3.74 \(\int \cot ^4(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx\)

Optimal. Leaf size=210 \[ \frac {\sqrt {a} (14 B+9 i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{8 d}-\frac {\sqrt {2} \sqrt {a} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {(6 B+i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}+\frac {(7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d} \]

[Out]

1/8*(9*I*A+14*B)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))*a^(1/2)/d-(I*A+B)*arctanh(1/2*(a+I*a*tan(d*x+c))^(1
/2)*2^(1/2)/a^(1/2))*2^(1/2)*a^(1/2)/d+1/8*(7*A-2*I*B)*cot(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/d-1/12*(I*A+6*B)*co
t(d*x+c)^2*(a+I*a*tan(d*x+c))^(1/2)/d-1/3*A*cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.75, antiderivative size = 210, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 7, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.194, Rules used = {3598, 3600, 3480, 206, 3599, 63, 208} \[ \frac {\sqrt {a} (14 B+9 i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{8 d}-\frac {\sqrt {2} \sqrt {a} (B+i A) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {(6 B+i A) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}+\frac {(7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Cot[c + d*x]^4*Sqrt[a + I*a*Tan[c + d*x]]*(A + B*Tan[c + d*x]),x]

[Out]

(Sqrt[a]*((9*I)*A + 14*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(8*d) - (Sqrt[2]*Sqrt[a]*(I*A + B)*ArcT
anh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d + ((7*A - (2*I)*B)*Cot[c + d*x]*Sqrt[a + I*a*Tan[c + d*x]
])/(8*d) - ((I*A + 6*B)*Cot[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(12*d) - (A*Cot[c + d*x]^3*Sqrt[a + I*a*Tan
[c + d*x]])/(3*d)

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 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]

Rule 3480

Int[Sqrt[(a_) + (b_.)*tan[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(2*a - x^2), x], x, Sq
rt[a + b*Tan[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 + b^2, 0]

Rule 3598

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*d - B*c)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(f
*(n + 1)*(c^2 + d^2)), x] - Dist[1/(a*(n + 1)*(c^2 + d^2)), Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n
+ 1)*Simp[A*(b*d*m - a*c*(n + 1)) - B*(b*c*m + a*d*(n + 1)) - a*(B*c - A*d)*(m + n + 1)*Tan[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[n, -1]

Rule 3599

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

Rule 3600

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

Rubi steps

\begin {align*} \int \cot ^4(c+d x) \sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \, dx &=-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {\int \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)} \left (\frac {1}{2} a (i A+6 B)-\frac {5}{2} a A \tan (c+d x)\right ) \, dx}{3 a}\\ &=-\frac {(i A+6 B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {\int \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{4} a^2 (7 A-2 i B)-\frac {3}{4} a^2 (i A+6 B) \tan (c+d x)\right ) \, dx}{6 a^2}\\ &=\frac {(7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {(i A+6 B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+\frac {\int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {3}{8} a^3 (9 i A+14 B)+\frac {3}{8} a^3 (7 A-2 i B) \tan (c+d x)\right ) \, dx}{6 a^3}\\ &=\frac {(7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {(i A+6 B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}+(A-i B) \int \sqrt {a+i a \tan (c+d x)} \, dx-\frac {(9 i A+14 B) \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx}{16 a}\\ &=\frac {(7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {(i A+6 B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {(2 a (i A+B)) \operatorname {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d}-\frac {(a (9 i A+14 B)) \operatorname {Subst}\left (\int \frac {1}{x \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{16 d}\\ &=-\frac {\sqrt {2} \sqrt {a} (i A+B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {(7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {(i A+6 B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}-\frac {(9 A-14 i B) \operatorname {Subst}\left (\int \frac {1}{i-\frac {i x^2}{a}} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{8 d}\\ &=\frac {\sqrt {a} (9 i A+14 B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{8 d}-\frac {\sqrt {2} \sqrt {a} (i A+B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}+\frac {(7 A-2 i B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{8 d}-\frac {(i A+6 B) \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{12 d}-\frac {A \cot ^3(c+d x) \sqrt {a+i a \tan (c+d x)}}{3 d}\\ \end {align*}

