3.299 \(\int \cos ^6(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=239 \[ -\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{3/2}}-\frac{3 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}-\frac{21 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac{35 i a^3}{128 d (a+i a \tan (c+d x))^{3/2}}+\frac{105 i a^2}{256 d \sqrt{a+i a \tan (c+d x)}}-\frac{105 i a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{256 \sqrt{2} d} \]

[Out]

(((-105*I)/256)*a^(3/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*d) + (((35*I)/128)*a^3
)/(d*(a + I*a*Tan[c + d*x])^(3/2)) - ((I/6)*a^6)/(d*(a - I*a*Tan[c + d*x])^3*(a + I*a*Tan[c + d*x])^(3/2)) - (
((3*I)/16)*a^5)/(d*(a - I*a*Tan[c + d*x])^2*(a + I*a*Tan[c + d*x])^(3/2)) - (((21*I)/64)*a^4)/(d*(a - I*a*Tan[
c + d*x])*(a + I*a*Tan[c + d*x])^(3/2)) + (((105*I)/256)*a^2)/(d*Sqrt[a + I*a*Tan[c + d*x]])

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Rubi [A]  time = 0.134344, antiderivative size = 239, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3487, 51, 63, 206} \[ -\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{3/2}}-\frac{3 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}-\frac{21 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac{35 i a^3}{128 d (a+i a \tan (c+d x))^{3/2}}+\frac{105 i a^2}{256 d \sqrt{a+i a \tan (c+d x)}}-\frac{105 i a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{256 \sqrt{2} d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(((-105*I)/256)*a^(3/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*d) + (((35*I)/128)*a^3
)/(d*(a + I*a*Tan[c + d*x])^(3/2)) - ((I/6)*a^6)/(d*(a - I*a*Tan[c + d*x])^3*(a + I*a*Tan[c + d*x])^(3/2)) - (
((3*I)/16)*a^5)/(d*(a - I*a*Tan[c + d*x])^2*(a + I*a*Tan[c + d*x])^(3/2)) - (((21*I)/64)*a^4)/(d*(a - I*a*Tan[
c + d*x])*(a + I*a*Tan[c + d*x])^(3/2)) + (((105*I)/256)*a^2)/(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) (a+i a \tan (c+d x))^{3/2} \, dx &=-\frac{\left (i a^7\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x)^4 (a+x)^{5/2}} \, 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))^{3/2}}-\frac{\left (3 i a^6\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x)^3 (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{4 d}\\ &=-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{3/2}}-\frac{3 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}-\frac{\left (21 i a^5\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x)^2 (a+x)^{5/2}} \, 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))^{3/2}}-\frac{3 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}-\frac{21 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}-\frac{\left (105 i a^4\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{128 d}\\ &=\frac{35 i a^3}{128 d (a+i a \tan (c+d x))^{3/2}}-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{3/2}}-\frac{3 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}-\frac{21 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}-\frac{\left (105 i a^3\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{256 d}\\ &=\frac{35 i a^3}{128 d (a+i a \tan (c+d x))^{3/2}}-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{3/2}}-\frac{3 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}-\frac{21 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac{105 i a^2}{256 d \sqrt{a+i a \tan (c+d x)}}-\frac{\left (105 i a^2\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt{a+x}} \, dx,x,i a \tan (c+d x)\right )}{512 d}\\ &=\frac{35 i a^3}{128 d (a+i a \tan (c+d x))^{3/2}}-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{3/2}}-\frac{3 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}-\frac{21 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac{105 i a^2}{256 d \sqrt{a+i a \tan (c+d x)}}-\frac{\left (105 i a^2\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{256 d}\\ &=-\frac{105 i a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{256 \sqrt{2} d}+\frac{35 i a^3}{128 d (a+i a \tan (c+d x))^{3/2}}-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 (a+i a \tan (c+d x))^{3/2}}-\frac{3 i a^5}{16 d (a-i a \tan (c+d x))^2 (a+i a \tan (c+d x))^{3/2}}-\frac{21 i a^4}{64 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac{105 i a^2}{256 d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.981902, size = 169, normalized size = 0.71 \[ \frac{a e^{-4 i (c+d x)} \cos ^2(c+d x) (\tan (c+d x)-i) \left (\sqrt{1+e^{2 i (c+d x)}} \left (-208 e^{2 i (c+d x)}+165 e^{4 i (c+d x)}+50 e^{6 i (c+d x)}+8 e^{8 i (c+d x)}-16\right )+315 e^{3 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right ) \sqrt{a+i a \tan (c+d x)}}{768 d \sqrt{1+e^{2 i (c+d x)}}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(a*(Sqrt[1 + E^((2*I)*(c + d*x))]*(-16 - 208*E^((2*I)*(c + d*x)) + 165*E^((4*I)*(c + d*x)) + 50*E^((6*I)*(c +
d*x)) + 8*E^((8*I)*(c + d*x))) + 315*E^((3*I)*(c + d*x))*ArcSinh[E^(I*(c + d*x))])*Cos[c + d*x]^2*(-I + Tan[c
+ d*x])*Sqrt[a + I*a*Tan[c + d*x]])/(768*d*E^((4*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))])

