\(\int \cos ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx\) [297]

Optimal result

Integrand size = 26, antiderivative size = 93 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {i a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} d}-\frac {i a^2 \sqrt {a+i a \tan (c+d x)}}{2 d (a-i a \tan (c+d x))} \]

[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))

Rubi [A] (verified)

Time = 0.11 (sec) , antiderivative size = 93, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.154, Rules used = {3568, 44, 65, 212} \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {i a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{2 \sqrt {2} d}-\frac {i a^2 \sqrt {a+i a \tan (c+d x)}}{2 d (a-i a \tan (c+d x))} \]

[In]

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

[Out]

((-1/2*I)*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
*Tan[c + d*x]])/(d*(a - I*a*Tan[c + d*x]))

Rule 44

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] && ILtQ[m, -1] &&  !IntegerQ[n] && LtQ[n, 0]

Rule 65

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 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 3568

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*} \text {integral}& = -\frac {\left (i a^3\right ) \text {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 ) \text {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 ) \text {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} \text {arctanh}\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*}

Mathematica [A] (verified)

Time = 0.17 (sec) , antiderivative size = 92, normalized size of antiderivative = 0.99 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {\sqrt {2} a^{3/2} \text {arctanh}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right ) (1-i \tan (c+d x))+2 a \sqrt {a+i a \tan (c+d x)}}{4 d (i+\tan (c+d x))} \]

[In]

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

[Out]

(Sqrt[2]*a^(3/2)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])]*(1 - I*Tan[c + d*x]) + 2*a*Sqrt[a + I*a
*Tan[c + d*x]])/(4*d*(I + 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. 575 vs. \(2 (73 ) = 146\).

Time = 15.41 (sec) , antiderivative size = 576, normalized size of antiderivative = 6.19

method	result	size
default	\(-\frac {\left (-\tan \left (d x +c \right )+i\right ) \sqrt {a \left (1+i \tan \left (d x +c \right )\right )}\, a \cos \left (d x +c \right ) \left (i \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\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 ) \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 )-i \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \left (\cos ^{2}\left (d x +c \right )\right )+i \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\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 ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \sin \left (d x +c \right )-i \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \cos \left (d x +c \right )-\operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\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 ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\cos ^{2}\left (d x +c \right )\right )-\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right )+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}}\, \operatorname {arctanh}\left (\frac {\sin \left (d x +c \right )}{\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 ) \cos \left (d x +c \right )-\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \arctan \left (\sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\right ) \sin \left (d x +c \right )+i \cos \left (d x +c \right )-\sin \left (d x +c \right ) \cos \left (d x +c \right )\right )}{2 d \left (i \cos \left (d x +c \right )+i-\sin \left (d x +c \right )\right )}\)	\(576\)

[In]

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

[Out]

-1/2/d*(-tan(d*x+c)+I)*(a*(1+I*tan(d*x+c)))^(1/2)*a*cos(d*x+c)*(I*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-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)*sin(d*x+c)-I*(-cos(d*x+c)/(cos(d*x+c)+
1))^(1/2)*arctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2+I*arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*
x+c)/(cos(d*x+c)+1))^(1/2))*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)-I*(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)
*arctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)-arctanh(sin(d*x+c)/(cos(d*x+c)+1)/(-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)^2-(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan((-cos
(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)*sin(d*x+c)+I*cos(d*x+c)^2-(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh
(sin(d*x+c)/(cos(d*x+c)+1)/(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)-(-cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a
rctan((-cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)+I*cos(d*x+c)-sin(d*x+c)*cos(d*x+c))/(I*cos(d*x+c)+I-sin(d
*x+c))

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. 244 vs. \(2 (68) = 136\).

Time = 0.24 (sec) , antiderivative size = 244, normalized size of antiderivative = 2.62 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=-\frac {\sqrt {\frac {1}{2}} \sqrt {-\frac {a^{3}}{d^{2}}} d \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - a^{2} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) - \sqrt {\frac {1}{2}} \sqrt {-\frac {a^{3}}{d^{2}}} d \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {-\frac {a^{3}}{d^{2}}} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} - a^{2} e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{a}\right ) - \sqrt {2} {\left (-i \, a e^{\left (3 i \, d x + 3 i \, c\right )} - i \, a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{4 \, d} \]

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

Sympy [F]

\[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}} \cos ^{2}{\left (c + d x \right )}\, dx \]

[In]

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

[Out]

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

Maxima [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 98, normalized size of antiderivative = 1.05 \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\frac {i \, {\left (\sqrt {2} a^{\frac {5}{2}} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right ) + \frac {16 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} a^{3}}{4 i \, a \tan \left (d x + c\right ) - 4 \, a}\right )}}{8 \, a d} \]

[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)

Giac [F(-1)]

Timed out. \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\text {Timed out} \]

[In]

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

[Out]

Timed out

Mupad [F(-1)]

Timed out. \[ \int \cos ^2(c+d x) (a+i a \tan (c+d x))^{3/2} \, dx=\int {\cos \left (c+d\,x\right )}^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2} \,d x \]

[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)