\(\int \frac {\cos (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx\) [345]

Optimal result

Integrand size = 24, antiderivative size = 122 \[ \int \frac {\cos (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {3 i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{4 \sqrt {2} \sqrt {a} d}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {3 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a d} \]

[Out]

3/8*I*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)/a^(1/2)+1/2*I*cos(d*x+c)/d/(a
+I*a*tan(d*x+c))^(1/2)-3/4*I*cos(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/a/d

Rubi [A] (verified)

Time = 0.18 (sec) , antiderivative size = 122, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3583, 3571, 3570, 212} \[ \int \frac {\cos (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {3 i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{4 \sqrt {2} \sqrt {a} d}-\frac {3 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a d}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}} \]

[In]

Int[Cos[c + d*x]/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(((3*I)/4)*ArcTanh[(Sqrt[a]*Sec[c + d*x])/(Sqrt[2]*Sqrt[a + I*a*Tan[c + d*x]])])/(Sqrt[2]*Sqrt[a]*d) + ((I/2)*
Cos[c + d*x])/(d*Sqrt[a + I*a*Tan[c + d*x]]) - (((3*I)/4)*Cos[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]])/(a*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 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*} \text {integral}& = \frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}+\frac {3 \int \cos (c+d x) \sqrt {a+i a \tan (c+d x)} \, dx}{4 a} \\ & = \frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {3 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a d}+\frac {3}{8} \int \frac {\sec (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx \\ & = \frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {3 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a d}+\frac {(3 i) \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 )}{4 d} \\ & = \frac {3 i \text {arctanh}\left (\frac {\sqrt {a} \sec (c+d x)}{\sqrt {2} \sqrt {a+i a \tan (c+d x)}}\right )}{4 \sqrt {2} \sqrt {a} d}+\frac {i \cos (c+d x)}{2 d \sqrt {a+i a \tan (c+d x)}}-\frac {3 i \cos (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 a d} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.60 (sec) , antiderivative size = 96, normalized size of antiderivative = 0.79 \[ \int \frac {\cos (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {\sec (c+d x) \left (3 i \sqrt {1+e^{2 i (c+d x)}} \text {arctanh}\left (\sqrt {1+e^{2 i (c+d x)}}\right )-i (1+\cos (2 (c+d x))+3 i \sin (2 (c+d x)))\right )}{8 d \sqrt {a+i a \tan (c+d x)}} \]

[In]

Integrate[Cos[c + d*x]/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(Sec[c + d*x]*((3*I)*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcTanh[Sqrt[1 + E^((2*I)*(c + d*x))]] - I*(1 + Cos[2*(c +
d*x)] + (3*I)*Sin[2*(c + d*x)])))/(8*d*Sqrt[a + I*a*Tan[c + d*x]])

Maple [B] (verified)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 349 vs. \(2 (97 ) = 194\).

Time = 9.74 (sec) , antiderivative size = 350, normalized size of antiderivative = 2.87

method	result	size
default	\(-\frac {2 i \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-3 i \cos \left (d x +c \right ) \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+2 i \cos \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-6 \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )-3 i \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right )+3 \arctan \left (\frac {i \sin \left (d x +c \right )-\cos \left (d x +c \right )-1}{2 \left (\cos \left (d x +c \right )+1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}\right ) \sin \left (d x +c \right )-6 \sin \left (d x +c \right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}}{8 d \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos \left (d x +c \right )+1\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}}\)	\(350\)

[In]

int(cos(d*x+c)/(a+I*a*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/8/d*(2*I*cos(d*x+c)^2*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-3*I*cos(d*x+c)*arctan(1/2*(I*sin(d*x+c)-cos(d*x+c)
-1)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+2*I*cos(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-6*(-c
os(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)*sin(d*x+c)-3*I*arctan(1/2*(I*sin(d*x+c)-cos(d*x+c)-1)/(cos(d*x+c)+1
)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+3*arctan(1/2*(I*sin(d*x+c)-cos(d*x+c)-1)/(cos(d*x+c)+1)/(-cos(d*x+c)/(co
s(d*x+c)+1))^(1/2))*sin(d*x+c)-6*sin(d*x+c)*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))/(-cos(d*x+c)/(cos(d*x+c)+1))^(
1/2)/(cos(d*x+c)+1)/(a*(1+I*tan(d*x+c)))^(1/2)

Fricas [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 245 vs. \(2 (91) = 182\).

