3.1.80 \(\int \cot ^3(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx\) [80]
Optimal. Leaf size=171 \[ \frac {a^{3/2} (11 A-12 i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 d}-\frac {2 \sqrt {2} a^{3/2} (A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {a (5 i A+4 B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {a A \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d} \]
[Out]
1/4*a^(3/2)*(11*A-12*I*B)*arctanh((a+I*a*tan(d*x+c))^(1/2)/a^(1/2))/d-2*a^(3/2)*(A-I*B)*arctanh(1/2*(a+I*a*tan
(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*2^(1/2)/d-1/4*a*(5*I*A+4*B)*cot(d*x+c)*(a+I*a*tan(d*x+c))^(1/2)/d-1/2*a*A*cot(
d*x+c)^2*(a+I*a*tan(d*x+c))^(1/2)/d
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Rubi [A]
time = 0.39, antiderivative size = 171, normalized size of antiderivative = 1.00, number of steps
used = 8, number of rules used = 8, integrand size = 36, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.222, Rules used = {3674, 3679,
3681, 3561, 212, 3680, 65, 214} \begin {gather*} \frac {a^{3/2} (11 A-12 i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 d}-\frac {2 \sqrt {2} a^{3/2} (A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {a (4 B+5 i A) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {a A \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]),x]
[Out]
(a^(3/2)*(11*A - (12*I)*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(4*d) - (2*Sqrt[2]*a^(3/2)*(A - I*B)*A
rcTanh[Sqrt[a + I*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (a*((5*I)*A + 4*B)*Cot[c + d*x]*Sqrt[a + I*a*Tan[c +
d*x]])/(4*d) - (a*A*Cot[c + d*x]^2*Sqrt[a + I*a*Tan[c + d*x]])/(2*d)
Rule 65
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 212
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 214
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-a/b, 2]/a)*ArcTanh[x/Rt[-a/b, 2]], x] /; FreeQ[{a, b},
x] && NegQ[a/b]
Rule 3561
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 3674
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] && E
qQ[a^2 + b^2, 0] && GtQ[m, 1] && LtQ[n, -1]
Rule 3679
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 3680
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 3681
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 ^3(c+d x) (a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \, dx &=-\frac {a A \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}+\frac {1}{2} \int \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)} \left (\frac {1}{2} a (5 i A+4 B)-\frac {1}{2} a (3 A-4 i B) \tan (c+d x)\right ) \, dx\\ &=-\frac {a (5 i A+4 B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {a A \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}+\frac {\int \cot (c+d x) \sqrt {a+i a \tan (c+d x)} \left (-\frac {1}{4} a^2 (11 A-12 i B)-\frac {1}{4} a^2 (5 i A+4 B) \tan (c+d x)\right ) \, dx}{2 a}\\ &=-\frac {a (5 i A+4 B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {a A \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}+\frac {1}{8} (-11 A+12 i B) \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt {a+i a \tan (c+d x)} \, dx-(2 a (i A+B)) \int \sqrt {a+i a \tan (c+d x)} \, dx\\ &=-\frac {a (5 i A+4 B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {a A \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}-\frac {\left (4 a^2 (A-i B)\right ) \text {Subst}\left (\int \frac {1}{2 a-x^2} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{d}-\frac {\left (a^2 (11 A-12 i B)\right ) \text {Subst}\left (\int \frac {1}{x \sqrt {a+i a x}} \, dx,x,\tan (c+d x)\right )}{8 d}\\ &=-\frac {2 \sqrt {2} a^{3/2} (A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {a (5 i A+4 B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {a A \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}+\frac {(a (11 i A+12 B)) \text {Subst}\left (\int \frac {1}{i-\frac {i x^2}{a}} \, dx,x,\sqrt {a+i a \tan (c+d x)}\right )}{4 d}\\ &=\frac {a^{3/2} (11 A-12 i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {a}}\right )}{4 d}-\frac {2 \sqrt {2} a^{3/2} (A-i B) \tanh ^{-1}\left (\frac {\sqrt {a+i a \tan (c+d x)}}{\sqrt {2} \sqrt {a}}\right )}{d}-\frac {a (5 i A+4 B) \cot (c+d x) \sqrt {a+i a \tan (c+d x)}}{4 d}-\frac {a A \cot ^2(c+d x) \sqrt {a+i a \tan (c+d x)}}{2 d}\\ \end {align*}
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Mathematica [B] Both result and optimal contain complex but leaf count is larger than twice
the leaf count of optimal. \(400\) vs. \(2(171)=342\).
