3.512 \(\int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx\)

Optimal. Leaf size=130 \[ -\frac {2 a^2 (5 A-7 i B)}{15 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (B+i A)}{d \sqrt {\cot (c+d x)}}+\frac {4 \sqrt [4]{-1} a^2 (A-i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}+\frac {2 i B \left (a^2 \cot (c+d x)+i a^2\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)} \]

[Out]

4*(-1)^(1/4)*a^2*(A-I*B)*arctanh((-1)^(3/4)*cot(d*x+c)^(1/2))/d-2/15*a^2*(5*A-7*I*B)/d/cot(d*x+c)^(3/2)+2/5*I*
B*(I*a^2+a^2*cot(d*x+c))/d/cot(d*x+c)^(5/2)+4*a^2*(I*A+B)/d/cot(d*x+c)^(1/2)

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Rubi [A]  time = 0.37, antiderivative size = 130, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3581, 3593, 3591, 3529, 3533, 208} \[ -\frac {2 a^2 (5 A-7 i B)}{15 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (B+i A)}{d \sqrt {\cot (c+d x)}}+\frac {4 \sqrt [4]{-1} a^2 (A-i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}+\frac {2 i B \left (a^2 \cot (c+d x)+i a^2\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)} \]

Antiderivative was successfully verified.

[In]

Int[((a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]))/Sqrt[Cot[c + d*x]],x]

[Out]

(4*(-1)^(1/4)*a^2*(A - I*B)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d - (2*a^2*(5*A - (7*I)*B))/(15*d*Cot[c +
d*x]^(3/2)) + (4*a^2*(I*A + B))/(d*Sqrt[Cot[c + d*x]]) + (((2*I)/5)*B*(I*a^2 + a^2*Cot[c + d*x]))/(d*Cot[c + d
*x]^(5/2))

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]

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(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*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[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 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 3591

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]

Rule 3593

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

Rubi steps

\begin {align*} \int \frac {(a+i a \tan (c+d x))^2 (A+B \tan (c+d x))}{\sqrt {\cot (c+d x)}} \, dx &=\int \frac {(i a+a \cot (c+d x))^2 (B+A \cot (c+d x))}{\cot ^{\frac {7}{2}}(c+d x)} \, dx\\ &=\frac {2 i B \left (i a^2+a^2 \cot (c+d x)\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {(i a+a \cot (c+d x)) \left (\frac {1}{2} a (5 i A+7 B)+\frac {1}{2} a (5 A-3 i B) \cot (c+d x)\right )}{\cot ^{\frac {5}{2}}(c+d x)} \, dx\\ &=-\frac {2 a^2 (5 A-7 i B)}{15 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {2 i B \left (i a^2+a^2 \cot (c+d x)\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {5 a^2 (i A+B)+5 a^2 (A-i B) \cot (c+d x)}{\cot ^{\frac {3}{2}}(c+d x)} \, dx\\ &=-\frac {2 a^2 (5 A-7 i B)}{15 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (i A+B)}{d \sqrt {\cot (c+d x)}}+\frac {2 i B \left (i a^2+a^2 \cot (c+d x)\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {2}{5} \int \frac {5 a^2 (A-i B)-5 a^2 (i A+B) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx\\ &=-\frac {2 a^2 (5 A-7 i B)}{15 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (i A+B)}{d \sqrt {\cot (c+d x)}}+\frac {2 i B \left (i a^2+a^2 \cot (c+d x)\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)}+\frac {\left (20 a^4 (A-i B)^2\right ) \operatorname {Subst}\left (\int \frac {1}{-5 a^2 (A-i B)-5 a^2 (i A+B) x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}\\ &=\frac {4 \sqrt [4]{-1} a^2 (A-i B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}-\frac {2 a^2 (5 A-7 i B)}{15 d \cot ^{\frac {3}{2}}(c+d x)}+\frac {4 a^2 (i A+B)}{d \sqrt {\cot (c+d x)}}+\frac {2 i B \left (i a^2+a^2 \cot (c+d x)\right )}{5 d \cot ^{\frac {5}{2}}(c+d x)}\\ \end {align*}

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Mathematica [A]  time = 7.78, size = 133, normalized size = 1.02 \[ \frac {a^2 \left (\sec ^2(c+d x) (-5 (A-2 i B) \sin (2 (c+d x))+(33 B+30 i A) \cos (2 (c+d x))+30 i A+27 B)-\frac {60 i (A-i B) \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )}{\sqrt {i \tan (c+d x)}}\right )}{15 d \sqrt {\cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[((a + I*a*Tan[c + d*x])^2*(A + B*Tan[c + d*x]))/Sqrt[Cot[c + d*x]],x]

[Out]

(a^2*(Sec[c + d*x]^2*((30*I)*A + 27*B + ((30*I)*A + 33*B)*Cos[2*(c + d*x)] - 5*(A - (2*I)*B)*Sin[2*(c + d*x)])
- ((60*I)*(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]]
))/(15*d*Sqrt[Cot[c + d*x]])

