3.55 \(\int \cos ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx\)
Optimal. Leaf size=78 \[ \frac{a^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{2 i \cos (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{d}-\frac{2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d} \]
[Out]
(a^4*ArcTanh[Sin[c + d*x]])/d - (((2*I)/3)*a*Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^3)/d + ((2*I)*Cos[c + d*x]*
(a^4 + I*a^4*Tan[c + d*x]))/d
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Rubi [A] time = 0.075611, antiderivative size = 78, normalized size of antiderivative = 1.,
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 =
{3496, 3770} \[ \frac{a^4 \tanh ^{-1}(\sin (c+d x))}{d}+\frac{2 i \cos (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{d}-\frac{2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^4,x]
[Out]
(a^4*ArcTanh[Sin[c + d*x]])/d - (((2*I)/3)*a*Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^3)/d + ((2*I)*Cos[c + d*x]*
(a^4 + I*a^4*Tan[c + d*x]))/d
Rule 3496
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]
Rule 3770
Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]
Rubi steps
\begin{align*} \int \cos ^3(c+d x) (a+i a \tan (c+d x))^4 \, dx &=-\frac{2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}-a^2 \int \cos (c+d x) (a+i a \tan (c+d x))^2 \, dx\\ &=-\frac{2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}+\frac{2 i \cos (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{d}+a^4 \int \sec (c+d x) \, dx\\ &=\frac{a^4 \tanh ^{-1}(\sin (c+d x))}{d}-\frac{2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^3}{3 d}+\frac{2 i \cos (c+d x) \left (a^4+i a^4 \tan (c+d x)\right )}{d}\\ \end{align*}
Mathematica [B] time = 0.485552, size = 246, normalized size = 3.15 \[ \frac{a^4 (\cos (c+d x)+i \sin (c+d x))^4 \left (6 i \sin (3 c) \sin (d x)-2 i \sin (c) \sin (3 d x)-2 \sin (c) \cos (3 d x)+6 \sin (3 c) \cos (d x)+\cos (3 c) (-6 \sin (d x)+6 i \cos (d x))+2 \cos (c) (\sin (3 d x)-i \cos (3 d x))-3 \cos (4 c) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+3 i \sin (4 c) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )+3 \cos (4 c) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )-3 i \sin (4 c) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )\right )}{3 d (\cos (d x)+i \sin (d x))^4} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^4,x]
[Out]
(a^4*(-3*Cos[4*c]*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]] + 3*Cos[4*c]*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2
]] - 2*Cos[3*d*x]*Sin[c] + 6*Cos[d*x]*Sin[3*c] + (3*I)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*Sin[4*c] - (3*
I)*Log[Cos[(c + d*x)/2] + Sin[(c + d*x)/2]]*Sin[4*c] + Cos[3*c]*((6*I)*Cos[d*x] - 6*Sin[d*x]) + (6*I)*Sin[3*c]
*Sin[d*x] - (2*I)*Sin[c]*Sin[3*d*x] + 2*Cos[c]*((-I)*Cos[3*d*x] + Sin[3*d*x]))*(Cos[c + d*x] + I*Sin[c + d*x])
^4)/(3*d*(Cos[d*x] + I*Sin[d*x])^4)