time = 5.85, size = 400, normalized size = 2.34 \begin {gather*} \frac {(a+i a \tan (c+d x))^{3/2} (A+B \tan (c+d x)) \left (\frac {-64 \sqrt {2} (A-i B) \sinh ^{-1}\left (e^{i (c+d x)}\right )-2 (11 A-12 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 (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)} \left (1+\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}\right )\right )-\log \left (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)} \left (1+\sqrt {2} \sqrt {1+e^{2 i (c+d x)}}\right )\right )\right )}{\left (\frac {e^{i (c+d x)}}{1+e^{2 i (c+d x)}}\right )^{3/2} \left (1+e^{2 i (c+d x)}\right )^{3/2}}+\frac {8 i \csc (c+d x) (2 A \csc (c+d x)+(5 i A+4 B) \sec (c+d x)) (i+\tan (c+d x))}{\sec ^{\frac {5}{2}}(c+d x)}\right )}{32 d \sec ^{\frac {5}{2}}(c+d x) (A \cos (c+d x)+B \sin (c+d x))} \end {gather*}
Warning: Unable to verify antiderivative.
[In]
Integrate[Cot[c + d*x]^3*(a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x]),x]
[Out]
((a + I*a*Tan[c + d*x])^(3/2)*(A + B*Tan[c + d*x])*((-64*Sqrt[2]*(A - I*B)*ArcSinh[E^(I*(c + d*x))] - 2*(11*A
- (12*I)*B)*(Log[(-1 + E^(I*(c + d*x)))^2] - Log[(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))])]))/((E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x))))^(3/2)*(1 + E^((2*I)*(c + d*x)))^(3/2)) +
((8*I)*Csc[c + d*x]*(2*A*Csc[c + d*x] + ((5*I)*A + 4*B)*Sec[c + d*x])*(I + Tan[c + d*x]))/Sec[c + d*x]^(5/2)))
/(32*d*Sec[c + d*x]^(5/2)*(A*Cos[c + d*x] + B*Sin[c + d*x]))
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Maple [B] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 1289 vs. \(2 (139 ) = 278\).
time = 0.59, size = 1290, normalized size = 7.54
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x,method=_RETURNVERBOSE)
[Out]
1/8/d*a*((I*sin(d*x+c)+cos(d*x+c))*a/cos(d*x+c))^(1/2)*(-10*A*cos(d*x+c)-8*B*cos(d*x+c)*sin(d*x+c)+16*I*A*(-2*
cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2*sin(d*x
+c)*2^(1/2)+16*I*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/
2)*sin(d*x+c)/cos(d*x+c))*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)+11*I*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1
/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2*sin(d*x+c)-4*A*cos(d*x+c)^2+16*A*2^(1/2)*arctanh(1/2*2^(1/
2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)
-16*B*2^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*si
n(d*x+c)-8*I*B*cos(d*x+c)^3+8*I*B*cos(d*x+c)+12*I*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln((sin(d*x+c)*(-2*co
s(d*x+c)/(cos(d*x+c)+1))^(1/2)-cos(d*x+c)+1)/sin(d*x+c))*cos(d*x+c)^2*sin(d*x+c)-16*I*A*(-2*cos(d*x+c)/(cos(d*
x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*sin(d*x+c)*2^(1/2)-16*I*B*(-2*cos(d*x+
c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))*sin(d
*x+c)*2^(1/2)-16*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/
2)*sin(d*x+c)/cos(d*x+c))*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)+16*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2
*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)^2*sin(d*x+c)*2^(1/2)+8*B*cos(d*x+c)^2*sin(d*x+c)-11*