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Mathematica [A]  time = 4.96, size = 414, normalized size = 1.97 \[ \frac {\sqrt {a+i a \tan (c+d x)} (A+B \tan (c+d x)) \left (-\frac {2 i \left ((9 A-14 i B) \left (\log \left (\left (-1+e^{i (c+d x)}\right )^2\right )-\log \left (\left (1+e^{i (c+d x)}\right )^2\right )+\log \left (-2 e^{i (c+d x)} \left (1+\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}\right )+3 e^{2 i (c+d x)}+2 \sqrt {2} \sqrt {1+e^{2 i (c+d x)}}+3\right )-\log \left (2 e^{i (c+d x)} \left (1+\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}\right )+3 e^{2 i (c+d x)}+2 \sqrt {2} \sqrt {1+e^{2 i (c+d x)}}+3\right )\right )+32 \sqrt {2} (A-i B) \sinh ^{-1}\left (e^{i (c+d x)}\right )\right )}{\sqrt {\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {1+e^{2 i (c+d x)}}}-\frac {4 \csc ^3(c+d x) (2 (6 B+i A) \sin (2 (c+d x))+(29 A-6 i B) \cos (2 (c+d x))-13 A+6 i B)}{3 \sqrt {\sec (c+d x)}}\right )}{64 d \sec ^{\frac {3}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Cot[c + d*x]^4*Sqrt[a + I*a*Tan[c + d*x]]*(A + B*Tan[c + d*x]),x]

[Out]

((((-2*I)*(32*Sqrt[2]*(A - I*B)*ArcSinh[E^(I*(c + d*x))] + (9*A - (14*I)*B)*(Log[(-1 + E^(I*(c + d*x)))^2] - L
og[(1 + E^(I*(c + d*x)))^2] + Log[3 + 3*E^((2*I)*(c + d*x)) + 2*Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))] - 2*E^(I
*(c + d*x))*(1 + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])] - Log[3 + 3*E^((2*I)*(c + d*x)) + 2*Sqrt[2]*Sqrt[1 +
E^((2*I)*(c + d*x))] + 2*E^(I*(c + d*x))*(1 + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))])])))/(Sqrt[E^(I*(c + d*x))
/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]) - (4*Csc[c + d*x]^3*(-13*A + (6*I)*B + (29*A - (6*I
)*B)*Cos[2*(c + d*x)] + 2*(I*A + 6*B)*Sin[2*(c + d*x)]))/(3*Sqrt[Sec[c + d*x]]))*Sqrt[a + I*a*Tan[c + d*x]]*(A
+ B*Tan[c + d*x]))/(64*d*Sec[c + d*x]^(3/2)*(A*Cos[c + d*x] + B*Sin[c + d*x]))

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fricas [B]  time = 0.87, size = 824, normalized size = 3.92 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")

[Out]

-1/96*(24*(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)*sqrt(-(8*A^2 - 16*I*
A*B - 8*B^2)*a/d^2)*log(((4*I*A + 4*B)*a*e^(I*d*x + I*c) + sqrt(2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(8*A^2 -
16*I*A*B - 8*B^2)*a/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/(I*A + B)) - 24*(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)*sqrt(-(8*A^2 - 16*I*A*B - 8*B^2)*a/d^2)*log(((4
*I*A + 4*B)*a*e^(I*d*x + I*c) - sqrt(2)*(d*e^(2*I*d*x + 2*I*c) + d)*sqrt(-(8*A^2 - 16*I*A*B - 8*B^2)*a/d^2)*sq
rt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/(I*A + B)) - 3*(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)*sqrt(-(81*A^2 - 252*I*A*B - 196*B^2)*a/d^2)*log(((432*I*A + 672*B)*a^2*e^(2
*I*d*x + 2*I*c) + (144*I*A + 224*B)*a^2 + 32*sqrt(2)*(a*d*e^(3*I*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))*sqrt(-(81
*A^2 - 252*I*A*B - 196*B^2)*a/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/(9*I*A + 14*B)) + 3
*(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)*sqrt(-(81*A^2 - 252*I*A*B - 1
96*B^2)*a/d^2)*log(((432*I*A + 672*B)*a^2*e^(2*I*d*x + 2*I*c) + (144*I*A + 224*B)*a^2 - 32*sqrt(2)*(a*d*e^(3*I
*d*x + 3*I*c) + a*d*e^(I*d*x + I*c))*sqrt(-(81*A^2 - 252*I*A*B - 196*B^2)*a/d^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) +
1)))*e^(-2*I*d*x - 2*I*c)/(9*I*A + 14*B)) - 4*sqrt(2)*((31*I*A + 18*B)*e^(7*I*d*x + 7*I*c) + (5*I*A + 6*B)*e^
(5*I*d*x + 5*I*c) + (I*A - 18*B)*e^(3*I*d*x + 3*I*c) + (27*I*A - 6*B)*e^(I*d*x + I*c))*sqrt(a/(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 \[ \int {\left (B \tan \left (d x + c\right ) + A\right )} \sqrt {i \, a \tan \left (d x + c\right ) + a} \cot \left (d x + c\right )^{4}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)*sqrt(I*a*tan(d*x + c) + a)*cot(d*x + c)^4, x)

