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

Optimal. Leaf size=93 \[ -\frac {i a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} d}-\frac {i a^2 \sqrt {a+i a \tan (c+d x)}}{2 d (a-i a \tan (c+d x))} \]

[Out]

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

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Rubi [A]  time = 0.08, antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3487, 51, 63, 206} \[ -\frac {i a^2 \sqrt {a+i a \tan (c+d x)}}{2 d (a-i a \tan (c+d x))}-\frac {i a^{3/2} \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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])

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]

Rubi steps

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

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Mathematica [A]  time = 0.64, size = 97, normalized size = 1.04 \[ -\frac {i a e^{-i (c+d x)} \sqrt {1+e^{2 i (c+d x)}} \left (e^{i (c+d x)} \sqrt {1+e^{2 i (c+d x)}}+\sinh ^{-1}\left (e^{i (c+d x)}\right )\right ) \sqrt {a+i a \tan (c+d x)}}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [B]  time = 0.46, size = 241, normalized size = 2.59 \[ \frac {\sqrt {\frac {1}{2}} \sqrt {-\frac {a^{3}}{d^{2}}} d \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (4 i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + 4 i \, d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + 4 \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) - \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{3}}{d^{2}}} d \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-4 i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - 4 i \, d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} + 4 \, a^{2} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) + \sqrt {2} {\left (-i \, a e^{\left (3 i \, d x + 3 i \, c\right )} - i \, a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} \cos \left (d x + c\right )^{2}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(d*x+c)^2*(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)^2, x)

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maple [B]  time = 1.27, size = 398, normalized size = 4.28 \[ -\frac {\left (-i \sin \left (d x +c \right ) \cos \left (d x +c \right ) \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sqrt {2}-\sin \left (d x +c \right ) \cos \left (d x +c \right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sqrt {2}-i \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sqrt {2}\, \sin \left (d x +c \right )-\sqrt {2}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \sin \left (d x +c \right )+8 i \left (\cos ^{4}\left (d x +c \right )\right )-8 \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )-4 i \left (\cos ^{3}\left (d x +c \right )\right )+4 \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )-4 i \left (\cos ^{2}\left (d x +c \right )\right )\right ) \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, a}{8 d \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )-1\right ) \cos \left (d x +c \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.80, size = 98, normalized size = 1.05 \[ \frac {i \, {\left (\sqrt {2} a^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {16 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{3}}{4 i \, a \tan \left (d x + c\right ) - 4 \, a}\right )}}{8 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/8*I*(sqrt(2)*a^(5/2)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(2)*sqrt(a) + sqrt(I*a*tan(d*x
+ c) + a))) + 16*sqrt(I*a*tan(d*x + c) + a)*a^3/(4*I*a*tan(d*x + c) - 4*a))/(a*d)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\cos \left (c+d\,x\right )}^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(c + d*x)^2*(a + a*tan(c + d*x)*1i)^(3/2),x)

[Out]

int(cos(c + d*x)^2*(a + a*tan(c + d*x)*1i)^(3/2), x)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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