A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln((sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-cos(d*x+c)+1)/sin(d
*x+c))*cos(d*x+c)^2*sin(d*x+c)+12*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1
))^(1/2))*cos(d*x+c)^2*sin(d*x+c)-11*I*A*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*(-2*cos(d*x+c)/(cos(d*
x+c)+1))^(1/2)*sin(d*x+c)-12*I*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*ln((sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c
)+1))^(1/2)-cos(d*x+c)+1)/sin(d*x+c))*sin(d*x+c)+14*A*cos(d*x+c)^3+14*I*A*cos(d*x+c)^2*sin(d*x+c)-10*I*A*cos(d
*x+c)*sin(d*x+c)-12*B*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*sin(
d*x+c)+11*A*ln((sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)-cos(d*x+c)+1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(
d*x+c)+1))^(1/2)*sin(d*x+c))/(-1+cos(d*x+c))/(I*sin(d*x+c)+cos(d*x+c)-1)/(cos(d*x+c)+1)
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Maxima [A]
time = 0.49, size = 203, normalized size = 1.19 \begin {gather*} \frac {{\left (\frac {8 \, \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 )}{\sqrt {a}} - \frac {{\left (11 \, A - 12 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 )}{\sqrt {a}} + \frac {2 \, {\left ({\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {3}{2}} {\left (5 \, A - 4 i \, B\right )} - \sqrt {i \, a \tan \left (d x + c\right ) + a} {\left (3 \, A - 4 i \, B\right )} a\right )}}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{2} - 2 \, {\left (i \, a \tan \left (d x + c\right ) + a\right )} a + a^{2}}\right )} a^{2}}{8 \, d} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")
[Out]
1/8*(8*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)))/sqrt(a) - (11*A - 12*I*B)*log((sqrt(I*a*tan(d*x + c) + a) - sqrt(a))/(sqrt(I*a*tan(d*x + c) + a
) + sqrt(a)))/sqrt(a) + 2*((I*a*tan(d*x + c) + a)^(3/2)*(5*A - 4*I*B) - sqrt(I*a*tan(d*x + c) + a)*(3*A - 4*I*
B)*a)/((I*a*tan(d*x + c) + a)^2 - 2*(I*a*tan(d*x + c) + a)*a + a^2))*a^2/d
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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice
the leaf count of optimal. 762 vs. \(2 (132) = 264\).
time = 2.75, size = 762, normalized size = 4.46 \begin {gather*} -\frac {16 \, \sqrt {2} \sqrt {\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{3}}{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 {4 \, {\left ({\left (-i \, A - B\right )} a^{2} e^{\left (i \, d x + i \, c\right )} - \sqrt {\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{3}}{d^{2}}} {\left (i \, d e^{\left (2 i \, d x + 2 i \, c\right )} + i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a}\right ) - 16 \, \sqrt {2} \sqrt {\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{3}}{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 {4 \, {\left ({\left (-i \, A - B\right )} a^{2} e^{\left (i \, d x + i \, c\right )} - \sqrt {\frac {{\left (A^{2} - 2 i \, A B - B^{2}\right )} a^{3}}{d^{2}}} {\left (-i \, d e^{\left (2 i \, d x + 2 i \, c\right )} - i \, d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{{\left (-i \, A - B\right )} a}\right ) + \sqrt {\frac {{\left (121 \, A^{2} - 264 i \, A B - 144 \, B^{2}\right )} a^{3}}{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 {16 \, {\left (3 \, {\left (-11 i \, A - 12 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-11 i \, A - 12 \, B\right )} a^{2} + 2 \, \sqrt {2} \sqrt {\frac {{\left (121 \, A^{2} - 264 i \, A B - 144 \, B^{2}\right )} a^{3}}{d^{2}}} {\left (i \, d e^{\left (3 i \, d x + 3 i \, c\right )} + i \, d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{11 i \, A + 12 \, B}\right ) - \sqrt {\frac {{\left (121 \, A^{2} - 264 i \, A B - 144 \, B^{2}\right )} a^{3}}{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 {16 \, {\left (3 \, {\left (-11 i \, A - 12 \, B\right )} a^{2} e^{\left (2 i \, d x + 2 i \, c\right )} + {\left (-11 i \, A - 12 \, B\right )} a^{2} + 2 \, \sqrt {2} \sqrt {\frac {{\left (121 \, A^{2} - 264 i \, A B - 144 \, B^{2}\right )} a^{3}}{d^{2}}} {\left (-i \, d e^{\left (3 i \, d x + 3 i \, c\right )} - i \, d e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}\right )} e^{\left (-2 i \, d x - 2 i \, c\right )}}{11 i \, A + 12 \, B}\right ) - 4 \, \sqrt {2} {\left ({\left (7 \, A - 4 i \, B\right )} a e^{\left (5 i \, d x + 5 i \, c\right )} + 4 \, A a e^{\left (3 i \, d x + 3 i \, c\right )} - {\left (3 \, A - 4 i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}}}{16 \, {\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 {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")
[Out]
-1/16*(16*sqrt(2)*sqrt((A^2 - 2*I*A*B - B^2)*a^3/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*lo
g(4*((-I*A - B)*a^2*e^(I*d*x + I*c) - sqrt((A^2 - 2*I*A*B - B^2)*a^3/d^2)*(I*d*e^(2*I*d*x + 2*I*c) + I*d)*sqrt
(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/((-I*A - B)*a)) - 16*sqrt(2)*sqrt((A^2 - 2*I*A*B - B^2)*a^3/d^
2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c) + d)*log(4*((-I*A - B)*a^2*e^(I*d*x + I*c) - sqrt((A^2 - 2
*I*A*B - B^2)*a^3/d^2)*(-I*d*e^(2*I*d*x + 2*I*c) - I*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-I*d*x - I*c)/((
-I*A - B)*a)) + sqrt((121*A^2 - 264*I*A*B - 144*B^2)*a^3/d^2)*(d*e^(4*I*d*x + 4*I*c) - 2*d*e^(2*I*d*x + 2*I*c)
+ d)*log(-16*(3*(-11*I*A - 12*B)*a^2*e^(2*I*d*x + 2*I*c) + (-11*I*A - 12*B)*a^2 + 2*sqrt(2)*sqrt((121*A^2 - 2
64*I*A*B - 144*B^2)*a^3/d^2)*(I*d*e^(3*I*d*x + 3*I*c) + I*d*e^(I*d*x + I*c))*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))
)*e^(-2*I*d*x - 2*I*c)/(11*I*A + 12*B)) - sqrt((121*A^2 - 264*I*A*B - 144*B^2)*a^3/d^2)*(d*e^(4*I*d*x + 4*I*c)
- 2*d*e^(2*I*d*x + 2*I*c) + d)*log(-16*(3*(-11*I*A - 12*B)*a^2*e^(2*I*d*x + 2*I*c) + (-11*I*A - 12*B)*a^2 + 2
*sqrt(2)*sqrt((121*A^2 - 264*I*A*B - 144*B^2)*a^3/d^2)*(-I*d*e^(3*I*d*x + 3*I*c) - I*d*e^(I*d*x + I*c))*sqrt(a
/(e^(2*I*d*x + 2*I*c) + 1)))*e^(-2*I*d*x - 2*I*c)/(11*I*A + 12*B)) - 4*sqrt(2)*((7*A - 4*I*B)*a*e^(5*I*d*x + 5
*I*c) + 4*A*a*e^(3*I*d*x + 3*I*c) - (3*A - 4*I*B)*a*e^(I*d*x + I*c))*sqrt(a/(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]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \left (i a \left (\tan {\left (c + d x \right )} - i\right )\right )^{\frac {3}{2}} \left (A + B \tan {\left (c + d x \right )}\right ) \cot ^{3}{\left (c + d x \right )}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)**3*(a+I*a*tan(d*x+c))**(3/2)*(A+B*tan(d*x+c)),x)
