∫ cot9 _ 2 (c + dx)(a + ia tan(c + dx))3(A + B tan(c + dx)) dx

3.513 \(\int \cot ^{\frac {9}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=171 \[ \frac {8 a^3 (23 A-21 i B) \cot ^{\frac {3}{2}}(c+d x)}{105 d}-\frac {2 (7 B+11 i A) \cot ^{\frac {3}{2}}(c+d x) \left (a^3 \cot (c+d x)+i a^3\right )}{35 d}+\frac {8 a^3 (B+i A) \sqrt {\cot (c+d x)}}{d}+\frac {8 \sqrt [4]{-1} a^3 (B+i A) \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (a \cot (c+d x)+i a)^2}{7 d} \]

[Out]

8*(-1)^(1/4)*a^3*(I*A+B)*arctanh((-1)^(3/4)*cot(d*x+c)^(1/2))/d+8/105*a^3*(23*A-21*I*B)*cot(d*x+c)^(3/2)/d-2/7

*a*A*cot(d*x+c)^(3/2)*(I*a+a*cot(d*x+c))^2/d-2/35*(11*I*A+7*B)*cot(d*x+c)^(3/2)*(I*a^3+a^3*cot(d*x+c))/d+8*a^3

*(I*A+B)*cot(d*x+c)^(1/2)/d

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Rubi [A] time = 0.53, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.167, Rules used = {3581, 3594, 3592, 3528, 3533, 208} \[ \frac {8 a^3 (23 A-21 i B) \cot ^{\frac {3}{2}}(c+d x)}{105 d}-\frac {2 (7 B+11 i A) \cot ^{\frac {3}{2}}(c+d x) \left (a^3 \cot (c+d x)+i a^3\right )}{35 d}+\frac {8 a^3 (B+i A) \sqrt {\cot (c+d x)}}{d}+\frac {8 \sqrt [4]{-1} a^3 (B+i A) \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (a \cot (c+d x)+i a)^2}{7 d} \]

Antiderivative was successfully veriﬁed.

[In]

Int[Cot[c + d*x]^(9/2)*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

(8*(-1)^(1/4)*a^3*(I*A + B)*ArcTanh[(-1)^(3/4)*Sqrt[Cot[c + d*x]]])/d + (8*a^3*(I*A + B)*Sqrt[Cot[c + d*x]])/d

+ (8*a^3*(23*A - (21*I)*B)*Cot[c + d*x]^(3/2))/(105*d) - (2*a*A*Cot[c + d*x]^(3/2)*(I*a + a*Cot[c + d*x])^2)/

(7*d) - (2*((11*I)*A + 7*B)*Cot[c + d*x]^(3/2)*(I*a^3 + a^3*Cot[c + d*x]))/(35*d)

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 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 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 3594

Int[((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e_

.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*B*(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^(n + 1))/(d*f

*(m + n)), x] + Dist[1/(d*(m + n)), Int[(a + b*Tan[e + f*x])^(m - 1)*(c + d*Tan[e + f*x])^n*Simp[a*A*d*(m + n)

+ B*(a*c*(m - 1) - b*d*(n + 1)) - (B*(b*c - a*d)*(m - 1) - d*(A*b + a*B)*(m + n))*Tan[e + f*x], x], x], x] /;

FreeQ[{a, b, c, d, e, f, A, B, n}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && GtQ[m, 1] && !LtQ[n, -1]