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maple [B]  time = 3.38, size = 1783, normalized size = 8.49 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x)

[Out]

-1/48/d*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(42*A*cos(d*x+c)-24*B*sin(d*x+c)*cos(d*x+c)^2+36*B*cos(
d*x+c)^3*sin(d*x+c)-46*A*cos(d*x+c)^2+62*A*cos(d*x+c)^4-12*B*cos(d*x+c)*sin(d*x+c)+27*A*(-2*cos(d*x+c)/(1+cos(
d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+42*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(-(-
(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))-36*I*B*cos(d*x+c)^4+12*I*B*cos(d*x+c
)^3+36*I*B*cos(d*x+c)^2-12*I*B*cos(d*x+c)-58*A*cos(d*x+c)^3+48*I*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan
h(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*cos(d*x+c)^4*2^(1/2)-48*I*B*(-2*cos(
d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*cos(d*x+c)^4*2^(1/2)-96*
I*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c
)*2^(1/2))*cos(d*x+c)^2*2^(1/2)+96*I*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d
*x+c)))^(1/2)*2^(1/2))*cos(d*x+c)^2*2^(1/2)+27*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/
(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^4+42*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+
c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*cos(d*x+c)^4-54*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(
1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^2-84*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(-(-(-2*cos(d
*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*cos(d*x+c)^2+48*A*(-2*cos(d*x+c)/(1+cos(d*x+c
)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2)+48*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)+62*I*A*cos(d*x+
c)^3*sin(d*x+c)-4*I*A*cos(d*x+c)^2*sin(d*x+c)+27*I*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(-(-(-2*cos(d*x+c)
/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))-42*I*A*cos(d*x+c)*sin(d*x+c)-42*I*B*(-2*cos(d*x+c)
/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))+48*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2
)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*cos(d*x+c)^4*2^(1/2)+48*B*(-2*cos(d*x+c)/(1+cos(d*x
+c)))^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*cos(d*x+c)^4*2^(1/
2)-96*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*cos(d*x+
c)^2*2^(1/2)-96*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*
x+c)/cos(d*x+c)*2^(1/2))*cos(d*x+c)^2*2^(1/2)+27*I*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*ln(-(-(-2*cos(d*x+c)
/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*cos(d*x+c)^4-42*I*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))
^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^4-54*I*A*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)
*ln(-(-(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*cos(d*x+c)^2+84*I*B*(-2*cos(d
*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2))*cos(d*x+c)^2+48*I*A*(-2*cos(d*x+c)/
(1+cos(d*x+c)))^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)-
48*I*B*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))*2^(1/2))/
(I*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c)^3

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maxima [A]  time = 0.70, size = 249, normalized size = 1.19 \[ \frac {i \, a^{3} {\left (\frac {2 \, {\left (3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} {\left (7 \, A - 2 i \, B\right )} - 40 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} A a + 3 \, \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (9 \, A + 2 i \, B\right )} a^{2}\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{3} a^{2} - 3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} a^{3} + 3 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a^{4} - a^{5}} + \frac {24 \, \sqrt {2} {\left (A - i \, B\right )} \log \left (-\frac {\sqrt {2} \sqrt {a} - \sqrt {i \, a \tan \left (d x + c\right ) + a}}{\sqrt {2} \sqrt {a} + \sqrt {i \, a \tan \left (d x + c\right ) + a}}\right )}{a^{\frac {5}{2}}} - \frac {3 \, {\left (9 \, A - 14 i \, B\right )} \log \left (\frac {\sqrt {i \, a \tan \left (d x + c\right ) + a} - \sqrt {a}}{\sqrt {i \, a \tan \left (d x + c\right ) + a} + \sqrt {a}}\right )}{a^{\frac {5}{2}}}\right )}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^4*(a+I*a*tan(d*x+c))^(1/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")

