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

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

[Out]

9/64*I*a^(5/2)*arctanh(1/2*sec(d*x+c)*a^(1/2)*2^(1/2)/(a+I*a*tan(d*x+c))^(1/2))/d*2^(1/2)+3/16*I*a^3*cos(d*x+c
)/d/(a+I*a*tan(d*x+c))^(1/2)-9/32*I*a^2*cos(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/d-3/20*I*a^2*cos(d*x+c)^3*(a+I*a*t
an(d*x+c))^(1/2)/d-9/70*I*a*cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(3/2)/d-1/7*I*cos(d*x+c)^7*(a+I*a*tan(d*x+c))^(5/2
)/d

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Rubi [A]  time = 0.34, antiderivative size = 231, normalized size of antiderivative = 1.00, 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 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 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 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 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]

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*}

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Mathematica [A]  time = 1.25, 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]

((-1/2240*I)*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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fricas [A]  time = 2.23, size = 300, normalized size = 1.30 \[ -\frac {{\left (315 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\frac {9 \, {\left (-i \, a^{3} + \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{16 \, d}\right ) - 315 \, \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{5}}{d^{2}}} d e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\frac {9 \, {\left (-i \, a^{3} - \sqrt {2} \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{5}}{d^{2}}} {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{16 \, d}\right ) - \sqrt {2} {\left (-10 i \, a^{2} e^{\left (10 i \, d x + 10 i \, c\right )} - 68 i \, a^{2} e^{\left (8 i \, d x + 8 i \, c\right )} - 214 i \, a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} - 544 i \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 353 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 35 i \, a^{2}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2240 \, d} \]

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^(2*I*d*x + 2*I*c)*log(-9/16*(-I*a^3 + sqrt(2)*sqrt(1/2)*sqrt(-a^5/d^
2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/d) - 315*sqrt(1/2)*sqrt(-a^
5/d^2)*d*e^(2*I*d*x + 2*I*c)*log(-9/16*(-I*a^3 - sqrt(2)*sqrt(1/2)*sqrt(-a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*
sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/d) - 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^(-2*I*d*x - 2*I*c)/d

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cos \left (d x + c\right )^{7}\,{d x} \]

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)

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maple [B]  time = 1.85, size = 1260, normalized size = 5.45 \[ \text {result too large to display} \]

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*(315*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(13/2)*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))*sin(d*x+c)*cos(d*x+c)^6*2^(1/2)+1890*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(13/2)*arctan
h(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))*sin(d*x+c)*cos(d*x+c)^5*2^(1/2)-1638
4*sin(d*x+c)*cos(d*x+c)^11-21504*sin(d*x+c)*cos(d*x+c)^9+26880*sin(d*x+c)*cos(d*x+c)^8+18432*sin(d*x+c)*cos(d*
x+c)^10-40320*sin(d*x+c)*cos(d*x+c)^7-81920*sin(d*x+c)*cos(d*x+c)^13+40960*sin(d*x+c)*cos(d*x+c)^12+81920*I*co
s(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*cos(d
*x+c)^8-40320*I*cos(d*x+c)^7-315*2^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*(-2*cos(d*x+
c)/(1+cos(d*x+c)))^(13/2)*sin(d*x+c)+3072*I*cos(d*x+c)^10+1890*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(13/2)*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))*sin(d*x+c)*cos(d*x+c)*2^(1/2)+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)))^(13/2)*sin(d*x+c)-315*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(13/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^
(1/2)*2^(1/2))*sin(d*x+c)*cos(d*x+c)^6*2^(1/2)-1890*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(13/2)*arctan(1/2*(-2*cos(d
*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*sin(d*x+c)*cos(d*x+c)^5*2^(1/2)-4725*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(13/2
)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*sin(d*x+c)*cos(d*x+c)^4*2^(1/2)-6300*(-2*cos(d*x+c)
/(1+cos(d*x+c)))^(13/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*sin(d*x+c)*cos(d*x+c)^3*2^(1/
2)-4725*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(13/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*sin(d*x
+c)*cos(d*x+c)^2*2^(1/2)-1890*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(13/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^
(1/2)*2^(1/2))*sin(d*x+c)*cos(d*x+c)*2^(1/2)+4725*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(13/2)*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))*sin(d*x+c)*cos(d*x+c)^4*2^(1/2)+6300*I*(-2*cos(d*x
+c)/(1+cos(d*x+c)))^(13/2)*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))*sin
(d*x+c)*cos(d*x+c)^3*2^(1/2)+4725*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(13/2)*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))*sin(d*x+c)*cos(d*x+c)^2*2^(1/2))*(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*a^2

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

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int {\cos \left (c+d\,x\right )}^7\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^7*(a + a*tan(c + d*x)*1i)^(5/2),x)

[Out]

int(cos(c + d*x)^7*(a + a*tan(c + d*x)*1i)^(5/2), 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)**7*(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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