3.504 \(\int \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx\)
Optimal. Leaf size=53 \[ -\frac{2 a A \sqrt{\cot (c+d x)}}{d}-\frac{2 \sqrt [4]{-1} a (A-i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d} \]
[Out]
(-2*(-1)^(1/4)*a*(A - I*B)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d - (2*a*A*Sqrt[Cot[c + d*x]])/d
________________________________________________________________________________________
Rubi [A] time = 0.150765, antiderivative size = 53, normalized size of antiderivative = 1.,
number of steps used = 4, number of rules used = 4, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.118, Rules used =
{3581, 3592, 3533, 208} \[ -\frac{2 a A \sqrt{\cot (c+d x)}}{d}-\frac{2 \sqrt [4]{-1} a (A-i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d} \]
Antiderivative was successfully verified.
[In]
Int[Cot[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])*(A + B*Tan[c + d*x]),x]
[Out]
(-2*(-1)^(1/4)*a*(A - I*B)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d - (2*a*A*Sqrt[Cot[c + d*x]])/d
Rule 3581
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 3592
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(B*d*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e
+ f*x])^m*Simp[A*c - B*d + (B*c + A*d)*Tan[e + f*x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b
*c - a*d, 0] && !LeQ[m, -1]
Rule 3533
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 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rubi steps
\begin{align*} \int \cot ^{\frac{3}{2}}(c+d x) (a+i a \tan (c+d x)) (A+B \tan (c+d x)) \, dx &=\int \frac{(i a+a \cot (c+d x)) (B+A \cot (c+d x))}{\sqrt{\cot (c+d x)}} \, dx\\ &=-\frac{2 a A \sqrt{\cot (c+d x)}}{d}+\int \frac{-a (A-i B)+a (i A+B) \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx\\ &=-\frac{2 a A \sqrt{\cot (c+d x)}}{d}+\frac{\left (2 a^2 (A-i B)^2\right ) \operatorname{Subst}\left (\int \frac{1}{a (A-i B)+a (i A+B) x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{d}\\ &=-\frac{2 \sqrt [4]{-1} a (A-i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}-\frac{2 a A \sqrt{\cot (c+d x)}}{d}\\ \end{align*}
Mathematica [A] time = 2.0035, size = 92, normalized size = 1.74 \[ \frac{2 a e^{-i c} (\cos (c)+i \sin (c)) \sqrt{\cot (c+d x)} \left (-A+(A-i B) \sqrt{i \tan (c+d x)} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )\right )}{d} \]
Antiderivative was successfully verified.
[In]
Integrate[Cot[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])*(A + B*Tan[c + d*x]),x]
[Out]
(2*a*Sqrt[Cot[c + d*x]]*(Cos[c] + I*Sin[c])*(-A + (A - I*B)*ArcTanh[Sqrt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2
*I)*(c + d*x)))]]*Sqrt[I*Tan[c + d*x]]))/(d*E^(I*c))
________________________________________________________________________________________
Maple [C] time = 0.417, size = 1425, normalized size = 26.9 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x)
[Out]
-a/d*2^(1/2)*(I*A*cos(d*x+c)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+
c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2
))-I*A*cos(d*x+c)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(
(cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2
))+I*B*cos(d*x+c)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*(
(cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2
))+I*A*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)
-1)/sin(d*x+c))^(1/2)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))-I*A*(-(cos(d*x+c)-1
-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*
EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-A*cos(d*x+c)*((cos(d*x+c)-1+si
n(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*El
lipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+I*B*(-(cos(d*x+c)-1-sin(d*x+c))/
sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticPi((-
(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+B*cos(d*x+c)*((cos(d*x+c)-1+sin(d*x+c))/sin
(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2
^(1/2))*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)-B*cos(d*x+c)*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2
)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-s
in(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-A*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c
)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/si
n(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))+B*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+
c))^(1/2)*EllipticF((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(-(cos(d*x+c)-1-sin(d*x+c))/sin
(d*x+c))^(1/2)-B*((cos(d*x+c)-1+sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1)/sin(d*x+c))^(1/2)*(-(cos(d*x+c)-
1-sin(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi((-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2)
)+A*cos(d*x+c)*2^(1/2))*(cos(d*x+c)/sin(d*x+c))^(3/2)*sin(d*x+c)/cos(d*x+c)^2
________________________________________________________________________________________
Maxima [B] time = 1.55758, size = 209, normalized size = 3.94 \begin{align*} -\frac{{\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 + \frac{8 \, A a}{\sqrt{\tan \left (d x + c\right )}}}{4 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x, algorithm="maxima")
[Out]
-1/4*((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)*l
og(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 + 8*A*a/sqrt(tan(d*x + c)))/d
________________________________________________________________________________________
Fricas [B] time = 1.41133, size = 803, normalized size = 15.15 \begin{align*} -\frac{8 \, A a \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}} - \sqrt{\frac{{\left (4 i \, A^{2} + 8 \, A B - 4 i \, B^{2}\right )} a^{2}}{d^{2}}} d \log \left (-\frac{{\left (2 \,{\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} -{\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt{\frac{{\left (4 i \, A^{2} + 8 \, A B - 4 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 ) + \sqrt{\frac{{\left (4 i \, A^{2} + 8 \, A B - 4 i \, B^{2}\right )} a^{2}}{d^{2}}} d \log \left (-\frac{{\left (2 \,{\left (A - i \, B\right )} a e^{\left (2 i \, d x + 2 i \, c\right )} -{\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt{\frac{{\left (4 i \, A^{2} + 8 \, A B - 4 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 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(3/2)*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x, algorithm="fricas")
[Out]
-1/4*(8*A*a*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)) - sqrt((4*I*A^2 + 8*A*B - 4*I*B^2)*a^2
/d^2)*d*log(-(2*(A - I*B)*a*e^(2*I*d*x + 2*I*c) - (I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt((4*I*A^2 + 8*A*B - 4*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))
+ sqrt((4*I*A^2 + 8*A*B - 4*I*B^2)*a^2/d^2)*d*log(-(2*(A - I*B)*a*e^(2*I*d*x + 2*I*c) - (-I*d*e^(2*I*d*x + 2*
I*c) + I*d)*sqrt((4*I*A^2 + 8*A*B - 4*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
________________________________________________________________________________________
Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)**(3/2)*(a+I*a*tan(d*x+c))*(A+B*tan(d*x+c)),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )} \cot \left (d x + c\right )^{\frac{3}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(3/2)*(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)*cot(d*x + c)^(3/2), x)