3.4.18 \(\int \cos ^7(c+d x) (a+i a \tan (c+d x))^{5/2} \, dx\) [318]

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

[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 = 0.25, 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 = {3578, 3583, 3571, 3570, 212} \begin {gather*} \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} \end {gather*}

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 212

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 3570

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 3571

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 3578

Int[((d_.)*sec[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(d*S
ec[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 3583

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 ) \text {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 = 1.28, size = 155, normalized size = 0.67 \begin {gather*} -\frac {i a^2 e^{-3 i (c+d x)} \left (-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)}} \tanh ^{-1}\left (\sqrt {1+e^{2 i (c+d x)}}\right )\right ) \sqrt {a+i a \tan (c+d x)}}{2240 d} \end {gather*}

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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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1259 vs. \(2 (188 ) = 376\).
time = 1.17, size = 1260, normalized size = 5.45

method	result	size
default	\(\text {Expression too large to display}\)	\(1260\)

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,method=_RETURNVERBOSE)

[Out]

-1/143360/d*(-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)+4725*I*cos(d*x+c)^4*sin(d*x+c)*(-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))*2^(1/2)+6300*I*cos(d*x+c)^3*sin(d*x+c)*(-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))*2^
(1/2)+4725*I*cos(d*x+c)^2*sin(d*x+c)*(-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))*2^(1/2)-21504*cos(d*x+c)^9*sin(d*x+c)+40960*sin(d*x+c)*cos(d*x+c)
^12+3072*I*cos(d*x+c)^10+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)^1
1+5376*I*cos(d*x+c)^9+13440*I*cos(d*x+c)^8-40320*I*cos(d*x+c)^7-81920*cos(d*x+c)^13*sin(d*x+c)-16384*sin(d*x+c
)*cos(d*x+c)^11+315*I*cos(d*x+c)^6*sin(d*x+c)*(-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))*2^(1/2)+1890*I*cos(d*x+c)^5*sin(d*x+c)*(-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))*2^(1/2)+2
6880*sin(d*x+c)*cos(d*x+c)^8+18432*cos(d*x+c)^10*sin(d*x+c)-40320*sin(d*x+c)*cos(d*x+c)^7-315*cos(d*x+c)^6*sin
(d*x+c)*(-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))*2^(1/2)
-1890*cos(d*x+c)^5*sin(d*x+c)*(-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))*2^(1/2)-4725*cos(d*x+c)^4*sin(d*x+c)*(-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))*2^(1/2)-6300*cos(d*x+c)^3*sin(d*x+c)*(-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))*2^(1/2)-4725*cos(d*x+c)^2*sin(d*x+c)*(-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))*2^(1/2)-1890*cos(d*x+c)*sin(d
*x+c)*(-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))*2^(1/2)+3
15*I*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(13/2)*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))*sin(d*x+c)+1890*I*cos(d*x+c)*sin(d*x+c)*(-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))*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)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.39, size = 300, normalized size = 1.30 \begin {gather*} -\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} \end {gather*}

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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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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 = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

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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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \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 \end {gather*}

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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