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Maple [B]  time = 0.365, size = 1086, normalized size = 4.5 \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))^(3/2),x)

[Out]

1/49152/d*a*(315*(-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)+1575*(-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)+3150*(-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)+3150*(-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)+1575*(-2*c
os(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)+315*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*co
s(d*x+c)/(cos(d*x+c)+1))^(11/2)*sin(d*x+c)+1575*I*2^(1/2)*(-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)+315*I*2^(1/2)*(-2*co
s(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)+9216*sin(d*x+c)*cos(d*x+c)^9+16384*sin(d*x+c)*cos(d*x+c)^11-8192*sin(d*x+c)*cos(d*x+
c)^10-2688*I*cos(d*x+c)^8-6720*I*cos(d*x+c)^7+20160*I*cos(d*x+c)^6-16384*I*cos(d*x+c)^12+8192*I*cos(d*x+c)^11-
1024*I*cos(d*x+c)^10-1536*I*cos(d*x+c)^9-10752*sin(d*x+c)*cos(d*x+c)^8+13440*sin(d*x+c)*cos(d*x+c)^7-20160*cos
(d*x+c)^6*sin(d*x+c)+315*(-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)+3150*I*2^(1/2)*(-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)^3*sin(d*x+c)+3150*I*2^(1/2)*(-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)+1575*I*2^(1/2)*(-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))*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/(I*s
in(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))^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 2.76755, size = 959, normalized size = 4.01 \begin{align*} \frac{{\left (315 \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{3}}{d^{2}}} d e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac{{\left (210 i \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} + 105 \, \sqrt{2}{\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{105 \, a}\right ) - 315 \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{3}}{d^{2}}} d e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (\frac{{\left (-210 i \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{3}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} + 105 \, \sqrt{2}{\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{105 \, a}\right ) + \sqrt{2}{\left (-8 i \, a e^{\left (10 i \, d x + 10 i \, c\right )} - 58 i \, a e^{\left (8 i \, d x + 8 i \, c\right )} - 215 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} + 43 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 224 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 16 i \, a\right )} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{1536 \, 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))^(3/2),x, algorithm="fricas")

[Out]

1/1536*(315*sqrt(1/2)*sqrt(-a^3/d^2)*d*e^(4*I*d*x + 4*I*c)*log(1/105*(210*I*sqrt(1/2)*sqrt(-a^3/d^2)*d*e^(2*I*
d*x + 2*I*c) + 105*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-
I*d*x - I*c)/a) - 315*sqrt(1/2)*sqrt(-a^3/d^2)*d*e^(4*I*d*x + 4*I*c)*log(1/105*(-210*I*sqrt(1/2)*sqrt(-a^3/d^2
)*d*e^(2*I*d*x + 2*I*c) + 105*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x +
I*c))*e^(-I*d*x - I*c)/a) + sqrt(2)*(-8*I*a*e^(10*I*d*x + 10*I*c) - 58*I*a*e^(8*I*d*x + 8*I*c) - 215*I*a*e^(6
*I*d*x + 6*I*c) + 43*I*a*e^(4*I*d*x + 4*I*c) + 224*I*a*e^(2*I*d*x + 2*I*c) + 16*I*a)*sqrt(a/(e^(2*I*d*x + 2*I*
c) + 1))*e^(I*d*x + I*c))*e^(-4*I*d*x - 4*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))**(3/2),x)

[Out]

Timed out

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Giac [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))^(3/2),x, algorithm="giac")

[Out]

Timed out