3.291 \(\int \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)} \, dx\)
Optimal. Leaf size=154 \[ -\frac{i \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}-\frac{5 i \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{8 d}+\frac{5 i a \cos (c+d x)}{12 d \sqrt{a+i a \tan (c+d x)}}+\frac{5 i \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{8 \sqrt{2} d} \]
[Out]
(((5*I)/8)*Sqrt[a]*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*d) + (((5*I)
/12)*a*Cos[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((5*I)/8)*Cos[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/d -
((I/3)*Cos[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/d
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Rubi [A] time = 0.223128, antiderivative size = 154, normalized size of antiderivative = 1.,
number of steps used = 5, 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} \[ -\frac{i \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}-\frac{5 i \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{8 d}+\frac{5 i a \cos (c+d x)}{12 d \sqrt{a+i a \tan (c+d x)}}+\frac{5 i \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{8 \sqrt{2} d} \]
Antiderivative was successfully verified.
[In]
Int[Cos[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]],x]
[Out]
(((5*I)/8)*Sqrt[a]*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*d) + (((5*I)
/12)*a*Cos[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((5*I)/8)*Cos[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/d -
((I/3)*Cos[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]])/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 ^3(c+d x) \sqrt{a+i a \tan (c+d x)} \, dx &=-\frac{i \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}+\frac{1}{6} (5 a) \int \frac{\cos (c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx\\ &=\frac{5 i a \cos (c+d x)}{12 d \sqrt{a+i a \tan (c+d x)}}-\frac{i \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}+\frac{5}{8} \int \cos (c+d x) \sqrt{a+i a \tan (c+d x)} \, dx\\ &=\frac{5 i a \cos (c+d x)}{12 d \sqrt{a+i a \tan (c+d x)}}-\frac{5 i \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{8 d}-\frac{i \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}+\frac{1}{16} (5 a) \int \frac{\sec (c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx\\ &=\frac{5 i a \cos (c+d x)}{12 d \sqrt{a+i a \tan (c+d x)}}-\frac{5 i \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{8 d}-\frac{i \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}+\frac{(5 i a) \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 )}{8 d}\\ &=\frac{5 i \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a} \sec (c+d x)}{\sqrt{2} \sqrt{a+i a \tan (c+d x)}}\right )}{8 \sqrt{2} d}+\frac{5 i a \cos (c+d x)}{12 d \sqrt{a+i a \tan (c+d x)}}-\frac{5 i \cos (c+d x) \sqrt{a+i a \tan (c+d x)}}{8 d}-\frac{i \cos ^3(c+d x) \sqrt{a+i a \tan (c+d x)}}{3 d}\\ \end{align*}
Mathematica [A] time = 0.462658, size = 126, normalized size = 0.82 \[ -\frac{i e^{-3 i (c+d x)} \left (11 e^{2 i (c+d x)}+16 e^{4 i (c+d x)}+2 e^{6 i (c+d x)}-15 e^{2 i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\sqrt{1+e^{2 i (c+d x)}}\right )-3\right ) \sqrt{a+i a \tan (c+d x)}}{48 d} \]
Antiderivative was successfully verified.