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fricas [B]  time = 0.55, size = 503, normalized size = 3.87 \[ -\frac {15 \, \sqrt {\frac {{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {{\left (4 \, {\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {\frac {{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, 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 )}}{{\left (2 i \, A + 2 \, B\right )} a^{2}}\right ) - 15 \, \sqrt {\frac {{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (d e^{\left (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )} \log \left (-\frac {{\left (4 \, {\left (A - i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {\frac {{\left (16 i \, A^{2} + 32 \, A B - 16 i \, B^{2}\right )} a^{4}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, 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 )}}{{\left (2 i \, A + 2 \, B\right )} a^{2}}\right ) - 8 \, {\left ({\left (35 \, A - 43 i \, B\right )} a^{2} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (25 \, A - 11 i \, B\right )} a^{2} e^{\left (4 i \, d x + 4 i \, c\right )} - {\left (35 \, A - 31 i \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} - {\left (25 \, A - 23 i \, B\right )} 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 (6 i \, d x + 6 i \, c\right )} + 3 \, d e^{\left (4 i \, d x + 4 i \, c\right )} + 3 \, d e^{\left (2 i \, d x + 2 i \, c\right )} + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/60*(15*sqrt((16*I*A^2 + 32*A*B - 16*I*B^2)*a^4/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*
e^(2*I*d*x + 2*I*c) + d)*log(-(4*(A - I*B)*a^2*e^(2*I*d*x + 2*I*c) - sqrt((16*I*A^2 + 32*A*B - 16*I*B^2)*a^4/d
^2)*(I*d*e^(2*I*d*x + 2*I*c) - I*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)/((2*I*A + 2*B)*a^2)) - 15*sqrt((16*I*A^2 + 32*A*B - 16*I*B^2)*a^4/d^2)*(d*e^(6*I*d*x + 6*I*c) + 3*d*e^
(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d*x + 2*I*c) + d)*log(-(4*(A - I*B)*a^2*e^(2*I*d*x + 2*I*c) - sqrt((16*I*A^2 +
32*A*B - 16*I*B^2)*a^4/d^2)*(-I*d*e^(2*I*d*x + 2*I*c) + I*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)/((2*I*A + 2*B)*a^2)) - 8*((35*A - 43*I*B)*a^2*e^(6*I*d*x + 6*I*c) + (25*A - 1
1*I*B)*a^2*e^(4*I*d*x + 4*I*c) - (35*A - 31*I*B)*a^2*e^(2*I*d*x + 2*I*c) - (25*A - 23*I*B)*a^2)*sqrt((I*e^(2*I
*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1)))/(d*e^(6*I*d*x + 6*I*c) + 3*d*e^(4*I*d*x + 4*I*c) + 3*d*e^(2*I*d
*x + 2*I*c) + d)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 1.77, size = 971, normalized size = 7.47 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*a^2/d*(-1+cos(d*x+c))*(-30*I*A*sin(d*x+c)*EllipticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2*2^
(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d
*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2+30*I*A*2^(1/2)*cos(d*x+c)^3-10*I*B*cos(d*x+c)*sin(d*x+c)*2^(1/2)+30*A*((
-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d*x+c))/si
n(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/
2*2^(1/2))-30*B*sin(d*x+c)*EllipticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*((-1+cos(d*x+
c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(
1/2)*cos(d*x+c)^2+30*B*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-si
n(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2*sin(d*x+c)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d
*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))-30*I*B*sin(d*x+c)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2
),1/2+1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*(-(-
sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2+30*I*A*sin(d*x+c)*EllipticPi((-(-sin(d*x+c)-1+cos(d*x+
c))/sin(d*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/si
n(d*x+c))^(1/2)*(-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2+10*I*B*2^(1/2)*sin(d*x+c)*cos(d*x+
c)^2-5*A*2^(1/2)*sin(d*x+c)*cos(d*x+c)^2-30*I*A*2^(1/2)*cos(d*x+c)^2+33*B*2^(1/2)*cos(d*x+c)^3+5*A*2^(1/2)*cos
(d*x+c)*sin(d*x+c)-33*B*2^(1/2)*cos(d*x+c)^2-3*B*2^(1/2)*cos(d*x+c)+3*B*2^(1/2))*(1+cos(d*x+c))^2/cos(d*x+c)^2
/sin(d*x+c)^4/(cos(d*x+c)/sin(d*x+c))^(1/2)*2^(1/2)

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maxima [A]  time = 1.71, size = 200, normalized size = 1.54 \[ -\frac {4 \, {\left (3 \, B a^{2} + \frac {5 \, {\left (A - 2 i \, B\right )} a^{2}}{\tan \left (d x + c\right )} + \frac {{\left (-30 i \, A - 30 \, B\right )} a^{2}}{\tan \left (d x + c\right )^{2}}\right )} \tan \left (d x + c\right )^{\frac {5}{2}} + 15 \, {\left (2 \, \sqrt {2} {\left (-\left (i - 1\right ) \, A - \left (i + 1\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} {\left (-\left (i - 1\right ) \, A - \left (i + 1\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + \sqrt {2} {\left (-\left (i + 1\right ) \, A + \left (i - 1\right ) \, B\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )\right )} a^{2}}{30 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/30*(4*(3*B*a^2 + 5*(A - 2*I*B)*a^2/tan(d*x + c) + (-30*I*A - 30*B)*a^2/tan(d*x + c)^2)*tan(d*x + c)^(5/2) +
15*(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^2)/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ - a^{2} \left (\int \left (- \frac {A}{\sqrt {\cot {\left (c + d x \right )}}}\right )\, dx + \int \frac {A \tan ^{2}{\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\, dx + \int \left (- \frac {B \tan {\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\right )\, dx + \int \frac {B \tan ^{3}{\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\, dx + \int \left (- \frac {2 i A \tan {\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\right )\, dx + \int \left (- \frac {2 i B \tan ^{2}{\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\right )\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-a**2*(Integral(-A/sqrt(cot(c + d*x)), x) + Integral(A*tan(c + d*x)**2/sqrt(cot(c + d*x)), x) + Integral(-B*ta
n(c + d*x)/sqrt(cot(c + d*x)), x) + Integral(B*tan(c + d*x)**3/sqrt(cot(c + d*x)), x) + Integral(-2*I*A*tan(c
+ d*x)/sqrt(cot(c + d*x)), x) + Integral(-2*I*B*tan(c + d*x)**2/sqrt(cot(c + d*x)), x))

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