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*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]])

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Rubi [A]  time = 0.14195, antiderivative size = 266, normalized size of antiderivative = 1., 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 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]

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])

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*}

Mathematica [A]  time = 0.609072, size = 159, normalized size = 0.6 \[ -\frac{i e^{-6 i (c+d x)} \left (-464 e^{2 i (c+d x)}-3184 e^{4 i (c+d x)}-1433 e^{6 i (c+d x)}+1645 e^{8 i (c+d x)}+350 e^{10 i (c+d x)}+40 e^{12 i (c+d x)}+3465 e^{5 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )-48\right ) \sqrt{a+i a \tan (c+d x)}}{15360 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/15360)*(-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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Maple [B]  time = 0.427, size = 1085, normalized size = 4.1 \begin{align*} \text{result too large to display} \end{align*}

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)/(cos(d*x+c)+1))^(11/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)
)*cos(d*x+c)^5*sin(d*x+c)*2^(1/2)+17325*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(11/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c
)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^4*sin(d*x+c)*2^(1/2)+34650*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(11/2)*arctan(1/
2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^3*sin(d*x+c)*2^(1/2)+34650*(-2*cos(d*x+c)/(cos(d*x+
c)+1))^(11/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)+17325*(
-2*cos(d*x+c)/(cos(d*x+c)+1))^(11/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)*sin(d
*x+c)*2^(1/2)+3465*I*2^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(
-2*cos(d*x+c)/(cos(d*x+c)+1))^(11/2)*sin(d*x+c)+101376*sin(d*x+c)*cos(d*x+c)^9+81920*sin(d*x+c)*cos(d*x+c)^11-
90112*sin(d*x+c)*cos(d*x+c)^10-81920*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)^9-29568*I*cos(d*x+c)^8-73920*I*cos(d*x+c)^7+221760*I*cos(d*x+c)^6-118272*sin(d*x+c)*cos(d*x+c)^8+147840*s
in(d*x+c)*cos(d*x+c)^7-221760*cos(d*x+c)^6*sin(d*x+c)+3465*I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(11/2)*arctanh(1/2
*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*cos(d*x+c)^5*sin(d*x+c)*2^(1/2)+17325*I*(
-2*cos(d*x+c)/(cos(d*x+c)+1))^(11/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d
*x+c))*cos(d*x+c)^4*sin(d*x+c)*2^(1/2)+34650*I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(11/2)*arctanh(1/2*2^(1/2)*(-2*c
os(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*cos(d*x+c)^3*sin(d*x+c)*2^(1/2)+34650*I*(-2*cos(d*x+c)/
(cos(d*x+c)+1))^(11/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*cos(d*x
+c)^2*sin(d*x+c)*2^(1/2)+17325*I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(11/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos
(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*cos(d*x+c)*sin(d*x+c)*2^(1/2)+3465*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(11
/2)*2^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*sin(d*x+c))*(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 [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

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]

Exception raised: ValueError

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Fricas [A]  time = 2.93715, size = 973, normalized size = 3.66 \begin{align*} \frac{{\left (3465 \, \sqrt{\frac{1}{2}} d \sqrt{-\frac{a}{d^{2}}} e^{\left (6 i \, d x + 6 i \, c\right )} \log \left (\frac{1}{231} \,{\left (462 i \, \sqrt{\frac{1}{2}} d \sqrt{-\frac{a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + 231 \, \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} 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 (6 i \, d x + 6 i \, c\right )} \log \left (\frac{1}{231} \,{\left (-462 i \, \sqrt{\frac{1}{2}} d \sqrt{-\frac{a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + 231 \, \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} 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 )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-6 i \, d x - 6 i \, c\right )}}{15360 \, d} \end{align*}

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^(6*I*d*x + 6*I*c)*log(1/231*(462*I*sqrt(1/2)*d*sqrt(-a/d^2)*e^(2*I*d*
x + 2*I*c) + 231*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d*
x - I*c)) - 3465*sqrt(1/2)*d*sqrt(-a/d^2)*e^(6*I*d*x + 6*I*c)*log(1/231*(-462*I*sqrt(1/2)*d*sqrt(-a/d^2)*e^(2*
I*d*x + 2*I*c) + 231*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-
I*d*x - I*c)) + sqrt(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 + 1
0*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^(I*d*x + I*c))*e^(-6*I*d*x - 6*I*c)/d

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{i \, a \tan \left (d x + c\right ) + a} \cos \left (d x + c\right )^{6}\,{d x} \end{align*}

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)