3.312 \(\int \cos ^6(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=210 \[ -\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 \sqrt{a+i a \tan (c+d x)}}-\frac{7 i a^5}{48 d (a-i a \tan (c+d x))^2 \sqrt{a+i a \tan (c+d x)}}-\frac{35 i a^4}{192 d (a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}+\frac{35 i a^3}{128 d \sqrt{a+i a \tan (c+d x)}}-\frac{35 i a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{128 \sqrt{2} d} \]

[Out]

(((-35*I)/128)*a^(5/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*d) + (((35*I)/128)*a^3)
/(d*Sqrt[a + I*a*Tan[c + d*x]]) - ((I/6)*a^6)/(d*(a - I*a*Tan[c + d*x])^3*Sqrt[a + I*a*Tan[c + d*x]]) - (((7*I
)/48)*a^5)/(d*(a - I*a*Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]]) - (((35*I)/192)*a^4)/(d*(a - I*a*Tan[c + d*
x])*Sqrt[a + I*a*Tan[c + d*x]])

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Rubi [A]  time = 0.119865, antiderivative size = 210, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3487, 51, 63, 206} \[ -\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 \sqrt{a+i a \tan (c+d x)}}-\frac{7 i a^5}{48 d (a-i a \tan (c+d x))^2 \sqrt{a+i a \tan (c+d x)}}-\frac{35 i a^4}{192 d (a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}+\frac{35 i a^3}{128 d \sqrt{a+i a \tan (c+d x)}}-\frac{35 i a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{128 \sqrt{2} d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^6*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(((-35*I)/128)*a^(5/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*d) + (((35*I)/128)*a^3)
/(d*Sqrt[a + I*a*Tan[c + d*x]]) - ((I/6)*a^6)/(d*(a - I*a*Tan[c + d*x])^3*Sqrt[a + I*a*Tan[c + d*x]]) - (((7*I
)/48)*a^5)/(d*(a - I*a*Tan[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]]) - (((35*I)/192)*a^4)/(d*(a - I*a*Tan[c + d*
x])*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))^{5/2} \, dx &=-\frac{\left (i a^7\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x)^4 (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 \sqrt{a+i a \tan (c+d x)}}-\frac{\left (7 i a^6\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x)^3 (a+x)^{3/2}} \, 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 \sqrt{a+i a \tan (c+d x)}}-\frac{7 i a^5}{48 d (a-i a \tan (c+d x))^2 \sqrt{a+i a \tan (c+d x)}}-\frac{\left (35 i a^5\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x)^2 (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{96 d}\\ &=-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 \sqrt{a+i a \tan (c+d x)}}-\frac{7 i a^5}{48 d (a-i a \tan (c+d x))^2 \sqrt{a+i a \tan (c+d x)}}-\frac{35 i a^4}{192 d (a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}-\frac{\left (35 i a^4\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{128 d}\\ &=\frac{35 i a^3}{128 d \sqrt{a+i a \tan (c+d x)}}-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 \sqrt{a+i a \tan (c+d x)}}-\frac{7 i a^5}{48 d (a-i a \tan (c+d x))^2 \sqrt{a+i a \tan (c+d x)}}-\frac{35 i a^4}{192 d (a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}-\frac{\left (35 i a^3\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt{a+x}} \, dx,x,i a \tan (c+d x)\right )}{256 d}\\ &=\frac{35 i a^3}{128 d \sqrt{a+i a \tan (c+d x)}}-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 \sqrt{a+i a \tan (c+d x)}}-\frac{7 i a^5}{48 d (a-i a \tan (c+d x))^2 \sqrt{a+i a \tan (c+d x)}}-\frac{35 i a^4}{192 d (a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}-\frac{\left (35 i a^3\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{128 d}\\ &=-\frac{35 i a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{128 \sqrt{2} d}+\frac{35 i a^3}{128 d \sqrt{a+i a \tan (c+d x)}}-\frac{i a^6}{6 d (a-i a \tan (c+d x))^3 \sqrt{a+i a \tan (c+d x)}}-\frac{7 i a^5}{48 d (a-i a \tan (c+d x))^2 \sqrt{a+i a \tan (c+d x)}}-\frac{35 i a^4}{192 d (a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.825375, size = 142, normalized size = 0.68 \[ -\frac{i a^2 e^{-2 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \left (\sqrt{1+e^{2 i (c+d x)}} \left (87 e^{2 i (c+d x)}+38 e^{4 i (c+d x)}+8 e^{6 i (c+d x)}-48\right )+105 e^{i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right ) \sqrt{a+i a \tan (c+d x)}}{768 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^6*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