[Out]
Integral((I*a*(tan(c + d*x) - I))**(3/2)*(A + B*tan(c + d*x))*cot(c + d*x)**3, x)
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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^3*(a+I*a*tan(d*x+c))^(3/2)*(A+B*tan(d*x+c)),x, algorithm="giac")
[Out]
integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(3/2)*cot(d*x + c)^3, x)
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Mupad [B]
time = 7.82, size = 2500, normalized size = 14.62 \begin {gather*} \text {Too large to display} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(c + d*x)^3*(A + B*tan(c + d*x))*(a + a*tan(c + d*x)*1i)^(3/2),x)
[Out]
2*atanh((3*d^4*(a + a*tan(c + d*x)*1i)^(1/2)*((249*A^2*a^3)/(128*d^2) - ((49*A^4*a^18)/(4*d^4) + (64*B^4*a^18)
/d^4 + (40*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*64i)/d^4 + (A^3*B*a^18*28i)/d^4)^(1/2)/(64*a^6) - (17*B^2*a^3)/(8*d
^2) - (A*B*a^3*65i)/(16*d^2))^(1/2)*((49*A^4*a^18)/(4*d^4) + (64*B^4*a^18)/d^4 + (40*A^2*B^2*a^18)/d^4 + (A*B^
3*a^18*64i)/d^4 + (A^3*B*a^18*28i)/d^4)^(1/2))/(2*((133*A^3*a^11*d)/16 - B^3*a^11*d*20i + 29*A*B^2*a^11*d + (A
^2*B*a^11*d*3i)/4 + (3*A*a^2*d^3*((49*A^4*a^18)/(4*d^4) + (64*B^4*a^18)/d^4 + (40*A^2*B^2*a^18)/d^4 + (A*B^3*a
^18*64i)/d^4 + (A^3*B*a^18*28i)/d^4)^(1/2))/8 - (B*a^2*d^3*((49*A^4*a^18)/(4*d^4) + (64*B^4*a^18)/d^4 + (40*A^
2*B^2*a^18)/d^4 + (A*B^3*a^18*64i)/d^4 + (A^3*B*a^18*28i)/d^4)^(1/2)*1i)/2)) + (7*A^2*a^6*d^2*(a + a*tan(c + d
*x)*1i)^(1/2)*((249*A^2*a^3)/(128*d^2) - ((49*A^4*a^18)/(4*d^4) + (64*B^4*a^18)/d^4 + (40*A^2*B^2*a^18)/d^4 +
(A*B^3*a^18*64i)/d^4 + (A^3*B*a^18*28i)/d^4)^(1/2)/(64*a^6) - (17*B^2*a^3)/(8*d^2) - (A*B*a^3*65i)/(16*d^2))^(
1/2))/(4*((133*A^3*a^8*d)/16 - B^3*a^8*d*20i + 29*A*B^2*a^8*d + (A^2*B*a^8*d*3i)/4 + (3*A*d^3*((49*A^4*a^18)/(
4*d^4) + (64*B^4*a^18)/d^4 + (40*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*64i)/d^4 + (A^3*B*a^18*28i)/d^4)^(1/2))/(8*a)
- (B*d^3*((49*A^4*a^18)/(4*d^4) + (64*B^4*a^18)/d^4 + (40*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*64i)/d^4 + (A^3*B*a
^18*28i)/d^4)^(1/2)*1i)/(2*a))) + (4*B^2*a^6*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*((249*A^2*a^3)/(128*d^2) - ((49
*A^4*a^18)/(4*d^4) + (64*B^4*a^18)/d^4 + (40*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*64i)/d^4 + (A^3*B*a^18*28i)/d^4)^
(1/2)/(64*a^6) - (17*B^2*a^3)/(8*d^2) - (A*B*a^3*65i)/(16*d^2))^(1/2))/((133*A^3*a^8*d)/16 - B^3*a^8*d*20i + 2
9*A*B^2*a^8*d + (A^2*B*a^8*d*3i)/4 + (3*A*d^3*((49*A^4*a^18)/(4*d^4) + (64*B^4*a^18)/d^4 + (40*A^2*B^2*a^18)/d
^4 + (A*B^3*a^18*64i)/d^4 + (A^3*B*a^18*28i)/d^4)^(1/2))/(8*a) - (B*d^3*((49*A^4*a^18)/(4*d^4) + (64*B^4*a^18)
/d^4 + (40*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*64i)/d^4 + (A^3*B*a^18*28i)/d^4)^(1/2)*1i)/(2*a)) + (A*B*a^6*d^2*(a
+ a*tan(c + d*x)*1i)^(1/2)*((249*A^2*a^3)/(128*d^2) - ((49*A^4*a^18)/(4*d^4) + (64*B^4*a^18)/d^4 + (40*A^2*B^
2*a^18)/d^4 + (A*B^3*a^18*64i)/d^4 + (A^3*B*a^18*28i)/d^4)^(1/2)/(64*a^6) - (17*B^2*a^3)/(8*d^2) - (A*B*a^3*65
