3.305 \(\int \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\)

Optimal. Leaf size=192 \[ \frac{7 i a^2 \cos (c+d x)}{24 d \sqrt{a+i a \tan (c+d x)}}+\frac{7 i a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{16 \sqrt{2} d}-\frac{i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}-\frac{7 i a \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{30 d}-\frac{7 i a \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{16 d} \]

[Out]

(((7*I)/16)*a^(3/2)*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*d) + (((7*I
)/24)*a^2*Cos[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((7*I)/16)*a*Cos[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]]
)/d - (((7*I)/30)*a*Cos[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d - ((I/5)*Cos[c + d*x]^5*(a + I*a*Tan[c + d*x]
)^(3/2))/d

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Rubi [A]  time = 0.270018, antiderivative size = 192, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ \frac{7 i a^2 \cos (c+d x)}{24 d \sqrt{a+i a \tan (c+d x)}}+\frac{7 i a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{16 \sqrt{2} d}-\frac{i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}-\frac{7 i a \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{30 d}-\frac{7 i a \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{16 d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

(((7*I)/16)*a^(3/2)*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*d) + (((7*I
)/24)*a^2*Cos[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((7*I)/16)*a*Cos[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]]
)/d - (((7*I)/30)*a*Cos[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d - ((I/5)*Cos[c + d*x]^5*(a + I*a*Tan[c + d*x]
)^(3/2))/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 ^5(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx &=-\frac{i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}+\frac{1}{10} (7 a) \int \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)} \, dx\\ &=-\frac{7 i a \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{30 d}-\frac{i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}+\frac{1}{12} \left (7 a^2\right ) \int \frac{\cos (c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx\\ &=\frac{7 i a^2 \cos (c+d x)}{24 d \sqrt{a+i a \tan (c+d x)}}-\frac{7 i a \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{30 d}-\frac{i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}+\frac{1}{16} (7 a) \int \cos (c+d x) \sqrt{a+i a \tan (c+d x)} \, dx\\ &=\frac{7 i a^2 \cos (c+d x)}{24 d \sqrt{a+i a \tan (c+d x)}}-\frac{7 i a \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{16 d}-\frac{7 i a \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{30 d}-\frac{i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}+\frac{1}{32} \left (7 a^2\right ) \int \frac{\sec (c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx\\ &=\frac{7 i a^2 \cos (c+d x)}{24 d \sqrt{a+i a \tan (c+d x)}}-\frac{7 i a \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{16 d}-\frac{7 i a \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{30 d}-\frac{i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}+\frac{\left (7 i a^2\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 )}{16 d}\\ &=\frac{7 i a^{3/2} \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{16 \sqrt{2} d}+\frac{7 i a^2 \cos (c+d x)}{24 d \sqrt{a+i a \tan (c+d x)}}-\frac{7 i a \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{16 d}-\frac{7 i a \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{30 d}-\frac{i \cos ^5(c+d x) (a+i a \tan (c+d x))^{3/2}}{5 d}\\ \end{align*}

Mathematica [A]  time = 1.37671, size = 160, normalized size = 0.83 \[ -\frac{i a e^{-3 i (c+d x)} \sqrt{\frac{a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \left (101 e^{2 i (c+d x)}+148 e^{4 i (c+d x)}+38 e^{6 i (c+d x)}+6 e^{8 i (c+d x)}-105 e^{2 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )-15\right )}{240 \sqrt{2} d} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^5*(a + I*a*Tan[c + d*x])^(3/2),x]

[Out]

((-I/240)*a*Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(-15 + 101*E^((2*I)*(c + d*x)) + 148*E^((4
*I)*(c + d*x)) + 38*E^((6*I)*(c + d*x)) + 6*E^((8*I)*(c + d*x)) - 105*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c
+ d*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]]))/(Sqrt[2]*d*E^((3*I)*(c + d*x)))

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Maple [B]  time = 0.313, size = 914, normalized size = 4.8 \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)^5*(a+I*a*tan(d*x+c))^(3/2),x)

[Out]

-1/7680/d*a*(256*I*cos(d*x+c)^8+1120*I*cos(d*x+c)^6-105*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))^(9/2)*2^(1/2)*cos(d*x+c)^4*sin(d*x+c)+3072*I*cos(d*x+c)^10-420*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))^(9/2)*2^(1/2)*cos(d*x+c)^3*sin(d*x+
c)+105*I*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2)*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))*sin(d*x+c)-630*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))^(9/2)*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)+420*I*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+
1))^(9/2)*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^(1/2)-420*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))^(9/2)*2^(1/2)*cos(d*x+c)*sin(d
*x+c)-105*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))^(9/2
)*sin(d*x+c)+105*I*cos(d*x+c)^4*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2)*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^(1/2)+448*I*cos(d*x+c)^7-3072*sin(d*x+c)*cos(d*x+c)^9-1536
*I*cos(d*x+c)^9+1536*sin(d*x+c)*cos(d*x+c)^8+630*I*cos(d*x+c)^2*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(9/2
)*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^(1/2)-1792*sin(d*x+c)*cos(
d*x+c)^7-3360*I*cos(d*x+c)^5+2240*cos(d*x+c)^6*sin(d*x+c)+420*I*cos(d*x+c)^3*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*
x+c)+1))^(9/2)*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^(1/2)-3360*co
s(d*x+c)^5*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)^4

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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)^5*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [B]  time = 2.39696, size = 892, normalized size = 4.65 \begin{align*} -\frac{{\left (105 \, \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 (14 i \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )} + 7 \, \sqrt{2}{\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\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 )}}{7 \, a}\right ) - 105 \, \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 (-14 i \, \sqrt{\frac{1}{2}} \sqrt{-\frac{a^{3}}{d^{2}}} d e^{\left (i \, d x + i \, c\right )} + 7 \, \sqrt{2}{\left (a e^{\left (2 i \, d x + 2 i \, c\right )} + a\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 )}}{7 \, a}\right ) - \sqrt{2}{\left (-6 i \, a e^{\left (8 i \, d x + 8 i \, c\right )} - 38 i \, a e^{\left (6 i \, d x + 6 i \, c\right )} - 148 i \, a e^{\left (4 i \, d x + 4 i \, c\right )} - 101 i \, a e^{\left (2 i \, d x + 2 i \, c\right )} + 15 i \, a\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 )}}{480 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

-1/480*(105*sqrt(1/2)*sqrt(-a^3/d^2)*d*e^(3*I*d*x + 3*I*c)*log(1/7*(14*I*sqrt(1/2)*sqrt(-a^3/d^2)*d*e^(I*d*x +
I*c) + 7*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-I*d*x - I
*c)/a) - 105*sqrt(1/2)*sqrt(-a^3/d^2)*d*e^(3*I*d*x + 3*I*c)*log(1/7*(-14*I*sqrt(1/2)*sqrt(-a^3/d^2)*d*e^(I*d*x
+ I*c) + 7*sqrt(2)*(a*e^(2*I*d*x + 2*I*c) + a)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-I*d*x -
I*c)/a) - sqrt(2)*(-6*I*a*e^(8*I*d*x + 8*I*c) - 38*I*a*e^(6*I*d*x + 6*I*c) - 148*I*a*e^(4*I*d*x + 4*I*c) - 10
1*I*a*e^(2*I*d*x + 2*I*c) + 15*I*a)*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)**5*(a+I*a*tan(d*x+c))**(3/2),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{3}{2}} \cos \left (d x + c\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^5*(a+I*a*tan(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((I*a*tan(d*x + c) + a)^(3/2)*cos(d*x + c)^5, x)