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

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

[Out]

-105/512*I*a^(3/2)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1/2)*2^(1/2)/a^(1/2))/d*2^(1/2)+105/256*I*a^2/d/(a+I*a*tan(
d*x+c))^(1/2)+35/128*I*a^3/d/(a+I*a*tan(d*x+c))^(3/2)-1/6*I*a^6/d/(a-I*a*tan(d*x+c))^3/(a+I*a*tan(d*x+c))^(3/2
)-3/16*I*a^5/d/(a-I*a*tan(d*x+c))^2/(a+I*a*tan(d*x+c))^(3/2)-21/64*I*a^4/d/(a-I*a*tan(d*x+c))/(a+I*a*tan(d*x+c
))^(3/2)

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Rubi [A]  time = 0.13, antiderivative size = 239, normalized size of antiderivative = 1.00, 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 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) (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*}

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Mathematica [A]  time = 1.11, 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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fricas [A]  time = 0.82, size = 310, normalized size = 1.30 \[ \frac {{\left (315 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{3}}{d^{2}}} d e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (512 i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + 512 i \, d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + 512 \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{128 \, a}\right ) - 315 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{3}}{d^{2}}} d e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-512 i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - 512 i \, d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + 512 \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{128 \, a}\right ) + \sqrt {2} {\left (-8 i \, a e^{\left (10 i \, d x + 10 i \, c\right )} - 58 i \, a e^{\left (8 i \, d x + 8 i \, c\right )} - 215 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} + 43 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} + 224 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 16 i \, a\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{1536 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

1/1536*(315*sqrt(1/2)*sqrt(-a^3/d^2)*d*e^(3*I*d*x + 3*I*c)*log(1/128*(sqrt(2)*sqrt(1/2)*(512*I*d*e^(2*I*d*x +
2*I*c) + 512*I*d)*sqrt(-a^3/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)) + 512*a^2*e^(I*d*x + I*c))*e^(-I*d*x - I*c)
/a) - 315*sqrt(1/2)*sqrt(-a^3/d^2)*d*e^(3*I*d*x + 3*I*c)*log(1/128*(sqrt(2)*sqrt(1/2)*(-512*I*d*e^(2*I*d*x + 2
*I*c) - 512*I*d)*sqrt(-a^3/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)) + 512*a^2*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^(-3*I*d*x -
3*I*c)/d

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giac [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))^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 1.51, size = 1086, normalized size = 4.54 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

1/49152/d*(315*I*2^(1/2)*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)/c
os(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(11/2)+1575*I*2^(1/2)*sin(d*x+c)*cos(d*x+c)^4*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)+315*(-
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
)+16384*sin(d*x+c)*cos(d*x+c)^11+9216*sin(d*x+c)*cos(d*x+c)^9-10752*sin(d*x+c)*cos(d*x+c)^8-8192*sin(d*x+c)*co
s(d*x+c)^10+315*sin(d*x+c)*cos(d*x+c)^5*(-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)+1575*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)+3150*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)+3150*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)+1575*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)+13440*sin(d*x+c)*cos(d*x+c)^7-20160*sin(d*x+c)*cos(d*x+c)^6+3150*I*2^(1/2)*sin(d*x+c)*cos(d*x+c)^3*ar
ctanh(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)-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)^9-2688*I*cos(d*x+c)^8-
6720*I*cos(d*x+c)^7+20160*I*cos(d*x+c)^6+3150*I*2^(1/2)*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)+1575*I*2^(1/2)*sin(d*x
+c)*cos(d*x+c)*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)+315*I*2^(1/2)*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)*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*a

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maxima [A]  time = 0.81, size = 212, normalized size = 0.89 \[ \frac {i \, {\left (315 \, \sqrt {2} a^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {4 \, {\left (315 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{4} a^{3} - 1680 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{4} + 2772 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{5} - 1152 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{6} - 256 \, a^{7}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {9}{2}} - 6 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {7}{2}} a + 12 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} a^{2} - 8 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} a^{3}}\right )}}{3072 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^6*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

1/3072*I*(315*sqrt(2)*a^(5/2)*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*(315*(I*a*tan(d*x + c) + a)^4*a^3 - 1680*(I*a*tan(d*x + c) + a)^3*a^4 + 2772*(I*a*tan(
d*x + c) + a)^2*a^5 - 1152*(I*a*tan(d*x + c) + a)*a^6 - 256*a^7)/((I*a*tan(d*x + c) + a)^(9/2) - 6*(I*a*tan(d*
x + c) + a)^(7/2)*a + 12*(I*a*tan(d*x + c) + a)^(5/2)*a^2 - 8*(I*a*tan(d*x + c) + a)^(3/2)*a^3))/(a*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (c+d\,x\right )}^6\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^6*(a + a*tan(c + d*x)*1i)^(3/2),x)

[Out]

int(cos(c + d*x)^6*(a + a*tan(c + d*x)*1i)^(3/2), 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))**(3/2),x)

[Out]

Timed out

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