\(\int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx\) [505]

Optimal result

Integrand size = 34, antiderivative size = 55 \[ \int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {2 \sqrt [4]{-1} a (i A+B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}+\frac {2 i a B}{d \sqrt {\cot (c+d x)}} \]

[Out]

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

Rubi [A] (verified)

Time = 0.21 (sec) , antiderivative size = 55, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.118, Rules used = {3662, 3672, 3614, 214} \[ \int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {2 \sqrt [4]{-1} a (B+i A) \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}+\frac {2 i a B}{d \sqrt {\cot (c+d x)}} \]

[In]

Int[Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])*(A + B*Tan[c + d*x]),x]

[Out]

(2*(-1)^(1/4)*a*(I*A + B)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d + ((2*I)*a*B)/(d*Sqrt[Cot[c + d*x]])

Rule 214

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]

Rule 3614

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2*(c^2/f), S
ubst[Int[1/(b*c - d*x^2), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && EqQ[c^2 + d^2, 0]

Rule 3662

Int[(cot[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((c_) + (d_.)*tan[(e_.)
+ (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[g^(m + n), Int[(g*Cot[e + f*x])^(p - m - n)*(b + a*Cot[e + f*x])^m*(d
+ c*Cot[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] &&  !IntegerQ[p] && IntegerQ[m] && IntegerQ
[n]

Rule 3672

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*c - a*d)*(A*b - a*B)*((a + b*Tan[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2
+ b^2))), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e + f*x])^(m + 1)*Simp[a*A*c + b*B*c + A*b*d - a*B*d - (A*b*
c - a*B*c - a*A*d - b*B*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0]
&& LtQ[m, -1] && NeQ[a^2 + b^2, 0]

Rubi steps \begin{align*} \text {integral}& = \int \frac {(i a+a \cot (c+d x)) (B+A \cot (c+d x))}{\cot ^{\frac {3}{2}}(c+d x)} \, dx \\ & = \frac {2 i a B}{d \sqrt {\cot (c+d x)}}+\int \frac {a (i A+B)+a (A-i B) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx \\ & = \frac {2 i a B}{d \sqrt {\cot (c+d x)}}+\frac {\left (2 a^2 (i A+B)^2\right ) \text {Subst}\left (\int \frac {1}{-a (i A+B)+a (A-i B) x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d} \\ & = \frac {2 \sqrt [4]{-1} a (i A+B) \text {arctanh}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}+\frac {2 i a B}{d \sqrt {\cot (c+d x)}} \\ \end{align*}

Mathematica [A] (verified)

Time = 0.73 (sec) , antiderivative size = 63, normalized size of antiderivative = 1.15 \[ \int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {2 a \left (i B-\frac {\sqrt [4]{-1} (A-i B) \arctan \left ((-1)^{3/4} \sqrt {\tan (c+d x)}\right )}{\sqrt {\tan (c+d x)}}\right )}{d \sqrt {\cot (c+d x)}} \]

[In]

Integrate[Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])*(A + B*Tan[c + d*x]),x]

[Out]

(2*a*(I*B - ((-1)^(1/4)*(A - I*B)*ArcTan[(-1)^(3/4)*Sqrt[Tan[c + d*x]]])/Sqrt[Tan[c + d*x]]))/(d*Sqrt[Cot[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. 421 vs. \(2 (45 ) = 90\).

Time = 0.36 (sec) , antiderivative size = 422, normalized size of antiderivative = 7.67

method	result	size
derivativedivides	\(\frac {a \sqrt {\frac {1}{\tan \left (d x +c \right )}}\, \sqrt {\tan \left (d x +c \right )}\, \left (2 i A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 i A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+i A \sqrt {2}\, \ln \left (-\frac {\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}{1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}\right )-i B \sqrt {2}\, \ln \left (-\frac {1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}{\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}\right )-2 i B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )-2 i B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+A \sqrt {2}\, \ln \left (-\frac {1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}{\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}\right )+2 A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+8 i B \sqrt {\tan \left (d x +c \right )}+2 B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+B \sqrt {2}\, \ln \left (-\frac {\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}{1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}\right )\right )}{4 d}\)	\(422\)
default	\(\frac {a \sqrt {\frac {1}{\tan \left (d x +c \right )}}\, \sqrt {\tan \left (d x +c \right )}\, \left (2 i A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 i A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+i A \sqrt {2}\, \ln \left (-\frac {\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}{1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}\right )-i B \sqrt {2}\, \ln \left (-\frac {1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}{\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}\right )-2 i B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )-2 i B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+A \sqrt {2}\, \ln \left (-\frac {1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}{\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}\right )+2 A \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 A \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+8 i B \sqrt {\tan \left (d x +c \right )}+2 B \sqrt {2}\, \arctan \left (1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+2 B \sqrt {2}\, \arctan \left (-1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}\right )+B \sqrt {2}\, \ln \left (-\frac {\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}-\tan \left (d x +c \right )-1}{1+\sqrt {2}\, \sqrt {\tan \left (d x +c \right )}+\tan \left (d x +c \right )}\right )\right )}{4 d}\)	\(422\)