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Maple [A] time = 0.052, size = 130, normalized size = 1.7 \begin{align*} -{\frac{7\,{a}^{4} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\,d}}-{\frac{{a}^{4}\sin \left ( dx+c \right ) }{3\,d}}+{\frac{{a}^{4}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}+{\frac{{\frac{4\,i}{3}}{a}^{4}\cos \left ( dx+c \right ) \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{d}}+{\frac{{\frac{8\,i}{3}}{a}^{4}\cos \left ( dx+c \right ) }{d}}-{\frac{{\frac{4\,i}{3}}{a}^{4} \left ( \cos \left ( dx+c \right ) \right ) ^{3}}{d}}+{\frac{\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{2}{a}^{4}}{3\,d}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^4,x)
[Out]
-7/3*a^4*sin(d*x+c)^3/d-1/3*a^4*sin(d*x+c)/d+1/d*a^4*ln(sec(d*x+c)+tan(d*x+c))+4/3*I/d*a^4*cos(d*x+c)*sin(d*x+
c)^2+8/3*I/d*a^4*cos(d*x+c)-4/3*I/d*a^4*cos(d*x+c)^3+1/3/d*sin(d*x+c)*cos(d*x+c)^2*a^4
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Maxima [A] time = 1.08211, size = 163, normalized size = 2.09 \begin{align*} -\frac{8 i \, a^{4} \cos \left (d x + c\right )^{3} + 12 \, a^{4} \sin \left (d x + c\right )^{3} + 8 i \,{\left (\cos \left (d x + c\right )^{3} - 3 \, \cos \left (d x + c\right )\right )} a^{4} +{\left (2 \, \sin \left (d x + c\right )^{3} - 3 \, \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, \log \left (\sin \left (d x + c\right ) - 1\right ) + 6 \, \sin \left (d x + c\right )\right )} a^{4} + 2 \,{\left (\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )\right )} a^{4}}{6 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^4,x, algorithm="maxima")
[Out]
-1/6*(8*I*a^4*cos(d*x + c)^3 + 12*a^4*sin(d*x + c)^3 + 8*I*(cos(d*x + c)^3 - 3*cos(d*x + c))*a^4 + (2*sin(d*x
+ c)^3 - 3*log(sin(d*x + c) + 1) + 3*log(sin(d*x + c) - 1) + 6*sin(d*x + c))*a^4 + 2*(sin(d*x + c)^3 - 3*sin(d
*x + c))*a^4)/d
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Fricas [A] time = 1.19146, size = 176, normalized size = 2.26 \begin{align*} \frac{-2 i \, a^{4} e^{\left (3 i \, d x + 3 i \, c\right )} + 6 i \, a^{4} e^{\left (i \, d x + i \, c\right )} + 3 \, a^{4} \log \left (e^{\left (i \, d x + i \, c\right )} + i\right ) - 3 \, a^{4} \log \left (e^{\left (i \, d x + i \, c\right )} - i\right )}{3 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^4,x, algorithm="fricas")
[Out]
1/3*(-2*I*a^4*e^(3*I*d*x + 3*I*c) + 6*I*a^4*e^(I*d*x + I*c) + 3*a^4*log(e^(I*d*x + I*c) + I) - 3*a^4*log(e^(I*
d*x + I*c) - I))/d
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Sympy [A] time = 0.735001, size = 110, normalized size = 1.41 \begin{align*} \frac{a^{4} \left (- \log{\left (e^{i d x} - i e^{- i c} \right )} + \log{\left (e^{i d x} + i e^{- i c} \right )}\right )}{d} + \begin{cases} \frac{- 2 i a^{4} d e^{3 i c} e^{3 i d x} + 6 i a^{4} d e^{i c} e^{i d x}}{3 d^{2}} & \text{for}\: 3 d^{2} \neq 0 \\x \left (2 a^{4} e^{3 i c} - 2 a^{4} e^{i c}\right ) & \text{otherwise} \end{cases} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)**3*(a+I*a*tan(d*x+c))**4,x)
[Out]
a**4*(-log(exp(I*d*x) - I*exp(-I*c)) + log(exp(I*d*x) + I*exp(-I*c)))/d + Piecewise(((-2*I*a**4*d*exp(3*I*c)*e
xp(3*I*d*x) + 6*I*a**4*d*exp(I*c)*exp(I*d*x))/(3*d**2), Ne(3*d**2, 0)), (x*(2*a**4*exp(3*I*c) - 2*a**4*exp(I*c
)), True))