[Out]

1/48*I*a^3*(2*(3*(I*a*tan(d*x + c) + a)^(5/2)*(7*A - 2*I*B) - 40*(I*a*tan(d*x + c) + a)^(3/2)*A*a + 3*sqrt(I*a
*tan(d*x + c) + a)*(9*A + 2*I*B)*a^2)/((I*a*tan(d*x + c) + a)^3*a^2 - 3*(I*a*tan(d*x + c) + a)^2*a^3 + 3*(I*a*
tan(d*x + c) + a)*a^4 - a^5) + 24*sqrt(2)*(A - I*B)*log(-(sqrt(2)*sqrt(a) - sqrt(I*a*tan(d*x + c) + a))/(sqrt(
2)*sqrt(a) + sqrt(I*a*tan(d*x + c) + a)))/a^(5/2) - 3*(9*A - 14*I*B)*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a)
)/(sqrt(I*a*tan(d*x + c) + a) + sqrt(a)))/a^(5/2))/d

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mupad [B]  time = 7.09, size = 735, normalized size = 3.50 \[ -\frac {\frac {\left (9\,A\,a^3+B\,a^3\,2{}\mathrm {i}\right )\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,1{}\mathrm {i}}{8\,d}+\frac {\left (7\,A\,a-B\,a\,2{}\mathrm {i}\right )\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}\,1{}\mathrm {i}}{8\,d}-\frac {A\,a^2\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{3/2}\,5{}\mathrm {i}}{3\,d}}{3\,a\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^2-3\,a^2\,\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )-{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^3+a^3}-\frac {\mathrm {atan}\left (\frac {47\,\sqrt {32}\,A^3\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{8\,\left (47{}\mathrm {i}\,d\,A^3\,a^5+51\,d\,A^2\,B\,a^5+64{}\mathrm {i}\,d\,A\,B^2\,a^5+68\,d\,B^3\,a^5\right )}-\frac {\sqrt {32}\,B^3\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,17{}\mathrm {i}}{2\,\left (47{}\mathrm {i}\,d\,A^3\,a^5+51\,d\,A^2\,B\,a^5+64{}\mathrm {i}\,d\,A\,B^2\,a^5+68\,d\,B^3\,a^5\right )}+\frac {8\,\sqrt {32}\,A\,B^2\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{47{}\mathrm {i}\,d\,A^3\,a^5+51\,d\,A^2\,B\,a^5+64{}\mathrm {i}\,d\,A\,B^2\,a^5+68\,d\,B^3\,a^5}-\frac {\sqrt {32}\,A^2\,B\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,51{}\mathrm {i}}{8\,\left (47{}\mathrm {i}\,d\,A^3\,a^5+51\,d\,A^2\,B\,a^5+64{}\mathrm {i}\,d\,A\,B^2\,a^5+68\,d\,B^3\,a^5\right )}\right )\,\left (B+A\,1{}\mathrm {i}\right )\,\sqrt {\frac {a}{32}}\,8{}\mathrm {i}}{d}+\frac {\sqrt {a}\,\mathrm {atan}\left (\frac {423\,A^3\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{8\,\left (\frac {423{}\mathrm {i}\,d\,A^3\,a^5}{8}+\frac {347\,d\,A^2\,B\,a^5}{4}+\frac {139{}\mathrm {i}\,d\,A\,B^2\,a^5}{2}+119\,d\,B^3\,a^5\right )}-\frac {B^3\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,119{}\mathrm {i}}{\frac {423{}\mathrm {i}\,d\,A^3\,a^5}{8}+\frac {347\,d\,A^2\,B\,a^5}{4}+\frac {139{}\mathrm {i}\,d\,A\,B^2\,a^5}{2}+119\,d\,B^3\,a^5}+\frac {139\,A\,B^2\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}}{2\,\left (\frac {423{}\mathrm {i}\,d\,A^3\,a^5}{8}+\frac {347\,d\,A^2\,B\,a^5}{4}+\frac {139{}\mathrm {i}\,d\,A\,B^2\,a^5}{2}+119\,d\,B^3\,a^5\right )}-\frac {A^2\,B\,a^{9/2}\,d\,\sqrt {a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}}\,347{}\mathrm {i}}{4\,\left (\frac {423{}\mathrm {i}\,d\,A^3\,a^5}{8}+\frac {347\,d\,A^2\,B\,a^5}{4}+\frac {139{}\mathrm {i}\,d\,A\,B^2\,a^5}{2}+119\,d\,B^3\,a^5\right )}\right )\,\left (14\,B+A\,9{}\mathrm {i}\right )\,1{}\mathrm {i}}{8\,d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cot(c + d*x)^4*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(1/2),x)

