3.285 \(\int \cos ^6(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx\)

Optimal. Leaf size=266 \[ -\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}-\frac {11 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {33 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {231 i a^3}{640 d (a+i a \tan (c+d x))^{5/2}}+\frac {77 i a^2}{256 d (a+i a \tan (c+d x))^{3/2}}+\frac {231 i a}{512 d \sqrt {a+i a \tan (c+d x)}}-\frac {231 i \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{512 \sqrt {2} d} \]

[Out]

-231/1024*I*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^(1/2)/d*2^(1/2)+231/512*I*a/d/(a+I*a*tan(d
*x+c))^(1/2)+231/640*I*a^3/d/(a+I*a*tan(d*x+c))^(5/2)-1/6*I*a^6/d/(a-I*a*tan(d*x+c))^3/(a+I*a*tan(d*x+c))^(5/2
)-11/48*I*a^5/d/(a-I*a*tan(d*x+c))^2/(a+I*a*tan(d*x+c))^(5/2)-33/64*I*a^4/d/(a-I*a*tan(d*x+c))/(a+I*a*tan(d*x+
c))^(5/2)+77/256*I*a^2/d/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]  time = 0.14, antiderivative size = 266, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3487, 51, 63, 206} \[ -\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}-\frac {11 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {33 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {231 i a^3}{640 d (a+i a \tan (c+d x))^{5/2}}+\frac {77 i a^2}{256 d (a+i a \tan (c+d x))^{3/2}}+\frac {231 i a}{512 d \sqrt {a+i a \tan (c+d x)}}-\frac {231 i \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{512 \sqrt {2} d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(((-231*I)/512)*Sqrt[a]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*d) + (((231*I)/640)*a^
3)/(d*(a + I*a*Tan[c + d*x])^(5/2)) - ((I/6)*a^6)/(d*(a - I*a*Tan[c + d*x])^3*(a + I*a*Tan[c + d*x])^(5/2)) -
(((11*I)/48)*a^5)/(d*(a - I*a*Tan[c + d*x])^2*(a + I*a*Tan[c + d*x])^(5/2)) - (((33*I)/64)*a^4)/(d*(a - I*a*Ta
n[c + d*x])*(a + I*a*Tan[c + d*x])^(5/2)) + (((77*I)/256)*a^2)/(d*(a + I*a*Tan[c + d*x])^(3/2)) + (((231*I)/51
2)*a)/(d*Sqrt[a + I*a*Tan[c + d*x]])

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3487

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 \cos ^6(c+d x) \sqrt {a+i a \tan (c+d x)} \, dx &=-\frac {\left (i a^7\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x)^4 (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}-\frac {\left (11 i a^6\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x)^3 (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{12 d}\\ &=-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}-\frac {11 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {\left (33 i a^5\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x)^2 (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{32 d}\\ &=-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}-\frac {11 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {33 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}-\frac {\left (231 i a^4\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{7/2}} \, dx,x,i a \tan (c+d x)\right )}{128 d}\\ &=\frac {231 i a^3}{640 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}-\frac {11 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {33 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}-\frac {\left (231 i a^3\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{256 d}\\ &=\frac {231 i a^3}{640 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}-\frac {11 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {33 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {77 i a^2}{256 d (a+i a \tan (c+d x))^{3/2}}-\frac {\left (231 i a^2\right ) \operatorname {Subst}\left (\int \frac {1}{(a-x) (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{512 d}\\ &=\frac {231 i a^3}{640 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}-\frac {11 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {33 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {77 i a^2}{256 d (a+i a \tan (c+d x))^{3/2}}+\frac {231 i a}{512 d \sqrt {a+i a \tan (c+d x)}}-\frac {(231 i a) \operatorname {Subst}\left (\int \frac {1}{(a-x) \sqrt {a+x}} \, dx,x,i a \tan (c+d x)\right )}{1024 d}\\ &=\frac {231 i a^3}{640 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}-\frac {11 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {33 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {77 i a^2}{256 d (a+i a \tan (c+d x))^{3/2}}+\frac {231 i a}{512 d \sqrt {a+i a \tan (c+d x)}}-\frac {(231 i a) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{512 d}\\ &=-\frac {231 i \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{512 \sqrt {2} d}+\frac {231 i a^3}{640 d (a+i a \tan (c+d x))^{5/2}}-\frac {i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{5/2}}-\frac {11 i a^5}{48 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{5/2}}-\frac {33 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{5/2}}+\frac {77 i a^2}{256 d (a+i a \tan (c+d x))^{3/2}}+\frac {231 i a}{512 d \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 0.78, size = 159, normalized size = 0.60 \[ -\frac {i e^{-6 i (c+d x)} \left (-464 e^{2 i (c+d x)}-3184 e^{4 i (c+d x)}-1433 e^{6 i (c+d x)}+1645 e^{8 i (c+d x)}+350 e^{10 i (c+d x)}+40 e^{12 i (c+d x)}+3465 e^{5 i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )-48\right ) \sqrt {a+i a \tan (c+d x)}}{15360 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

