\(\int \cos ^3(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx\) [328]

Optimal result

Integrand size = 26, antiderivative size = 71 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\frac {8 i a^2 \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^{5/2}}{d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.16 (sec) , antiderivative size = 71, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.077, Rules used = {3575, 3574} \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\frac {8 i a^2 \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^{5/2}}{d} \]

[In]

Int[Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(((8*I)/3)*a^2*Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^(3/2))/d - ((2*I)*a*Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])
^(5/2))/d

Rule 3574

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] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 + b^2
, 0] && EqQ[Simplify[m/2 + n - 1], 0]

Rule 3575

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^{5/2}}{d}-(4 a) \int \cos ^3(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx \\ & = \frac {8 i a^2 \cos ^3(c+d x) (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac {2 i a \cos ^3(c+d x) (a+i a \tan (c+d x))^{5/2}}{d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.96 (sec) , antiderivative size = 86, normalized size of antiderivative = 1.21 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\frac {2 a^3 \cos (c+d x) (i \cos (c+d x)+3 \sin (c+d x)) (\cos (c+4 d x)+i \sin (c+4 d x)) \sqrt {a+i a \tan (c+d x)}}{3 d (\cos (d x)+i \sin (d x))^3} \]

[In]

Integrate[Cos[c + d*x]^3*(a + I*a*Tan[c + d*x])^(7/2),x]

[Out]

(2*a^3*Cos[c + d*x]*(I*Cos[c + d*x] + 3*Sin[c + d*x])*(Cos[c + 4*d*x] + I*Sin[c + 4*d*x])*Sqrt[a + I*a*Tan[c +
d*x]])/(3*d*(Cos[d*x] + I*Sin[d*x])^3)

Maple [A] (verified)

Time = 34.54 (sec) , antiderivative size = 88, normalized size of antiderivative = 1.24

method	result	size
default	\(\frac {2 \left (\tan \left (d x +c \right )-i\right )^{3} \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a^{3} \left (\cos ^{4}\left (d x +c \right )\right ) \left (3 i \sin \left (d x +c \right )-\cos \left (d x +c \right )\right ) \left (2 i \cos \left (d x +c \right ) \sin \left (d x +c \right )-2 \left (\cos ^{2}\left (d x +c \right )\right )+1\right )}{3 d}\)	\(88\)

[In]

int(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^(7/2),x,method=_RETURNVERBOSE)

[Out]

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

Fricas [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 59, normalized size of antiderivative = 0.83 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\frac {\sqrt {2} {\left (-i \, a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + i \, a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, a^{3}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{3 \, d} \]

[In]

integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/3*sqrt(2)*(-I*a^3*e^(4*I*d*x + 4*I*c) + I*a^3*e^(2*I*d*x + 2*I*c) + 2*I*a^3)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1
))/d

Sympy [F(-1)]

Timed out. \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\text {Timed out} \]

[In]

integrate(cos(d*x+c)**3*(a+I*a*tan(d*x+c))**(7/2),x)

[Out]

Timed out

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 504 vs. \(2 (57) = 114\).

Time = 0.40 (sec) , antiderivative size = 504, normalized size of antiderivative = 7.10 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=-\frac {2 \, {\left (-i \, a^{\frac {7}{2}} - \frac {6 \, a^{\frac {7}{2}} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {5 i \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {24 \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {10 i \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {36 \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac {10 i \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {24 \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {5 i \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac {6 \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + \frac {i \, a^{\frac {7}{2}} \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}\right )} {\left (-\frac {2 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - 1\right )}^{\frac {7}{2}}}{-3 \, d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}} {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - 1\right )}^{\frac {7}{2}} {\left (-\frac {4 i \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + \frac {3 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {8 i \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {14 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {14 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac {8 i \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac {3 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {4 i \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac {\sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - 1\right )}} \]

[In]

integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

-2*(-I*a^(7/2) - 6*a^(7/2)*sin(d*x + c)/(cos(d*x + c) + 1) + 5*I*a^(7/2)*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 +
24*a^(7/2)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 10*I*a^(7/2)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 36*a^(7/2
)*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 10*I*a^(7/2)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + 24*a^(7/2)*sin(d*x
+ c)^7/(cos(d*x + c) + 1)^7 - 5*I*a^(7/2)*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 6*a^(7/2)*sin(d*x + c)^9/(cos(
d*x + c) + 1)^9 + I*a^(7/2)*sin(d*x + c)^10/(cos(d*x + c) + 1)^10)*(-2*I*sin(d*x + c)/(cos(d*x + c) + 1) + sin
(d*x + c)^2/(cos(d*x + c) + 1)^2 - 1)^(7/2)/(d*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(7/2)*(sin(d*x + c)/(cos(
d*x + c) + 1) - 1)^(7/2)*(12*I*sin(d*x + c)/(cos(d*x + c) + 1) - 9*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 24*I*
sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 42*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 42*sin(d*x + c)^6/(cos(d*x + c)
+ 1)^6 - 24*I*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 9*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 - 12*I*sin(d*x + c)
^9/(cos(d*x + c) + 1)^9 + 3*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 + 3))

Giac [F]

\[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\int { {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} \cos \left (d x + c\right )^{3} \,d x } \]

[In]

integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^(7/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(7/2)*cos(d*x + c)^3, x)

Mupad [B] (verification not implemented)

Time = 0.93 (sec) , antiderivative size = 85, normalized size of antiderivative = 1.20 \[ \int \cos ^3(c+d x) (a+i a \tan (c+d x))^{7/2} \, dx=\frac {a^3\,\sqrt {\frac {a\,\left (\cos \left (2\,c+2\,d\,x\right )+1+\sin \left (2\,c+2\,d\,x\right )\,1{}\mathrm {i}\right )}{\cos \left (2\,c+2\,d\,x\right )+1}}\,\left (\sin \left (c+d\,x\right )+\sin \left (3\,c+3\,d\,x\right )+\cos \left (c+d\,x\right )\,3{}\mathrm {i}-\cos \left (3\,c+3\,d\,x\right )\,1{}\mathrm {i}\right )}{3\,d} \]

[In]

int(cos(c + d*x)^3*(a + a*tan(c + d*x)*1i)^(7/2),x)

[Out]

(a^3*((a*(cos(2*c + 2*d*x) + sin(2*c + 2*d*x)*1i + 1))/(cos(2*c + 2*d*x) + 1))^(1/2)*(cos(c + d*x)*3i + sin(c
+ d*x) - cos(3*c + 3*d*x)*1i + sin(3*c + 3*d*x)))/(3*d)