Rubi steps

\begin {align*} \int \cot ^{\frac {9}{2}}(c+d x) (a+i a \tan (c+d x))^3 (A+B \tan (c+d x)) \, dx &=\int \sqrt {\cot (c+d x)} (i a+a \cot (c+d x))^3 (B+A \cot (c+d x)) \, dx\\ &=-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (i a+a \cot (c+d x))^2}{7 d}-\frac {2}{7} \int \sqrt {\cot (c+d x)} (i a+a \cot (c+d x))^2 \left (\frac {1}{2} a (3 A-7 i B)-\frac {1}{2} a (11 i A+7 B) \cot (c+d x)\right ) \, dx\\ &=-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (i a+a \cot (c+d x))^2}{7 d}-\frac {2 (11 i A+7 B) \cot ^{\frac {3}{2}}(c+d x) \left (i a^3+a^3 \cot (c+d x)\right )}{35 d}+\frac {4}{35} \int \sqrt {\cot (c+d x)} (i a+a \cot (c+d x)) \left (-2 a^2 (6 i A+7 B)-a^2 (23 A-21 i B) \cot (c+d x)\right ) \, dx\\ &=\frac {8 a^3 (23 A-21 i B) \cot ^{\frac {3}{2}}(c+d x)}{105 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (i a+a \cot (c+d x))^2}{7 d}-\frac {2 (11 i A+7 B) \cot ^{\frac {3}{2}}(c+d x) \left (i a^3+a^3 \cot (c+d x)\right )}{35 d}+\frac {4}{35} \int \sqrt {\cot (c+d x)} \left (35 a^3 (A-i B)-35 a^3 (i A+B) \cot (c+d x)\right ) \, dx\\ &=\frac {8 a^3 (i A+B) \sqrt {\cot (c+d x)}}{d}+\frac {8 a^3 (23 A-21 i B) \cot ^{\frac {3}{2}}(c+d x)}{105 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (i a+a \cot (c+d x))^2}{7 d}-\frac {2 (11 i A+7 B) \cot ^{\frac {3}{2}}(c+d x) \left (i a^3+a^3 \cot (c+d x)\right )}{35 d}+\frac {4}{35} \int \frac {35 a^3 (i A+B)+35 a^3 (A-i B) \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx\\ &=\frac {8 a^3 (i A+B) \sqrt {\cot (c+d x)}}{d}+\frac {8 a^3 (23 A-21 i B) \cot ^{\frac {3}{2}}(c+d x)}{105 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (i a+a \cot (c+d x))^2}{7 d}-\frac {2 (11 i A+7 B) \cot ^{\frac {3}{2}}(c+d x) \left (i a^3+a^3 \cot (c+d x)\right )}{35 d}+\frac {\left (280 a^6 (i A+B)^2\right ) \operatorname {Subst}\left (\int \frac {1}{-35 a^3 (i A+B)+35 a^3 (A-i B) x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}\\ &=\frac {8 \sqrt [4]{-1} a^3 (i A+B) \tanh ^{-1}\left ((-1)^{3/4} \sqrt {\cot (c+d x)}\right )}{d}+\frac {8 a^3 (i A+B) \sqrt {\cot (c+d x)}}{d}+\frac {8 a^3 (23 A-21 i B) \cot ^{\frac {3}{2}}(c+d x)}{105 d}-\frac {2 a A \cot ^{\frac {3}{2}}(c+d x) (i a+a \cot (c+d x))^2}{7 d}-\frac {2 (11 i A+7 B) \cot ^{\frac {3}{2}}(c+d x) \left (i a^3+a^3 \cot (c+d x)\right )}{35 d}\\ \end {align*}

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Mathematica [A] time = 9.79, size = 161, normalized size = 0.94 \[ \frac {a^3 \sqrt {\cot (c+d x)} \left (-1680 i (A-i B) \sqrt {i \tan (c+d x)} \tanh ^{-1}\left (\sqrt {\frac {-1+e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}}\right )-\left (\csc ^3(c+d x) ((-95 A+105 i B) \cos (c+d x)+5 (31 A-21 i B) \cos (3 (c+d x))+42 \sin (c+d x) ((21 B+23 i A) \cos (2 (c+d x))-17 i A-19 B))\right )\right )}{210 d} \]

Antiderivative was successfully veriﬁed.

[In]

Integrate[Cot[c + d*x]^(9/2)*(a + I*a*Tan[c + d*x])^3*(A + B*Tan[c + d*x]),x]

[Out]

(a^3*Sqrt[Cot[c + d*x]]*(-(Csc[c + d*x]^3*((-95*A + (105*I)*B)*Cos[c + d*x] + 5*(31*A - (21*I)*B)*Cos[3*(c + d

*x)] + 42*((-17*I)*A - 19*B + ((23*I)*A + 21*B)*Cos[2*(c + d*x)])*Sin[c + d*x])) - (1680*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]]))/(210*d)