((-1/15360*I)*(-48 - 464*E^((2*I)*(c + d*x)) - 3184*E^((4*I)*(c + d*x)) - 1433*E^((6*I)*(c + d*x)) + 1645*E^((
8*I)*(c + d*x)) + 350*E^((10*I)*(c + d*x)) + 40*E^((12*I)*(c + d*x)) + 3465*E^((5*I)*(c + d*x))*Sqrt[1 + E^((2
*I)*(c + d*x))]*ArcSinh[E^(I*(c + d*x))])*Sqrt[a + I*a*Tan[c + d*x]])/(d*E^((6*I)*(c + d*x)))

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fricas [A]  time = 0.58, size = 296, normalized size = 1.11 \[ \frac {{\left (3465 \, \sqrt {\frac {1}{2}} d \sqrt {-\frac {a}{d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {1}{256} \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (1024 i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + 1024 i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {a}{d^{2}}} + 1024 \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 3465 \, \sqrt {\frac {1}{2}} d \sqrt {-\frac {a}{d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {1}{256} \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-1024 i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - 1024 i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {-\frac {a}{d^{2}}} + 1024 \, a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-40 i \, e^{\left (12 i \, d x + 12 i \, c\right )} - 350 i \, e^{\left (10 i \, d x + 10 i \, c\right )} - 1645 i \, e^{\left (8 i \, d x + 8 i \, c\right )} + 1433 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 3184 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 464 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 48 i\right )}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{15360 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/15360*(3465*sqrt(1/2)*d*sqrt(-a/d^2)*e^(5*I*d*x + 5*I*c)*log(1/256*(sqrt(2)*sqrt(1/2)*(1024*I*d*e^(2*I*d*x +
2*I*c) + 1024*I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-a/d^2) + 1024*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c))
- 3465*sqrt(1/2)*d*sqrt(-a/d^2)*e^(5*I*d*x + 5*I*c)*log(1/256*(sqrt(2)*sqrt(1/2)*(-1024*I*d*e^(2*I*d*x + 2*I*
c) - 1024*I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(-a/d^2) + 1024*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) + sq
rt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-40*I*e^(12*I*d*x + 12*I*c) - 350*I*e^(10*I*d*x + 10*I*c) - 1645*I*e^
(8*I*d*x + 8*I*c) + 1433*I*e^(6*I*d*x + 6*I*c) + 3184*I*e^(4*I*d*x + 4*I*c) + 464*I*e^(2*I*d*x + 2*I*c) + 48*I
))*e^(-5*I*d*x - 5*I*c)/d

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {i \, a \tan \left (d x + c\right ) + a} \cos \left (d x + c\right )^{6}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(I*a*tan(d*x + c) + a)*cos(d*x + c)^6, x)