[In]
Integrate[Cos[c + d*x]^3*Sqrt[a + I*a*Tan[c + d*x]],x]
[Out]
((-I/48)*(-3 + 11*E^((2*I)*(c + d*x)) + 16*E^((4*I)*(c + d*x)) + 2*E^((6*I)*(c + d*x)) - 15*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 = 0.387, size = 569, normalized size = 3.7 \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)^3*(a+I*a*tan(d*x+c))^(1/2),x)
[Out]
-1/192/d*(15*I*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*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))*cos(d*x+c)^2*sin(d*x+c)+30*I*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(5/2)*arctan
h(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))*cos(d*x+c)*sin(d*x+c)-15*2^(1/2)*arc
tan(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))^(5/2)*cos(d*x+c)^2*sin(d*
x+c)+15*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))^(5/2)*sin(d*x+c)-30*2^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*co
s(d*x+c)/(cos(d*x+c)+1))^(5/2)*cos(d*x+c)*sin(d*x+c)-15*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))^(5/2)*sin(d*x+c)+64*I*cos(d*x+c)^6+16*I*cos(d*x+c)^5-64*cos(d*x+c)^5
*sin(d*x+c)+40*I*cos(d*x+c)^4+80*sin(d*x+c)*cos(d*x+c)^4-120*I*cos(d*x+c)^3-120*cos(d*x+c)^3*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)^2
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Maxima [B] time = 2.2328, size = 1261, normalized size = 8.19 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
1/192*((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(3/4)*(-8*I*sqrt(2)*cos(3/2*arctan2(
sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 8*sqrt(2)*sin(3/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)))
*sqrt(a) + (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*((12*I*sqrt(2)*cos(2*d*x +
2*c) + 12*sqrt(2)*sin(2*d*x + 2*c) - 48*I*sqrt(2))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) -
12*(sqrt(2)*cos(2*d*x + 2*c) - I*sqrt(2)*sin(2*d*x + 2*c) - 4*sqrt(2))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(
2*d*x + 2*c) + 1)))*sqrt(a) - (30*sqrt(2)*arctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c
) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^
2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + 1) - 30*sqrt(2)*a
rctan2((cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*sin(1/2*arctan2(sin(2*d*x + 2*
c), cos(2*d*x + 2*c) + 1)), (cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*a
rctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) - 1) - 15*I*sqrt(2)*log(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x +
2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 + sqrt(cos(2*d*x +
2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)
)^2 + 2*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2
*c), cos(2*d*x + 2*c) + 1)) + 1) + 15*I*sqrt(2)*log(sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x
+ 2*c) + 1)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 + sqrt(cos(2*d*x + 2*c)^2 + sin(2*d*x
+ 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1))^2 - 2*(cos(2*d*x +
2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c
) + 1)) + 1))*sqrt(a))/d
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Fricas [B] time = 2.38646, size = 813, normalized size = 5.28 \begin{align*} -\frac{{\left (15 \, \sqrt{\frac{1}{2}} d \sqrt{-\frac{a}{d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac{1}{5} \,{\left (10 i \, \sqrt{\frac{1}{2}} d \sqrt{-\frac{a}{d^{2}}} e^{\left (i \, d x + i \, c\right )} + 5 \, \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - 15 \, \sqrt{\frac{1}{2}} d \sqrt{-\frac{a}{d^{2}}} e^{\left (3 i \, d x + 3 i \, c\right )} \log \left (\frac{1}{5} \,{\left (-10 i \, \sqrt{\frac{1}{2}} d \sqrt{-\frac{a}{d^{2}}} e^{\left (i \, d x + i \, c\right )} + 5 \, \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (e^{\left (2 i \, d x + 2 i \, c\right )} + 1\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}\right ) - \sqrt{2} \sqrt{\frac{a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}{\left (-2 i \, e^{\left (6 i \, d x + 6 i \, c\right )} - 16 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - 11 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 3 i\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-3 i \, d x - 3 i \, c\right )}}{48 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
-1/48*(15*sqrt(1/2)*d*sqrt(-a/d^2)*e^(3*I*d*x + 3*I*c)*log(1/5*(10*I*sqrt(1/2)*d*sqrt(-a/d^2)*e^(I*d*x + I*c)
+ 5*sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - 1
5*sqrt(1/2)*d*sqrt(-a/d^2)*e^(3*I*d*x + 3*I*c)*log(1/5*(-10*I*sqrt(1/2)*d*sqrt(-a/d^2)*e^(I*d*x + I*c) + 5*sqr
t(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(e^(2*I*d*x + 2*I*c) + 1)*e^(I*d*x + I*c))*e^(-I*d*x - I*c)) - sqrt(2)*
sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-2*I*e^(6*I*d*x + 6*I*c) - 16*I*e^(4*I*d*x + 4*I*c) - 11*I*e^(2*I*d*x + 2*I
*c) + 3*I)*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)**3*(a+I*a*tan(d*x+c))**(1/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{i \, a \tan \left (d x + c\right ) + a} \cos \left (d x + c\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cos(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate(sqrt(I*a*tan(d*x + c) + a)*cos(d*x + c)^3, x)