3.318 \(\int \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\)

Optimal. Leaf size=231 \[ -\frac{3 i a^2 \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{20 d}-\frac{9 i a^2 \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{32 d}+\frac{3 i a^3 \cos (c+d x)}{16 d \sqrt{a+i a \tan (c+d x)}}+\frac{9 i a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{32 \sqrt{2} d}-\frac{i \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}-\frac{9 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{70 d} \]

[Out]

(((9*I)/32)*a^(5/2)*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*d) + (((3*I
)/16)*a^3*Cos[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((9*I)/32)*a^2*Cos[c + d*x]*Sqrt[a + I*a*Tan[c + d*x
]])/d - (((3*I)/20)*a^2*Cos[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d - (((9*I)/70)*a*Cos[c + d*x]^5*(a + I*a*T
an[c + d*x])^(3/2))/d - ((I/7)*Cos[c + d*x]^7*(a + I*a*Tan[c + d*x])^(5/2))/d

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Rubi [A]  time = 0.338896, antiderivative size = 231, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used = {3497, 3502, 3490, 3489, 206} \[ -\frac{3 i a^2 \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{20 d}-\frac{9 i a^2 \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{32 d}+\frac{3 i a^3 \cos (c+d x)}{16 d \sqrt{a+i a \tan (c+d x)}}+\frac{9 i a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{32 \sqrt{2} d}-\frac{i \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}-\frac{9 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{70 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^7*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

(((9*I)/32)*a^(5/2)*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*d) + (((3*I
)/16)*a^3*Cos[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((9*I)/32)*a^2*Cos[c + d*x]*Sqrt[a + I*a*Tan[c + d*x
]])/d - (((3*I)/20)*a^2*Cos[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d - (((9*I)/70)*a*Cos[c + d*x]^5*(a + I*a*T
an[c + d*x])^(3/2))/d - ((I/7)*Cos[c + d*x]^7*(a + I*a*Tan[c + d*x])^(5/2))/d

Rule 3497

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(d*
Sec[e + f*x])^m*(a + b*Tan[e + f*x])^n)/(a*f*m), x] + Dist[(a*(m + n))/(m*d^2), Int[(d*Sec[e + f*x])^(m + 2)*(
a + b*Tan[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && GtQ[n, 0] && LtQ[m, -
1] && IntegersQ[2*m, 2*n]

Rule 3502

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^n)/(b*f*(m + 2*n)), x] + Dist[Simplify[m + n]/(a*(m + 2*n)), Int[(d*Sec[
e + f*x])^m*(a + b*Tan[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, m}, x] && EqQ[a^2 + b^2, 0] && LtQ[n
, 0] && NeQ[m + 2*n, 0] && IntegersQ[2*m, 2*n]

Rule 3490

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(d
*Sec[e + f*x])^m*(a + b*Tan[e + f*x])^n)/(a*f*m), x] + Dist[a/(2*d^2), Int[(d*Sec[e + f*x])^(m + 2)*(a + b*Tan
[e + f*x])^(n - 1), x], x] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 + b^2, 0] && EqQ[m/2 + n, 0] && GtQ[n, 0]

Rule 3489

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(-2*a)/(b*f), Subst[
Int[1/(2 - a*x^2), x], x, Sec[e + f*x]/Sqrt[a + b*Tan[e + f*x]]], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 + b^
2, 0]

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 ^7(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx &=-\frac{i \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}+\frac{1}{14} (9 a) \int \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\\ &=-\frac{9 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{70 d}-\frac{i \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}+\frac{1}{20} \left (9 a^2\right ) \int \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)} \, dx\\ &=-\frac{3 i a^2 \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{20 d}-\frac{9 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{70 d}-\frac{i \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}+\frac{1}{8} \left (3 a^3\right ) \int \frac{\cos (c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx\\ &=\frac{3 i a^3 \cos (c+d x)}{16 d \sqrt{a+i a \tan (c+d x)}}-\frac{3 i a^2 \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{20 d}-\frac{9 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{70 d}-\frac{i \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}+\frac{1}{32} \left (9 a^2\right ) \int \cos (c+d x) \sqrt{a+i a \tan (c+d x)} \, dx\\ &=\frac{3 i a^3 \cos (c+d x)}{16 d \sqrt{a+i a \tan (c+d x)}}-\frac{9 i a^2 \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{32 d}-\frac{3 i a^2 \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{20 d}-\frac{9 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{70 d}-\frac{i \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}+\frac{1}{64} \left (9 a^3\right ) \int \frac{\sec (c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx\\ &=\frac{3 i a^3 \cos (c+d x)}{16 d \sqrt{a+i a \tan (c+d x)}}-\frac{9 i a^2 \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{32 d}-\frac{3 i a^2 \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{20 d}-\frac{9 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{70 d}-\frac{i \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}+\frac{\left (9 i a^3\right ) \operatorname{Subst}\left (\int \frac{1}{2-a x^2} \, dx,x,\frac{\sec (c+d x)}{\sqrt{a+i a \tan (c+d x)}}\right )}{32 d}\\ &=\frac{9 i a^{5/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{32 \sqrt{2} d}+\frac{3 i a^3 \cos (c+d x)}{16 d \sqrt{a+i a \tan (c+d x)}}-\frac{9 i a^2 \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{32 d}-\frac{3 i a^2 \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{20 d}-\frac{9 i a \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{70 d}-\frac{i \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2}}{7 d}\\ \end{align*}