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Giac [B] time = 1.65592, size = 1754, normalized size = 22.49 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^4,x, algorithm="giac")
[Out]
1/768*(1110*a^4*e^(12*I*d*x + 6*I*c)*log(I*e^(I*d*x + I*c) + 1) + 6660*a^4*e^(10*I*d*x + 4*I*c)*log(I*e^(I*d*x
+ I*c) + 1) + 16650*a^4*e^(8*I*d*x + 2*I*c)*log(I*e^(I*d*x + I*c) + 1) + 16650*a^4*e^(4*I*d*x - 2*I*c)*log(I*
e^(I*d*x + I*c) + 1) + 6660*a^4*e^(2*I*d*x - 4*I*c)*log(I*e^(I*d*x + I*c) + 1) + 22200*a^4*e^(6*I*d*x)*log(I*e
^(I*d*x + I*c) + 1) + 1110*a^4*e^(-6*I*c)*log(I*e^(I*d*x + I*c) + 1) + 1875*a^4*e^(12*I*d*x + 6*I*c)*log(I*e^(
I*d*x + I*c) - 1) + 11250*a^4*e^(10*I*d*x + 4*I*c)*log(I*e^(I*d*x + I*c) - 1) + 28125*a^4*e^(8*I*d*x + 2*I*c)*
log(I*e^(I*d*x + I*c) - 1) + 28125*a^4*e^(4*I*d*x - 2*I*c)*log(I*e^(I*d*x + I*c) - 1) + 11250*a^4*e^(2*I*d*x -
4*I*c)*log(I*e^(I*d*x + I*c) - 1) + 37500*a^4*e^(6*I*d*x)*log(I*e^(I*d*x + I*c) - 1) + 1875*a^4*e^(-6*I*c)*lo
g(I*e^(I*d*x + I*c) - 1) - 1110*a^4*e^(12*I*d*x + 6*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 6660*a^4*e^(10*I*d*x +
4*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 16650*a^4*e^(8*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 16650*a^4*e^(
4*I*d*x - 2*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 6660*a^4*e^(2*I*d*x - 4*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 2220
0*a^4*e^(6*I*d*x)*log(-I*e^(I*d*x + I*c) + 1) - 1110*a^4*e^(-6*I*c)*log(-I*e^(I*d*x + I*c) + 1) - 1875*a^4*e^(
12*I*d*x + 6*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 11250*a^4*e^(10*I*d*x + 4*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 2
8125*a^4*e^(8*I*d*x + 2*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 28125*a^4*e^(4*I*d*x - 2*I*c)*log(-I*e^(I*d*x + I*c
) - 1) - 11250*a^4*e^(2*I*d*x - 4*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 37500*a^4*e^(6*I*d*x)*log(-I*e^(I*d*x + I
*c) - 1) - 1875*a^4*e^(-6*I*c)*log(-I*e^(I*d*x + I*c) - 1) - 3*a^4*e^(12*I*d*x + 6*I*c)*log(I*e^(I*d*x) + e^(-
I*c)) - 18*a^4*e^(10*I*d*x + 4*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 45*a^4*e^(8*I*d*x + 2*I*c)*log(I*e^(I*d*x) +
e^(-I*c)) - 45*a^4*e^(4*I*d*x - 2*I*c)*log(I*e^(I*d*x) + e^(-I*c)) - 18*a^4*e^(2*I*d*x - 4*I*c)*log(I*e^(I*d*
x) + e^(-I*c)) - 60*a^4*e^(6*I*d*x)*log(I*e^(I*d*x) + e^(-I*c)) - 3*a^4*e^(-6*I*c)*log(I*e^(I*d*x) + e^(-I*c))
+ 3*a^4*e^(12*I*d*x + 6*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 18*a^4*e^(10*I*d*x + 4*I*c)*log(-I*e^(I*d*x) + e^
(-I*c)) + 45*a^4*e^(8*I*d*x + 2*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 45*a^4*e^(4*I*d*x - 2*I*c)*log(-I*e^(I*d*x
) + e^(-I*c)) + 18*a^4*e^(2*I*d*x - 4*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) + 60*a^4*e^(6*I*d*x)*log(-I*e^(I*d*x)
+ e^(-I*c)) + 3*a^4*e^(-6*I*c)*log(-I*e^(I*d*x) + e^(-I*c)) - 512*I*a^4*e^(15*I*d*x + 9*I*c) - 1536*I*a^4*e^(1
3*I*d*x + 7*I*c) + 1536*I*a^4*e^(11*I*d*x + 5*I*c) + 12800*I*a^4*e^(9*I*d*x + 3*I*c) + 23040*I*a^4*e^(7*I*d*x
+ I*c) + 19968*I*a^4*e^(5*I*d*x - I*c) + 8704*I*a^4*e^(3*I*d*x - 3*I*c) + 1536*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))