6*I*c)*log(I*e^(I*d*x + I*c) - 1) + 516390*a^4*e^(10*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) - 1) + 1290975*a^4*e
^(8*I*d*x + 2*I*c)*log(I*e^(I*d*x + I*c) - 1) + 1290975*a^4*e^(4*I*d*x - 2*I*c)*log(I*e^(I*d*x + I*c) - 1) + 5
16390*a^4*e^(2*I*d*x - 4*I*c)*log(I*e^(I*d*x + I*c) - 1) + 1721300*a^4*e^(6*I*d*x)*log(I*e^(I*d*x + I*c) - 1)
+ 86065*a^4*e^(-6*I*c)*log(I*e^(I*d*x + I*c) - 1) - 89950*a^4*e^(12*I*d*x + 6*I*c)*log(-I*e^(I*d*x + I*c) + 1)
- 539700*a^4*e^(10*I*d*x + 4*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 1349250*a^4*e^(8*I*d*x + 2*I*c)*log(-I*e^(I*d
*x + I*c) + 1) - 1349250*a^4*e^(4*I*d*x - 2*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 539700*a^4*e^(2*I*d*x - 4*I*c)*
log(-I*e^(I*d*x + I*c) + 1) - 1799000*a^4*e^(6*I*d*x)*log(-I*e^(I*d*x + I*c) + 1) - 89950*a^4*e^(-6*I*c)*log(-
I*e^(I*d*x + I*c) + 1) - 86065*a^4*e^(12*I*d*x + 6*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 516390*a^4*e^(10*I*d*x +
4*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 1290975*a^4*e^(8*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 1290975*a^
4*e^(4*I*d*x - 2*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 516390*a^4*e^(2*I*d*x - 4*I*c)*log(-I*e^(I*d*x + I*c) - 1)
- 1721300*a^4*e^(6*I*d*x)*log(-I*e^(I*d*x + I*c) - 1) - 86065*a^4*e^(-6*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 38
85*a^4*e^(12*I*d*x + 6*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 23310*a^4*e^(10*I*d*x + 4*I*c)*log(I*e^(I*d*x) + e^(
-I*c)) - 58275*a^4*e^(8*I*d*x + 2*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 58275*a^4*e^(4*I*d*x - 2*I*c)*log(I*e^(I*
d*x) + e^(-I*c)) - 23310*a^4*e^(2*I*d*x - 4*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 77700*a^4*e^(6*I*d*x)*log(I*e^(
I*d*x) + e^(-I*c)) - 3885*a^4*e^(-6*I*c)*log(I*e^(I*d*x) + e^(-I*c)) + 3885*a^4*e^(12*I*d*x + 6*I*c)*log(-I*e^
(I*d*x) + e^(-I*c)) + 23310*a^4*e^(10*I*d*x + 4*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 58275*a^4*e^(8*I*d*x + 2*I
*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 58275*a^4*e^(4*I*d*x - 2*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 23310*a^4*e^(2
*I*d*x - 4*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 77700*a^4*e^(6*I*d*x)*log(-I*e^(I*d*x) + e^(-I*c)) + 3885*a^4*e
^(-6*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 2560*I*a^4*e^(19*I*d*x + 13*I*c) - 26112*I*a^4*e^(17*I*d*x + 11*I*c)
- 120832*I*a^4*e^(15*I*d*x + 9*I*c) - 337920*I*a^4*e^(13*I*d*x + 7*I*c) - 629760*I*a^4*e^(11*I*d*x + 5*I*c) -
803840*I*a^4*e^(9*I*d*x + 3*I*c) - 694272*I*a^4*e^(7*I*d*x + I*c) - 387072*I*a^4*e^(5*I*d*x - I*c) - 125440*I*
a^4*e^(3*I*d*x - 3*I*c) - 17920*I*a^4*e^(I*d*x - 5*I*c))/(d*e^(12*I*d*x + 6*I*c) + 6*d*e^(10*I*d*x + 4*I*c) +
15*d*e^(8*I*d*x + 2*I*c) + 15*d*e^(4*I*d*x - 2*I*c) + 6*d*e^(2*I*d*x - 4*I*c) + 20*d*e^(6*I*d*x) + d*e^(-6*I*c
))

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Mupad [B]
time = 4.27, size = 186, normalized size = 1.82 \begin {gather*} -\frac {2\,a^4\,\left (35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,105{}\mathrm {i}-210\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,210{}\mathrm {i}+147\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,49{}\mathrm {i}-12\right )}{35\,d\,\left (-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,7{}\mathrm {i}+21\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,35{}\mathrm {i}-35\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,21{}\mathrm {i}+7\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1{}\mathrm {i}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^7*(a + a*tan(c + d*x)*1i)^4,x)

[Out]

-(2*a^4*(tan(c/2 + (d*x)/2)*49i + 147*tan(c/2 + (d*x)/2)^2 - tan(c/2 + (d*x)/2)^3*210i - 210*tan(c/2 + (d*x)/2
)^4 + tan(c/2 + (d*x)/2)^5*105i + 35*tan(c/2 + (d*x)/2)^6 - 12))/(35*d*(7*tan(c/2 + (d*x)/2) - tan(c/2 + (d*x)
/2)^2*21i - 35*tan(c/2 + (d*x)/2)^3 + tan(c/2 + (d*x)/2)^4*35i + 21*tan(c/2 + (d*x)/2)^5 - tan(c/2 + (d*x)/2)^
6*7i - tan(c/2 + (d*x)/2)^7 + 1i))

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