Time = 0.25 (sec) , antiderivative size = 245, normalized size of antiderivative = 2.01 \[ \int \frac {\cos (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\frac {{\left (-3 i \, \sqrt {\frac {1}{2}} a d \sqrt {\frac {1}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\frac {3 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, d}\right ) + 3 i \, \sqrt {\frac {1}{2}} a d \sqrt {\frac {1}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (-\frac {3 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {1}{a d^{2}}} - i\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, d}\right ) + \sqrt {2} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} {\left (-2 i \, e^{\left (4 i \, d x + 4 i \, c\right )} - i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i\right )}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{8 \, a d} \]

[In]

integrate(cos(d*x+c)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/8*(-3*I*sqrt(1/2)*a*d*sqrt(1/(a*d^2))*e^(2*I*d*x + 2*I*c)*log(-3/2*(sqrt(2)*sqrt(1/2)*(I*d*e^(2*I*d*x + 2*I*
c) + I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt(1/(a*d^2)) - I)*e^(-I*d*x - I*c)/d) + 3*I*sqrt(1/2)*a*d*sqrt(
1/(a*d^2))*e^(2*I*d*x + 2*I*c)*log(-3/2*(sqrt(2)*sqrt(1/2)*(-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(a/(e^(2*I*d*x
+ 2*I*c) + 1))*sqrt(1/(a*d^2)) - I)*e^(-I*d*x - I*c)/d) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-2*I*e^(
4*I*d*x + 4*I*c) - I*e^(2*I*d*x + 2*I*c) + I))*e^(-2*I*d*x - 2*I*c)/(a*d)

Sympy [F]

\[ \int \frac {\cos (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {\cos {\left (c + d x \right )}}{\sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )}}\, dx \]

[In]

integrate(cos(d*x+c)/(a+I*a*tan(d*x+c))**(1/2),x)

[Out]

Integral(cos(c + d*x)/sqrt(I*a*(tan(c + d*x) - I)), x)

Maxima [B] (verification not implemented)

Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 837 vs. \(2 (91) = 182\).

Time = 0.44 (sec) , antiderivative size = 837, normalized size of antiderivative = 6.86 \[ \int \frac {\cos (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\text {Too large to display} \]

[In]

integrate(cos(d*x+c)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

-1/32*(4*(cos(2*d*x + 2*c)^2 + sin(2*d*x + 2*c)^2 + 2*cos(2*d*x + 2*c) + 1)^(1/4)*((-I*sqrt(2)*cos(2*d*x + 2*c
) - sqrt(2)*sin(2*d*x + 2*c) + 2*I*sqrt(2))*cos(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c) + 1)) + (sqrt(2
)*cos(2*d*x + 2*c) - I*sqrt(2)*sin(2*d*x + 2*c) - 2*sqrt(2))*sin(1/2*arctan2(sin(2*d*x + 2*c), cos(2*d*x + 2*c
) + 1)))*sqrt(a) + 3*(2*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) - 2*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) - 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) + 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))*sqr
t(a))/(a*d)

Giac [F]

\[ \int \frac {\cos (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int { \frac {\cos \left (d x + c\right )}{\sqrt {i \, a \tan \left (d x + c\right ) + a}} \,d x } \]

[In]

integrate(cos(d*x+c)/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

integrate(cos(d*x + c)/sqrt(I*a*tan(d*x + c) + a), x)

Mupad [F(-1)]

Timed out. \[ \int \frac {\cos (c+d x)}{\sqrt {a+i a \tan (c+d x)}} \, dx=\int \frac {\cos \left (c+d\,x\right )}{\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}} \,d x \]

[In]

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

[Out]

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