[In]

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

[Out]

1/4*a/d*(1/tan(d*x+c))^(1/2)*tan(d*x+c)^(1/2)*(2*I*A*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*I*A*2^(1/2)*
arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))+I*A*2^(1/2)*ln(-(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1)/(1+2^(1/2)*tan(d*x
+c)^(1/2)+tan(d*x+c)))-I*B*2^(1/2)*ln(-(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(2^(1/2)*tan(d*x+c)^(1/2)-tan(d
*x+c)-1))-2*I*B*2^(1/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))-2*I*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))+A
*2^(1/2)*ln(-(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/(2^(1/2)*tan(d*x+c)^(1/2)-tan(d*x+c)-1))+2*A*2^(1/2)*arct
an(1+2^(1/2)*tan(d*x+c)^(1/2))+2*A*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))+8*I*B*tan(d*x+c)^(1/2)+2*B*2^(1
/2)*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))+2*B*2^(1/2)*arctan(-1+2^(1/2)*tan(d*x+c)^(1/2))+B*2^(1/2)*ln(-(2^(1/2)*
tan(d*x+c)^(1/2)-tan(d*x+c)-1)/(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(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. 364 vs. \(2 (43) = 86\).

Time = 0.25 (sec) , antiderivative size = 364, normalized size of antiderivative = 6.62 \[ \int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (-\frac {2 \, {\left ({\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{2}}{d^{2}}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (i \, A + B\right )} a}\right ) - {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{2}}{d^{2}}} \log \left (-\frac {2 \, {\left ({\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\right )} \sqrt {-\frac {{\left (i \, A^{2} + 2 \, A B - i \, B^{2}\right )} a^{2}}{d^{2}}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{{\left (i \, A + B\right )} a}\right ) + 4 \, {\left (B a e^{\left (2 i \, d x + 2 i \, c\right )} - B a\right )} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}}{2 \, {\left (d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

[In]

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

[Out]

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

Sympy [F]

\[ \int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=i a \left (\int \left (- i A \sqrt {\cot {\left (c + d x \right )}}\right )\, dx + \int A \tan {\left (c + d x \right )} \sqrt {\cot {\left (c + d x \right )}}\, dx + \int B \tan ^{2}{\left (c + d x \right )} \sqrt {\cot {\left (c + d x \right )}}\, dx + \int \left (- i B \tan {\left (c + d x \right )} \sqrt {\cot {\left (c + d x \right )}}\right )\, dx\right ) \]

[In]

integrate(cot(d*x+c)**(1/2)*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x)

[Out]

I*a*(Integral(-I*A*sqrt(cot(c + d*x)), x) + Integral(A*tan(c + d*x)*sqrt(cot(c + d*x)), x) + Integral(B*tan(c
+ d*x)**2*sqrt(cot(c + d*x)), x) + Integral(-I*B*tan(c + d*x)*sqrt(cot(c + d*x)), 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. 155 vs. \(2 (43) = 86\).

Time = 0.76 (sec) , antiderivative size = 155, normalized size of antiderivative = 2.82 \[ \int \sqrt {\cot (c+d x)} (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx=\frac {8 i \, B a \sqrt {\tan \left (d x + c\right )} + {\left (2 \, \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )\right )} a}{4 \, d} \]

[In]

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

[Out]

1/4*(8*I*B*a*sqrt(tan(d*x + c)) + (2*sqrt(2)*(-(I + 1)*A + (I - 1)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan
(d*x + c)))) + 2*sqrt(2)*(-(I + 1)*A + (I - 1)*B)*arctan(-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) - sqrt
(2)*((I - 1)*A + (I + 1)*B)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) + sqrt(2)*((I - 1)*A + (I + 1
)*B)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))*a)/d

Giac [F]

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

[In]

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

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)*sqrt(cot(d*x + c)), x)

Mupad [F(-1)]

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

[In]

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

[Out]

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