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maple [B]  time = 1.43, size = 1085, normalized size = 4.08 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^(1/2),x)

[Out]

1/491520/d*(3465*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)
)*2^(1/2)*sin(d*x+c)-81920*I*cos(d*x+c)^12-8192*I*cos(d*x+c)^11-11264*I*cos(d*x+c)^10-16896*I*cos(d*x+c)^9-295
68*I*cos(d*x+c)^8-73920*I*cos(d*x+c)^7+221760*I*cos(d*x+c)^6+81920*sin(d*x+c)*cos(d*x+c)^11+101376*sin(d*x+c)*
cos(d*x+c)^9-118272*sin(d*x+c)*cos(d*x+c)^8-90112*sin(d*x+c)*cos(d*x+c)^10+3465*sin(d*x+c)*cos(d*x+c)^5*(-2*co
s(d*x+c)/(1+cos(d*x+c)))^(11/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)+17325*sin(d*x
+c)*cos(d*x+c)^4*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)
)*2^(1/2)+34650*sin(d*x+c)*cos(d*x+c)^3*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos
(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)+34650*sin(d*x+c)*cos(d*x+c)^2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*arctan(1/
2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)+17325*sin(d*x+c)*cos(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+
c)))^(11/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)+3465*I*(-2*cos(d*x+c)/(1+cos(d*x+
c)))^(11/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)*sin(d*x+c)
+147840*sin(d*x+c)*cos(d*x+c)^7-221760*sin(d*x+c)*cos(d*x+c)^6+17325*I*sin(d*x+c)*cos(d*x+c)*arctanh(1/2*(-2*c
os(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*2^(1/2)+3
465*I*sin(d*x+c)*cos(d*x+c)^5*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*
(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*2^(1/2)+17325*I*sin(d*x+c)*cos(d*x+c)^4*arctanh(1/2*(-2*cos(d*x+c)/(1+co
s(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*2^(1/2)+34650*I*sin(d*x+
c)*cos(d*x+c)^3*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)
/(1+cos(d*x+c)))^(11/2)*2^(1/2)+34650*I*sin(d*x+c)*cos(d*x+c)^2*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/
2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)*2^(1/2))*(a*(I*sin(d*x+c)+cos(d*x+c))/
cos(d*x+c))^(1/2)/(I*sin(d*x+c)+cos(d*x+c)-1)/cos(d*x+c)^5

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maxima [A]  time = 0.59, size = 230, normalized size = 0.86 \[ \frac {i \, {\left (3465 \, \sqrt {2} a^{\frac {3}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (3465 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{5} a^{2} - 18480 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{3} + 30492 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{4} - 12672 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{5} - 2816 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{6} - 1536 \, a^{7}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {11}{2}} - 6 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} a + 12 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a^{2} - 8 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{3}}\right )}}{30720 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

1/30720*I*(3465*sqrt(2)*a^(3/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*
a*tan(d*x + c) + a))) + 4*(3465*(I*a*tan(d*x + c) + a)^5*a^2 - 18480*(I*a*tan(d*x + c) + a)^4*a^3 + 30492*(I*a
*tan(d*x + c) + a)^3*a^4 - 12672*(I*a*tan(d*x + c) + a)^2*a^5 - 2816*(I*a*tan(d*x + c) + a)*a^6 - 1536*a^7)/((
I*a*tan(d*x + c) + a)^(11/2) - 6*(I*a*tan(d*x + c) + a)^(9/2)*a + 12*(I*a*tan(d*x + c) + a)^(7/2)*a^2 - 8*(I*a
*tan(d*x + c) + a)^(5/2)*a^3))/(a*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (c+d\,x\right )}^6\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6*(a + a*tan(c + d*x)*1i)^(1/2),x)

[Out]

int(cos(c + d*x)^6*(a + a*tan(c + d*x)*1i)^(1/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)**6*(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Timed out

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