3.724 \(\int \cot ^{\frac{7}{2}}(c+d x) (a+i a \tan (c+d x))^2 \, dx\)
Optimal. Leaf size=91 \[ -\frac{2 a^2 \cot ^{\frac{5}{2}}(c+d x)}{5 d}-\frac{4 i a^2 \cot ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{4 a^2 \sqrt{\cot (c+d x)}}{d}+\frac{4 \sqrt [4]{-1} a^2 \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d} \]
[Out]
(4*(-1)^(1/4)*a^2*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d + (4*a^2*Sqrt[Cot[c + d*x]])/d - (((4*I)/3)*a^2*Co
t[c + d*x]^(3/2))/d - (2*a^2*Cot[c + d*x]^(5/2))/(5*d)
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Rubi [A] time = 0.172378, antiderivative size = 91, normalized size of antiderivative = 1.,
number of steps used = 6, number of rules used = 5, integrand size = 26, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.192, Rules used =
{3673, 3543, 3528, 3533, 208} \[ -\frac{2 a^2 \cot ^{\frac{5}{2}}(c+d x)}{5 d}-\frac{4 i a^2 \cot ^{\frac{3}{2}}(c+d x)}{3 d}+\frac{4 a^2 \sqrt{\cot (c+d x)}}{d}+\frac{4 \sqrt [4]{-1} a^2 \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d} \]
Antiderivative was successfully verified.
[In]
Int[Cot[c + d*x]^(7/2)*(a + I*a*Tan[c + d*x])^2,x]
[Out]
(4*(-1)^(1/4)*a^2*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d + (4*a^2*Sqrt[Cot[c + d*x]])/d - (((4*I)/3)*a^2*Co
t[c + d*x]^(3/2))/d - (2*a^2*Cot[c + d*x]^(5/2))/(5*d)
Rule 3673
Int[(cot[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Cot[e + f*x])^(m - n*p)*(b + a*Cot[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] && !IntegerQ[m] && IntegersQ[n, p]
Rule 3543
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^2, x_Symbol] :> Simp[
(d^2*(a + b*Tan[e + f*x])^(m + 1))/(b*f*(m + 1)), x] + Int[(a + b*Tan[e + f*x])^m*Simp[c^2 - d^2 + 2*c*d*Tan[e
+ f*x], x], x] /; FreeQ[{a, b, c, d, e, f, m}, x] && NeQ[b*c - a*d, 0] && !LeQ[m, -1] && !(EqQ[m, 2] && EqQ
[a, 0])
Rule 3528
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(d
*(a + b*Tan[e + f*x])^m)/(f*m), x] + Int[(a + b*Tan[e + f*x])^(m - 1)*Simp[a*c - b*d + (b*c + a*d)*Tan[e + f*x
], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && GtQ[m, 0]
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{7}{2}}(c+d x) (a+i a \tan (c+d x))^2 \, dx &=\int \cot ^{\frac{3}{2}}(c+d x) (i a+a \cot (c+d x))^2 \, dx\\ &=-\frac{2 a^2 \cot ^{\frac{5}{2}}(c+d x)}{5 d}+\int \cot ^{\frac{3}{2}}(c+d x) \left (-2 a^2+2 i a^2 \cot (c+d x)\right ) \, dx\\ &=-\frac{4 i a^2 \cot ^{\frac{3}{2}}(c+d x)}{3 d}-\frac{2 a^2 \cot ^{\frac{5}{2}}(c+d x)}{5 d}+\int \sqrt{\cot (c+d x)} \left (-2 i a^2-2 a^2 \cot (c+d x)\right ) \, dx\\ &=\frac{4 a^2 \sqrt{\cot (c+d x)}}{d}-\frac{4 i a^2 \cot ^{\frac{3}{2}}(c+d x)}{3 d}-\frac{2 a^2 \cot ^{\frac{5}{2}}(c+d x)}{5 d}+\int \frac{2 a^2-2 i a^2 \cot (c+d x)}{\sqrt{\cot (c+d x)}} \, dx\\ &=\frac{4 a^2 \sqrt{\cot (c+d x)}}{d}-\frac{4 i a^2 \cot ^{\frac{3}{2}}(c+d x)}{3 d}-\frac{2 a^2 \cot ^{\frac{5}{2}}(c+d x)}{5 d}+\frac{\left (8 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{-2 a^2-2 i a^2 x^2} \, dx,x,\sqrt{\cot (c+d x)}\right )}{d}\\ &=\frac{4 \sqrt [4]{-1} a^2 \tanh ^{-1}\left ((-1)^{3/4} \sqrt{\cot (c+d x)}\right )}{d}+\frac{4 a^2 \sqrt{\cot (c+d x)}}{d}-\frac{4 i a^2 \cot ^{\frac{3}{2}}(c+d x)}{3 d}-\frac{2 a^2 \cot ^{\frac{5}{2}}(c+d x)}{5 d}\\ \end{align*}
Mathematica [A] time = 3.3961, size = 104, normalized size = 1.14 \[ -\frac{a^2 \sqrt{\cot (c+d x)} \left (60 \sqrt{i \tan (c+d x)} \tanh ^{-1}\left (\sqrt{\frac{-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )+\csc ^2(c+d x) (10 i \sin (2 (c+d x))+33 \cos (2 (c+d x))-27)\right )}{15 d} \]
Antiderivative was successfully verified.