[Out]

(a^(1/2)*atan((423*A^3*a^(9/2)*d*(a + a*tan(c + d*x)*1i)^(1/2))/(8*((A^3*a^5*d*423i)/8 + 119*B^3*a^5*d + (A*B^
2*a^5*d*139i)/2 + (347*A^2*B*a^5*d)/4)) - (B^3*a^(9/2)*d*(a + a*tan(c + d*x)*1i)^(1/2)*119i)/((A^3*a^5*d*423i)
/8 + 119*B^3*a^5*d + (A*B^2*a^5*d*139i)/2 + (347*A^2*B*a^5*d)/4) + (139*A*B^2*a^(9/2)*d*(a + a*tan(c + d*x)*1i
)^(1/2))/(2*((A^3*a^5*d*423i)/8 + 119*B^3*a^5*d + (A*B^2*a^5*d*139i)/2 + (347*A^2*B*a^5*d)/4)) - (A^2*B*a^(9/2
)*d*(a + a*tan(c + d*x)*1i)^(1/2)*347i)/(4*((A^3*a^5*d*423i)/8 + 119*B^3*a^5*d + (A*B^2*a^5*d*139i)/2 + (347*A
^2*B*a^5*d)/4)))*(A*9i + 14*B)*1i)/(8*d) - (atan((47*32^(1/2)*A^3*a^(9/2)*d*(a + a*tan(c + d*x)*1i)^(1/2))/(8*
(A^3*a^5*d*47i + 68*B^3*a^5*d + A*B^2*a^5*d*64i + 51*A^2*B*a^5*d)) - (32^(1/2)*B^3*a^(9/2)*d*(a + a*tan(c + d*
x)*1i)^(1/2)*17i)/(2*(A^3*a^5*d*47i + 68*B^3*a^5*d + A*B^2*a^5*d*64i + 51*A^2*B*a^5*d)) + (8*32^(1/2)*A*B^2*a^
(9/2)*d*(a + a*tan(c + d*x)*1i)^(1/2))/(A^3*a^5*d*47i + 68*B^3*a^5*d + A*B^2*a^5*d*64i + 51*A^2*B*a^5*d) - (32
^(1/2)*A^2*B*a^(9/2)*d*(a + a*tan(c + d*x)*1i)^(1/2)*51i)/(8*(A^3*a^5*d*47i + 68*B^3*a^5*d + A*B^2*a^5*d*64i +
51*A^2*B*a^5*d)))*(A*1i + B)*(a/32)^(1/2)*8i)/d - (((9*A*a^3 + B*a^3*2i)*(a + a*tan(c + d*x)*1i)^(1/2)*1i)/(8
*d) + ((7*A*a - B*a*2i)*(a + a*tan(c + d*x)*1i)^(5/2)*1i)/(8*d) - (A*a^2*(a + a*tan(c + d*x)*1i)^(3/2)*5i)/(3*
d))/(3*a*(a + a*tan(c + d*x)*1i)^2 - 3*a^2*(a + a*tan(c + d*x)*1i) - (a + a*tan(c + d*x)*1i)^3 + a^3)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {i a \left (\tan {\left (c + d x \right )} - i\right )} \left (A + B \tan {\left (c + d x \right )}\right ) \cot ^{4}{\left (c + d x \right )}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)**4*(a+I*a*tan(d*x+c))**(1/2)*(A+B*tan(d*x+c)),x)

[Out]

Integral(sqrt(I*a*(tan(c + d*x) - I))*(A + B*tan(c + d*x))*cot(c + d*x)**4, x)

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