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fricas [B] time = 0.55, size = 503, normalized size = 2.94 \[ \frac {105 \, \sqrt {\frac {{\left (-64 i \, A^{2} - 128 \, A B + 64 i \, B^{2}\right )} a^{6}}{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 (8 \, {\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + \sqrt {\frac {{\left (-64 i \, A^{2} - 128 \, A B + 64 i \, B^{2}\right )} a^{6}}{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 )}}{{\left (4 i \, A + 4 \, B\right )} a^{3}}\right ) - 105 \, \sqrt {\frac {{\left (-64 i \, A^{2} - 128 \, A B + 64 i \, B^{2}\right )} a^{6}}{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 (8 \, {\left (A - i \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} - \sqrt {\frac {{\left (-64 i \, A^{2} - 128 \, A B + 64 i \, B^{2}\right )} a^{6}}{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 )}}{{\left (4 i \, A + 4 \, B\right )} a^{3}}\right ) + {\left ({\left (5104 i \, A + 4368 \, B\right )} a^{3} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (-10336 i \, A - 10752 \, B\right )} a^{3} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (8816 i \, A + 9072 \, B\right )} a^{3} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-2624 i \, A - 2688 \, B\right )} a^{3}\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}}}{420 \, {\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(cot(d*x+c)^(9/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

1/420*(105*sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*a^6/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(-(8*(A - I*B)*a^3*e^(2*I*d*x + 2*I*c) + sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*

a^6/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)/((4*I*A + 4*B)*a^3)) - 105*sqrt((-64*I*A^2 - 128*A*B + 64*I*B^2)*a^6/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(-(8*(A - I*B)*a^3*e^(2*I*d*x + 2*I*c) - sqrt((-64*I*A

^2 - 128*A*B + 64*I*B^2)*a^6/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)/((4*I*A + 4*B)*a^3)) + ((5104*I*A + 4368*B)*a^3*e^(6*I*d*x + 6*I*c) + (-1033

6*I*A - 10752*B)*a^3*e^(4*I*d*x + 4*I*c) + (8816*I*A + 9072*B)*a^3*e^(2*I*d*x + 2*I*c) + (-2624*I*A - 2688*B)*

a^3)*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 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} \cot \left (d x + c\right )^{\frac {9}{2}}\,{d x} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(9/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^3*cot(d*x + c)^(9/2), x)

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maple [C] time = 2.20, size = 3132, normalized size = 18.32 \[ \text {output too large to display} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^(9/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x)

[Out]

-1/105*a^3/d*(-420*A*(-(-sin(d*x+c)-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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*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))*sin(d*x+c)+155*A*2^(1/2)*cos(d*x+c)^4-140*A*2^(1/2)*cos(d*x+c)^2-420*A*cos(d*x+c)^3*sin(d*x+c)*Ellip

ticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(-(-sin(d*x+c)-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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+420*A*cos(d*x+c)^3*sin(d*x+

c)*(-(-sin(d*x+c)-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)*((-1+cos(d*x+c

))/sin(d*x+c))^(1/2)*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))-420*B*co

s(d*x+c)^3*sin(d*x+c)*(-(-sin(d*x+c)-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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*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))+420*I*A*sin(d*x+c)*(-(-sin(d*x+c)-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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*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))-420*I*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))*(-

(-sin(d*x+c)-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)*((-1+cos(d*x+c))/si

n(d*x+c))^(1/2)+420*I*B*sin(d*x+c)*(-(-sin(d*x+c)-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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*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))+420*B*(-(-sin(d*x+c)-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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*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))*sin(d*x+c)+441*B*cos(d*x+c)^3*sin(d*x+c)*2^(1/2)-105*I*B*cos(d*x+c)^4*2^(1/2)+105*I*B*cos(d

*x+c)^2*2^(1/2)-420*B*2^(1/2)*cos(d*x+c)*sin(d*x+c)+420*A*(-(-sin(d*x+c)-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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticF((-(-sin(d*x+c)-1+cos(d*x

+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*sin(d*x+c)+420*I*B*cos(d*x+c)*sin(d*x+c)*(-(-sin(d*x+c)-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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*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))-420*I*A*cos(d*x+c)^2*sin(d*x+c)*(-(-sin(d*x+

c)-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)*((-1+cos(d*x+c))/sin(d*x+c))^