[In]
Integrate[Cot[c + d*x]^(7/2)*(a + I*a*Tan[c + d*x])^2,x]
[Out]
-(a^2*Sqrt[Cot[c + d*x]]*(Csc[c + d*x]^2*(-27 + 33*Cos[2*(c + d*x)] + (10*I)*Sin[2*(c + d*x)]) + 60*ArcTanh[Sq
rt[(-1 + E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]]*Sqrt[I*Tan[c + d*x]]))/(15*d)
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Maple [C] time = 0.321, size = 1482, normalized size = 16.3 \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)^(7/2)*(a+I*a*tan(d*x+c))^2,x)
[Out]
-1/15*a^2/d*2^(1/2)*(30*I*(-(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))*
cos(d*x+c)^3-30*I*(-(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
))*cos(d*x+c)^3+30*I*(-(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))*cos(d
*x+c)^2-30*(-(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))*cos(
d*x+c)^3-30*I*(-(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))*c
os(d*x+c)^2-30*I*(-(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))*cos(d*x+c
)-30*(-(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))*cos(d*x+c)
^2+30*I*(-(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))*cos(d*x
+c)+10*I*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)-30*I*(-(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))+30*(-(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))*cos(d*x+c)+30*I*(-(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))+33*2^(1/2)*cos(d*x+c)^3+30*(-(cos(d*x+c)-1-sin(d*x+c))/sin(d*x+c))^(1/2)*((cos(d*x+c)-1+sin(d*x+c))/s
in(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))-30*2^(1/2)*cos(d*x+c))*(cos(d*x+c)/sin(d*x+c))^(7/2)*sin(d*x+c)/cos(d*x+c)^4
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Maxima [B] time = 1.5782, size = 213, normalized size = 2.34 \begin{align*} -\frac{15 \,{\left (-\left (2 i - 2\right ) \, \sqrt{2} \arctan \left (\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} + \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) - \left (2 i - 2\right ) \, \sqrt{2} \arctan \left (-\frac{1}{2} \, \sqrt{2}{\left (\sqrt{2} - \frac{2}{\sqrt{\tan \left (d x + c\right )}}\right )}\right ) + \left (i + 1\right ) \, \sqrt{2} \log \left (\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right ) - \left (i + 1\right ) \, \sqrt{2} \log \left (-\frac{\sqrt{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{1}{\tan \left (d x + c\right )} + 1\right )\right )} a^{2} - \frac{120 \, a^{2}}{\sqrt{\tan \left (d x + c\right )}} + \frac{40 i \, a^{2}}{\tan \left (d x + c\right )^{\frac{3}{2}}} + \frac{12 \, a^{2}}{\tan \left (d x + c\right )^{\frac{5}{2}}}}{30 \, d} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^2,x, algorithm="maxima")
[Out]
-1/30*(15*(-(2*I - 2)*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) - (2*I - 2)*sqrt(2)*arctan(
-1/2*sqrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) + (I + 1)*sqrt(2)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x +
c) + 1) - (I + 1)*sqrt(2)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1))*a^2 - 120*a^2/sqrt(tan(d*x +
c)) + 40*I*a^2/tan(d*x + c)^(3/2) + 12*a^2/tan(d*x + c)^(5/2))/d
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Fricas [B] time = 1.44629, size = 937, normalized size = 10.3 \begin{align*} -\frac{15 \, \sqrt{\frac{16 i \, a^{4}}{d^{2}}}{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac{{\left (4 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt{\frac{16 i \, a^{4}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, a^{2}}\right ) - 15 \, \sqrt{\frac{16 i \, a^{4}}{d^{2}}}{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (\frac{{\left (4 i \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt{\frac{16 i \, a^{4}}{d^{2}}}{\left (d e^{\left (2 i \, d x + 2 i \, c\right )} - d\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}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{2 \, a^{2}}\right ) - 8 \,{\left (43 \, a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - 54 \, a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + 23 \, a^{2}\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}}}{60 \,{\left (d e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^2,x, algorithm="fricas")
[Out]
-1/60*(15*sqrt(16*I*a^4/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log(1/2*(4*I*a^2*e^(2*I*d*x
+ 2*I*c) + sqrt(16*I*a^4/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*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)/a^2) - 15*sqrt(16*I*a^4/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) +
d)*log(1/2*(4*I*a^2*e^(2*I*d*x + 2*I*c) - sqrt(16*I*a^4/d^2)*(d*e^(2*I*d*x + 2*I*c) - d)*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)/a^2) - 8*(43*a^2*e^(4*I*d*x + 4*I*c) - 54*a^2*e^
(2*I*d*x + 2*I*c) + 23*a^2)*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))/(d*e^(4*I*d*x + 4*I*c
) - 2*d*e^(2*I*d*x + 2*I*c) + d)
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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)**(7/2)*(a+I*a*tan(d*x+c))**2,x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} \cot \left (d x + c\right )^{\frac{7}{2}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(7/2)*(a+I*a*tan(d*x+c))^2,x, algorithm="giac")
[Out]
integrate((I*a*tan(d*x + c) + a)^2*cot(d*x + c)^(7/2), x)