i)/(16*d^2))^(1/2)*2i)/((133*A^3*a^8*d)/16 - B^3*a^8*d*20i + 29*A*B^2*a^8*d + (A^2*B*a^8*d*3i)/4 + (3*A*d^3*((
49*A^4*a^18)/(4*d^4) + (64*B^4*a^18)/d^4 + (40*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*64i)/d^4 + (A^3*B*a^18*28i)/d^4
)^(1/2))/(8*a) - (B*d^3*((49*A^4*a^18)/(4*d^4) + (64*B^4*a^18)/d^4 + (40*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*64i)/
d^4 + (A^3*B*a^18*28i)/d^4)^(1/2)*1i)/(2*a)))*((249*A^2*a^3)/(128*d^2) - ((49*A^4*a^18)/(4*d^4) + (64*B^4*a^18
)/d^4 + (40*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*64i)/d^4 + (A^3*B*a^18*28i)/d^4)^(1/2)/(64*a^6) - (17*B^2*a^3)/(8*
d^2) - (A*B*a^3*65i)/(16*d^2))^(1/2) - (((3*A*a^3 - B*a^3*4i)*(a + a*tan(c + d*x)*1i)^(1/2))/(4*d) - ((5*A*a^2
- B*a^2*4i)*(a + a*tan(c + d*x)*1i)^(3/2))/(4*d))/((a + a*tan(c + d*x)*1i)^2 - 2*a*(a + a*tan(c + d*x)*1i) +
a^2) + 2*atanh((7*A^2*a^6*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*(((49*A^4*a^18)/(4*d^4) + (64*B^4*a^18)/d^4 + (40*
A^2*B^2*a^18)/d^4 + (A*B^3*a^18*64i)/d^4 + (A^3*B*a^18*28i)/d^4)^(1/2)/(64*a^6) + (249*A^2*a^3)/(128*d^2) - (1
7*B^2*a^3)/(8*d^2) - (A*B*a^3*65i)/(16*d^2))^(1/2))/(4*((133*A^3*a^8*d)/16 - B^3*a^8*d*20i + 29*A*B^2*a^8*d +
(A^2*B*a^8*d*3i)/4 - (3*A*d^3*((49*A^4*a^18)/(4*d^4) + (64*B^4*a^18)/d^4 + (40*A^2*B^2*a^18)/d^4 + (A*B^3*a^18
*64i)/d^4 + (A^3*B*a^18*28i)/d^4)^(1/2))/(8*a) + (B*d^3*((49*A^4*a^18)/(4*d^4) + (64*B^4*a^18)/d^4 + (40*A^2*B
^2*a^18)/d^4 + (A*B^3*a^18*64i)/d^4 + (A^3*B*a^18*28i)/d^4)^(1/2)*1i)/(2*a))) - (3*d^4*(a + a*tan(c + d*x)*1i)
^(1/2)*(((49*A^4*a^18)/(4*d^4) + (64*B^4*a^18)/d^4 + (40*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*64i)/d^4 + (A^3*B*a^1
8*28i)/d^4)^(1/2)/(64*a^6) + (249*A^2*a^3)/(128*d^2) - (17*B^2*a^3)/(8*d^2) - (A*B*a^3*65i)/(16*d^2))^(1/2)*((
49*A^4*a^18)/(4*d^4) + (64*B^4*a^18)/d^4 + (40*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*64i)/d^4 + (A^3*B*a^18*28i)/d^4
)^(1/2))/(2*((133*A^3*a^11*d)/16 - B^3*a^11*d*20i + 29*A*B^2*a^11*d + (A^2*B*a^11*d*3i)/4 - (3*A*a^2*d^3*((49*
A^4*a^18)/(4*d^4) + (64*B^4*a^18)/d^4 + (40*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*64i)/d^4 + (A^3*B*a^18*28i)/d^4)^(
1/2))/8 + (B*a^2*d^3*((49*A^4*a^18)/(4*d^4) + (64*B^4*a^18)/d^4 + (40*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*64i)/d^4
+ (A^3*B*a^18*28i)/d^4)^(1/2)*1i)/2)) + (4*B^2*a^6*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*(((49*A^4*a^18)/(4*d^4)
+ (64*B^4*a^18)/d^4 + (40*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*64i)/d^4 + (A^3*B*a^18*28i)/d^4)^(1/2)/(64*a^6) + (2
49*A^2*a^3)/(128*d^2) - (17*B^2*a^3)/(8*d^2) - (A*B*a^3*65i)/(16*d^2))^(1/2))/((133*A^3*a^8*d)/16 - B^3*a^8*d*
20i + 29*A*B^2*a^8*d + (A^2*B*a^8*d*3i)/4 - (3*A*d^3*((49*A^4*a^18)/(4*d^4) + (64*B^4*a^18)/d^4 + (40*A^2*B^2*
a^18)/d^4 + (A*B^3*a^18*64i)/d^4 + (A^3*B*a^18*28i)/d^4)^(1/2))/(8*a) + (B*d^3*((49*A^4*a^18)/(4*d^4) + (64*B^
4*a^18)/d^4 + (40*A^2*B^2*a^18)/d^4 + (A*B^3*a^18*64i)/d^4 + (A^3*B*a^18*28i)/d^4)^(1/2)*1i)/(2*a)) + (A*B*a^6
*d^2*(a + a*tan(c + d*x)*1i)^(1/2)*(((49*A^4*a^...
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