3.337 \(\int \frac{\cos ^2(c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx\)

Optimal. Leaf size=146 \[ -\frac{i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac{5 i a}{12 d (a+i a \tan (c+d x))^{3/2}}+\frac{5 i}{8 d \sqrt{a+i a \tan (c+d x)}}-\frac{5 i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{8 \sqrt{2} \sqrt{a} d} \]

[Out]

(((-5*I)/8)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*Sqrt[a]*d) + (((5*I)/12)*a)/(d*(a
+ I*a*Tan[c + d*x])^(3/2)) - ((I/2)*a^2)/(d*(a - I*a*Tan[c + d*x])*(a + I*a*Tan[c + d*x])^(3/2)) + ((5*I)/8)/(
d*Sqrt[a + I*a*Tan[c + d*x]])

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Rubi [A]  time = 0.0981909, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3487, 51, 63, 206} \[ -\frac{i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac{5 i a}{12 d (a+i a \tan (c+d x))^{3/2}}+\frac{5 i}{8 d \sqrt{a+i a \tan (c+d x)}}-\frac{5 i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{8 \sqrt{2} \sqrt{a} d} \]

Antiderivative was successfully verified.

[In]

Int[Cos[c + d*x]^2/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

(((-5*I)/8)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*Sqrt[a]*d) + (((5*I)/12)*a)/(d*(a
+ I*a*Tan[c + d*x])^(3/2)) - ((I/2)*a^2)/(d*(a - I*a*Tan[c + d*x])*(a + I*a*Tan[c + d*x])^(3/2)) + ((5*I)/8)/(
d*Sqrt[a + I*a*Tan[c + d*x]])

Rule 3487

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[1/(a^(m - 2)*b
*f), Subst[Int[(a - x)^(m/2 - 1)*(a + x)^(n + m/2 - 1), x], x, b*Tan[e + f*x]], x] /; FreeQ[{a, b, e, f, n}, x
] && EqQ[a^2 + b^2, 0] && IntegerQ[m/2]

Rule 51

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n + 1
))/((b*c - a*d)*(m + 1)), x] - Dist[(d*(m + n + 2))/((b*c - a*d)*(m + 1)), Int[(a + b*x)^(m + 1)*(c + d*x)^n,
x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] &&  !(LtQ[n, -1] && (EqQ[a, 0] || (NeQ[
c, 0] && LtQ[m - n, 0] && IntegerQ[n]))) && IntLinearQ[a, b, c, d, m, n, x]

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

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 \frac{\cos ^2(c+d x)}{\sqrt{a+i a \tan (c+d x)}} \, dx &=-\frac{\left (i a^3\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x)^2 (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac{i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}-\frac{\left (5 i a^2\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{5/2}} \, dx,x,i a \tan (c+d x)\right )}{4 d}\\ &=\frac{5 i a}{12 d (a+i a \tan (c+d x))^{3/2}}-\frac{i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}-\frac{(5 i a) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{8 d}\\ &=\frac{5 i a}{12 d (a+i a \tan (c+d x))^{3/2}}-\frac{i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac{5 i}{8 d \sqrt{a+i a \tan (c+d x)}}-\frac{(5 i) \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt{a+x}} \, dx,x,i a \tan (c+d x)\right )}{16 d}\\ &=\frac{5 i a}{12 d (a+i a \tan (c+d x))^{3/2}}-\frac{i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac{5 i}{8 d \sqrt{a+i a \tan (c+d x)}}-\frac{(5 i) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{8 d}\\ &=-\frac{5 i \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{8 \sqrt{2} \sqrt{a} d}+\frac{5 i a}{12 d (a+i a \tan (c+d x))^{3/2}}-\frac{i a^2}{2 d (a-i a \tan (c+d x)) (a+i a \tan (c+d x))^{3/2}}+\frac{5 i}{8 d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.4769, size = 126, normalized size = 0.86 \[ -\frac{i e^{-2 i (c+d x)} \left (\sqrt{1+e^{2 i (c+d x)}} \left (-14 e^{2 i (c+d x)}+3 e^{4 i (c+d x)}-2\right )+15 e^{3 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{24 d \sqrt{1+e^{2 i (c+d x)}} \sqrt{a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[c + d*x]^2/Sqrt[a + I*a*Tan[c + d*x]],x]

[Out]

((-I/24)*(Sqrt[1 + E^((2*I)*(c + d*x))]*(-2 - 14*E^((2*I)*(c + d*x)) + 3*E^((4*I)*(c + d*x))) + 15*E^((3*I)*(c
+ d*x))*ArcSinh[E^(I*(c + d*x))]))/(d*E^((2*I)*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*Sqrt[a + I*a*Tan[c +
d*x]])

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Maple [B]  time = 0.334, size = 341, normalized size = 2.3 \begin{align*}{\frac{1}{96\,ad}\sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \right ) }}} \left ( 15\,i\sqrt{2}\arctan \left ({\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i-\sin \left ( dx+c \right ) \right ) }{2\,\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\cos \left ( dx+c \right ) +32\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}+15\,i\sqrt{2}\arctan \left ({\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i-\sin \left ( dx+c \right ) \right ) }{2\,\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}+15\,\sqrt{2}\arctan \left ( 1/2\,{\frac{\sqrt{2} \left ( i\cos \left ( dx+c \right ) -i-\sin \left ( dx+c \right ) \right ) }{\sin \left ( dx+c \right ) }{\frac{1}{\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}}}} \right ) \sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}\sin \left ( dx+c \right ) +32\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +20\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}+60\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96/d/a*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(15*I*2^(1/2)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d
*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+32*I*c
os(d*x+c)^4+15*I*2^(1/2)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+
1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)+15*2^(1/2)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(
d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+32*cos(d*x+c)^3*s
in(d*x+c)+20*I*cos(d*x+c)^2+60*cos(d*x+c)*sin(d*x+c))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 2.46901, size = 837, normalized size = 5.73 \begin{align*} \frac{{\left (-15 i \, \sqrt{\frac{1}{2}} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left ({\left (2 \, \sqrt{\frac{1}{2}} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + \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 i \, \sqrt{\frac{1}{2}} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (4 i \, d x + 4 i \, c\right )} \log \left (-{\left (2 \, \sqrt{\frac{1}{2}} a d \sqrt{\frac{1}{a d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} - \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 (-3 i \, e^{\left (6 i \, d x + 6 i \, c\right )} + 11 i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 16 i \, e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i\right )} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-4 i \, d x - 4 i \, c\right )}}{48 \, a d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/48*(-15*I*sqrt(1/2)*a*d*sqrt(1/(a*d^2))*e^(4*I*d*x + 4*I*c)*log((2*sqrt(1/2)*a*d*sqrt(1/(a*d^2))*e^(2*I*d*x
+ 2*I*c) + 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)) + 15*I*sqrt(1/2)*a*d*sqrt(1/(a*d^2))*e^(4*I*d*x + 4*I*c)*log(-(2*sqrt(1/2)*a*d*sqrt(1/(a*d^2))*e^(2*I*d*x
+ 2*I*c) - 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)) + sqrt(2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*(-3*I*e^(6*I*d*x + 6*I*c) + 11*I*e^(4*I*d*x + 4*I*c) + 16*I*e^
(2*I*d*x + 2*I*c) + 2*I)*e^(I*d*x + I*c))*e^(-4*I*d*x - 4*I*c)/(a*d)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos ^{2}{\left (c + d x \right )}}{\sqrt{a \left (i \tan{\left (c + d x \right )} + 1\right )}}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\cos \left (d x + c\right )^{2}}{\sqrt{i \, a \tan \left (d x + c\right ) + a}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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