(1/2)*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))+420*I*B*cos(d*x+c)^2*si

n(d*x+c)*EllipticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*(-(-sin(d*x+c)-1+cos(d*x+c))/si

n(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-420*I*B*cos(d

*x+c)^2*sin(d*x+c)*(-(-sin(d*x+c)-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

)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*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))-420*A*(-(-sin(d*x+c)-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)*((-

1+cos(d*x+c))/sin(d*x+c))^(1/2)*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

))*cos(d*x+c)*sin(d*x+c)+420*A*(-(-sin(d*x+c)-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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticF((-(-sin(d*x+c)-1+cos(d*x+c))/sin(d*x+c))^(1/2),1/2*

2^(1/2))*cos(d*x+c)*sin(d*x+c)+420*B*(-(-sin(d*x+c)-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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*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))*cos(d*x+c)*sin(d*x+c)+420*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))/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))-420*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)*(-(-sin(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))-420*A*((-1+cos(d*x+c))/s

in(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*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))+483*I*A*cos(d*x+

c)^3*sin(d*x+c)*2^(1/2)-420*I*A*cos(d*x+c)*sin(d*x+c)*2^(1/2)+420*I*A*cos(d*x+c)*sin(d*x+c)*(-(-sin(d*x+c)-1+c

os(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*

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))-420*I*B*cos(d*x+c)*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))*(-(-sin(d*x+c)-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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-420*I*A*cos(d*x+c)^3*

sin(d*x+c)*(-(-sin(d*x+c)-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)*((-1+c

os(d*x+c))/sin(d*x+c))^(1/2)*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))+

420*I*B*cos(d*x+c)^3*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))*(-(-sin(

d*x+c)-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)*((-1+cos(d*x+c))/sin(d*x+

c))^(1/2)-420*I*B*cos(d*x+c)^3*sin(d*x+c)*(-(-sin(d*x+c)-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)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*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)))*(cos(d*x+c)/sin(d*x+c))^(9/2)*sin(d*x+c)/cos(d*x+c)^5*2^(1/2)

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maxima [A] time = 1.66, size = 218, normalized size = 1.27 \[ -\frac {105 \, {\left (\sqrt {2} {\left (\left (2 i + 2\right ) \, A - \left (2 i - 2\right ) \, B\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \sqrt {2} {\left (\left (2 i + 2\right ) \, A - \left (2 i - 2\right ) \, B\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - \sqrt {2} {\left (\left (i - 1\right ) \, A + \left (i + 1\right ) \, B\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right )\right )} a^{3} + \frac {2 \, {\left (-420 i \, A - 420 \, B\right )} a^{3}}{\sqrt {\tan \left (d x + c\right )}} - \frac {70 \, {\left (4 \, A - 3 i \, B\right )} a^{3}}{\tan \left (d x + c\right )^{\frac {3}{2}}} + \frac {2 \, {\left (63 i \, A + 21 \, B\right )} a^{3}}{\tan \left (d x + c\right )^{\frac {5}{2}}} + \frac {30 \, A a^{3}}{\tan \left (d x + c\right )^{\frac {7}{2}}}}{105 \, d} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(9/2)*(a+I*a*tan(d*x+c))^3*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

-1/105*(105*(sqrt(2)*((2*I + 2)*A - (2*I - 2)*B)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + sqrt(2

)*((2*I + 2)*A - (2*I - 2)*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)/sq

rt(tan(d*x + c)) + 1/tan(d*x + c) + 1))*a^3 + 2*(-420*I*A - 420*B)*a^3/sqrt(tan(d*x + c)) - 70*(4*A - 3*I*B)*a

^3/tan(d*x + c)^(3/2) + 2*(63*I*A + 21*B)*a^3/tan(d*x + c)^(5/2) + 30*A*a^3/tan(d*x + c)^(7/2))/d

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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\mathrm {cot}\left (c+d\,x\right )}^{9/2}\,\left (A+B\,\mathrm {tan}\left (c+d\,x\right )\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3 \,d x \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^(9/2)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^3,x)

[Out]

int(cot(c + d*x)^(9/2)*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^3, x)

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sympy [F(-1)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**(9/2)*(a+I*a*tan(d*x+c))**3*(A+B*tan(d*x+c)),x)

[Out]

Timed out

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