Mathematica [A]  time = 1.04712, size = 155, normalized size = 0.67 \[ -\frac{i a^2 e^{-3 i (c+d x)} \left (353 e^{2 i (c+d x)}+544 e^{4 i (c+d x)}+214 e^{6 i (c+d x)}+68 e^{8 i (c+d x)}+10 e^{10 i (c+d x)}-315 e^{2 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )-35\right ) \sqrt{a+i a \tan (c+d x)}}{2240 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^7*(a + I*a*Tan[c + d*x])^(5/2),x]

[Out]

((-I/2240)*a^2*(-35 + 353*E^((2*I)*(c + d*x)) + 544*E^((4*I)*(c + d*x)) + 214*E^((6*I)*(c + d*x)) + 68*E^((8*I
)*(c + d*x)) + 10*E^((10*I)*(c + d*x)) - 315*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[1
+ E^((2*I)*(c + d*x))]])*Sqrt[a + I*a*Tan[c + d*x]])/(d*E^((3*I)*(c + d*x)))

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Maple [B]  time = 0.484, size = 1260, normalized size = 5.5 \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)^7*(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

-1/143360/d*a^2*(1890*I*2^(1/2)*cos(d*x+c)^5*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(13/2)*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))+315*I*2^(1/2)*cos(d*x+c)^6*sin(d*x+c)*(-2*cos(
d*x+c)/(cos(d*x+c)+1))^(13/2)*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))-
81920*sin(d*x+c)*cos(d*x+c)^13+40960*sin(d*x+c)*cos(d*x+c)^12-21504*sin(d*x+c)*cos(d*x+c)^9-16384*sin(d*x+c)*c
os(d*x+c)^11+18432*sin(d*x+c)*cos(d*x+c)^10+26880*sin(d*x+c)*cos(d*x+c)^8-40320*sin(d*x+c)*cos(d*x+c)^7+81920*
I*cos(d*x+c)^14-40960*I*cos(d*x+c)^13-24576*I*cos(d*x+c)^12+2048*I*cos(d*x+c)^11+5376*I*cos(d*x+c)^9+13440*I*c
os(d*x+c)^8-40320*I*cos(d*x+c)^7+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*cos(d*x+c)/(cos(d*x+c)+1))^(13/2)*sin(d*x+c)-315*2^(1/2)*cos(d*x+c)^6*sin(d*x+c)*arctan(1
/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(13/2)-1890*2^(1/2)*cos(d*x+c)
^5*sin(d*x+c)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(13/2)-4
725*2^(1/2)*cos(d*x+c)^4*sin(d*x+c)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(c
os(d*x+c)+1))^(13/2)-6300*2^(1/2)*cos(d*x+c)^3*sin(d*x+c)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1
/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(13/2)-4725*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)*arctan(1/2*2^(1/2)*(-2*cos(d*x
+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(13/2)-1890*2^(1/2)*cos(d*x+c)*sin(d*x+c)*arctan(1/2
*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(13/2)+3072*I*cos(d*x+c)^10-315*
2^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(13/2)*sin(d*x
+c)+4725*I*2^(1/2)*cos(d*x+c)^4*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(13/2)*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))+6300*I*2^(1/2)*cos(d*x+c)^3*sin(d*x+c)*(-2*cos(d*x+c)/(cos(
d*x+c)+1))^(13/2)*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))+4725*I*2^(1/
2)*cos(d*x+c)^2*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(13/2)*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))+1890*I*2^(1/2)*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(13/2)*
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)))*(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)^6

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Maxima [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)^7*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.04606, size = 971, normalized size = 4.2 \begin{align*} -\frac{{\left (315 \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{5}}{d^{2}}} d e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac{{\left (18 i \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )} + 9 \, \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 )}}{9 \, a^{2}}\right ) - 315 \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{5}}{d^{2}}} d e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac{{\left (-18 i \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{5}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )} + 9 \, \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 )}}{9 \, a^{2}}\right ) - \sqrt{2}{\left (-10 i \, a^{2} e^{\left (10 i \, d x + 10 i \, c\right )} - 68 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 214 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 544 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 353 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 35 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 (-3 i \, d x - 3 i \, c\right )}}{2240 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/2240*(315*sqrt(1/2)*sqrt(-a^5/d^2)*d*e^(3*I*d*x + 3*I*c)*log(1/9*(18*I*sqrt(1/2)*sqrt(-a^5/d^2)*d*e^(I*d*x
+ I*c) + 9*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) - 315*sqrt(1/2)*sqrt(-a^5/d^2)*d*e^(3*I*d*x + 3*I*c)*log(1/9*(-18*I*sqrt(1/2)*sqrt(-a^5/d^2)*d*e
^(I*d*x + I*c) + 9*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)*(-10*I*a^2*e^(10*I*d*x + 10*I*c) - 68*I*a^2*e^(8*I*d*x + 8*I*c) - 214*I*a^2*e^
(6*I*d*x + 6*I*c) - 544*I*a^2*e^(4*I*d*x + 4*I*c) - 353*I*a^2*e^(2*I*d*x + 2*I*c) + 35*I*a^2)*sqrt(a/(e^(2*I*d
*x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-3*I*d*x - 3*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)**7*(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cos \left (d x + c\right )^{7}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^7*(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(5/2)*cos(d*x + c)^7, x)