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

Optimal. Leaf size=120 \[ -\frac{i a^2}{2 d (a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}+\frac{3 i a}{4 d \sqrt{a+i a \tan (c+d x)}}-\frac{3 i \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} d} \]

[Out]

(((-3*I)/4)*Sqrt[a]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*d) + (((3*I)/4)*a)/(d*Sqrt
[a + I*a*Tan[c + d*x]]) - ((I/2)*a^2)/(d*(a - I*a*Tan[c + d*x])*Sqrt[a + I*a*Tan[c + d*x]])

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Rubi [A]  time = 0.0981145, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.154, Rules used = {3487, 51, 63, 206} \[ -\frac{i a^2}{2 d (a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}+\frac{3 i a}{4 d \sqrt{a+i a \tan (c+d x)}}-\frac{3 i \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(((-3*I)/4)*Sqrt[a]*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/(Sqrt[2]*d) + (((3*I)/4)*a)/(d*Sqrt
[a + I*a*Tan[c + d*x]]) - ((I/2)*a^2)/(d*(a - I*a*Tan[c + d*x])*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 \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)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{d}\\ &=-\frac{i a^2}{2 d (a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}-\frac{\left (3 i a^2\right ) \operatorname{Subst}\left (\int \frac{1}{(a-x) (a+x)^{3/2}} \, dx,x,i a \tan (c+d x)\right )}{4 d}\\ &=\frac{3 i a}{4 d \sqrt{a+i a \tan (c+d x)}}-\frac{i a^2}{2 d (a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}-\frac{(3 i a) \operatorname{Subst}\left (\int \frac{1}{(a-x) \sqrt{a+x}} \, dx,x,i a \tan (c+d x)\right )}{8 d}\\ &=\frac{3 i a}{4 d \sqrt{a+i a \tan (c+d x)}}-\frac{i a^2}{2 d (a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}-\frac{(3 i a) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{4 d}\\ &=-\frac{3 i \sqrt{a} \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} d}+\frac{3 i a}{4 d \sqrt{a+i a \tan (c+d x)}}-\frac{i a^2}{2 d (a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}

Mathematica [A]  time = 0.43948, size = 105, normalized size = 0.88 \[ -\frac{i e^{-2 i (c+d x)} \left (-e^{2 i (c+d x)}+e^{4 i (c+d x)}+3 e^{i (c+d x)} \sqrt{1+e^{2 i (c+d x)}} \sinh ^{-1}\left (e^{i (c+d x)}\right )-2\right ) \sqrt{a+i a \tan (c+d x)}}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-I/8)*(-2 - E^((2*I)*(c + d*x)) + E^((4*I)*(c + d*x)) + 3*E^(I*(c + d*x))*Sqrt[1 + E^((2*I)*(c + d*x))]*ArcS
inh[E^(I*(c + d*x))])*Sqrt[a + I*a*Tan[c + d*x]])/(d*E^((2*I)*(c + d*x)))

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Maple [B]  time = 0.39, size = 397, normalized size = 3.3 \begin{align*}{\frac{1}{16\,d \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) -1 \right ) \cos \left ( dx+c \right ) } \left ( 3\,i\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{{\frac{3}{2}}}{\it Artanh} \left ({\frac{\sqrt{2}\sin \left ( dx+c \right ) }{2\,\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \sqrt{2}+3\,i\sqrt{2}{\it Artanh} \left ({\frac{\sqrt{2}\sin \left ( dx+c \right ) }{2\,\cos \left ( dx+c \right ) }\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{{\frac{3}{2}}}\sin \left ( dx+c \right ) +3\,\cos \left ( dx+c \right ) \sin \left ( dx+c \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\arctan \left ( 1/2\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \right ) \sqrt{2}+3\,\sqrt{2}\arctan \left ( 1/2\,\sqrt{2}\sqrt{-2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}}} \right ) \left ( -2\,{\frac{\cos \left ( dx+c \right ) }{\cos \left ( dx+c \right ) +1}} \right ) ^{3/2}\sin \left ( dx+c \right ) -8\,i \left ( \cos \left ( dx+c \right ) \right ) ^{4}-4\,i \left ( \cos \left ( dx+c \right ) \right ) ^{3}+8\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}\sin \left ( dx+c \right ) +12\,i \left ( \cos \left ( dx+c \right ) \right ) ^{2}-12\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) \right ) \sqrt{{\frac{a \left ( i\sin \left ( dx+c \right ) +\cos \left ( dx+c \right ) \right ) }{\cos \left ( dx+c \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]

1/16/d*(3*I*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*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)+3*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))^(3/2)*sin(d*x+c)+3*cos(d*x+c)*sin(d*x+c)*(-2*cos
(d*x+c)/(cos(d*x+c)+1))^(3/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)+3*2^(1/2)*arcta
n(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))^(3/2)*sin(d*x+c)-8*I*cos(d*
x+c)^4-4*I*cos(d*x+c)^3+8*cos(d*x+c)^3*sin(d*x+c)+12*I*cos(d*x+c)^2-12*cos(d*x+c)^2*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)

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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 = 2.69812, size = 772, normalized size = 6.43 \begin{align*} \frac{{\left (3 \, \sqrt{\frac{1}{2}} d \sqrt{-\frac{a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac{1}{3} \,{\left (6 i \, \sqrt{\frac{1}{2}} d \sqrt{-\frac{a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \, \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 ) - 3 \, \sqrt{\frac{1}{2}} d \sqrt{-\frac{a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} \log \left (\frac{1}{3} \,{\left (-6 i \, \sqrt{\frac{1}{2}} d \sqrt{-\frac{a}{d^{2}}} e^{\left (2 i \, d x + 2 i \, c\right )} + 3 \, \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 (-i \, e^{\left (4 i \, d x + 4 i \, c\right )} + 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 (-2 i \, d x - 2 i \, c\right )}}{8 \, 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]

1/8*(3*sqrt(1/2)*d*sqrt(-a/d^2)*e^(2*I*d*x + 2*I*c)*log(1/3*(6*I*sqrt(1/2)*d*sqrt(-a/d^2)*e^(2*I*d*x + 2*I*c)
+ 3*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)) - 3
*sqrt(1/2)*d*sqrt(-a/d^2)*e^(2*I*d*x + 2*I*c)*log(1/3*(-6*I*sqrt(1/2)*d*sqrt(-a/d^2)*e^(2*I*d*x + 2*I*c) + 3*s
qrt(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))*(-I*e^(4*I*d*x + 4*I*c) + I*e^(2*I*d*x + 2*I*c) + 2*I)*e^(I*d*x + I*c))*e^
(-2*I*d*x - 2*I*c)/d

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

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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 )^{2}\,{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(sqrt(I*a*tan(d*x + c) + a)*cos(d*x + c)^2, x)