((-I/768)*a^2*Sqrt[1 + E^((2*I)*(c + d*x))]*(Sqrt[1 + E^((2*I)*(c + d*x))]*(-48 + 87*E^((2*I)*(c + d*x)) + 38*
E^((4*I)*(c + d*x)) + 8*E^((6*I)*(c + d*x))) + 105*E^(I*(c + d*x))*ArcSinh[E^(I*(c + d*x))])*Sqrt[a + I*a*Tan[
c + d*x]])/(d*E^((2*I)*(c + d*x)))

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Maple [B]  time = 0.402, size = 1088, normalized size = 5.2 \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))^(5/2),x)

[Out]

1/24576/d*a^2*(5120*I*cos(d*x+c)^10-512*I*cos(d*x+c)^9-896*I*cos(d*x+c)^8-2240*I*cos(d*x+c)^7+6720*I*cos(d*x+c
)^6+105*(-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)+525*(-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)+1050*(-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)+1050*(-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)+525*(-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)+5
25*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)*si
n(d*x+c)/cos(d*x+c))*sin(d*x+c)*cos(d*x+c)+1050*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))*sin(d*x+c)*cos(d*x+c)^2+3072*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+105*I*2^(1/2)*arctanh(1/2*2^(1/2)*(-2*co
s(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)-16384*I
*cos(d*x+c)^12+8192*I*cos(d*x+c)^11-3584*sin(d*x+c)*cos(d*x+c)^8+4480*sin(d*x+c)*cos(d*x+c)^7-6720*cos(d*x+c)^
6*sin(d*x+c)+105*(-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)+105*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)/(c
os(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d*x+c)*cos(d*x+c)^5+525*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))*sin(d*x+c)*cos(d*x+c)
^4+1050*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))*sin(d*x+c)*cos(d*x+c)^3)*(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))^(5/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [B]  time = 3.17728, size = 936, normalized size = 4.46 \begin{align*} \frac{{\left (105 \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac{{\left (70 i \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} + 35 \, \sqrt{2}{\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\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 )}}{35 \, a^{2}}\right ) - 105 \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac{{\left (-70 i \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} + 35 \, \sqrt{2}{\left (a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + a^{2}\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 )}}{35 \, a^{2}}\right ) + \sqrt{2}{\left (-8 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 46 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 125 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 39 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 48 i \, a^{2}\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 (-2 i \, d x - 2 i \, c\right )}}{768 \, 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))^(5/2),x, algorithm="fricas")

[Out]

1/768*(105*sqrt(1/2)*sqrt(-a^5/d^2)*d*e^(2*I*d*x + 2*I*c)*log(1/35*(70*I*sqrt(1/2)*sqrt(-a^5/d^2)*d*e^(2*I*d*x
+ 2*I*c) + 35*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-
I*d*x - I*c)/a^2) - 105*sqrt(1/2)*sqrt(-a^5/d^2)*d*e^(2*I*d*x + 2*I*c)*log(1/35*(-70*I*sqrt(1/2)*sqrt(-a^5/d^2
)*d*e^(2*I*d*x + 2*I*c) + 35*sqrt(2)*(a^2*e^(2*I*d*x + 2*I*c) + a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*
x + I*c))*e^(-I*d*x - I*c)/a^2) + sqrt(2)*(-8*I*a^2*e^(8*I*d*x + 8*I*c) - 46*I*a^2*e^(6*I*d*x + 6*I*c) - 125*I
*a^2*e^(4*I*d*x + 4*I*c) - 39*I*a^2*e^(2*I*d*x + 2*I*c) + 48*I*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x
+ I*c))*e^(-2*I*d*x - 2*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))**(5/2),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))^(5/2),x, algorithm="giac")

[Out]

Timed out