3.85 \(\int \cot (c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx\)
Optimal. Leaf size=147 \[ -\frac{2 a^2 (A-2 i B) \sqrt{a+i a \tan (c+d x)}}{d}+\frac{4 \sqrt{2} a^{5/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{2 a^{5/2} A \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{d}+\frac{2 i a B (a+i a \tan (c+d x))^{3/2}}{3 d} \]
[Out]
(-2*a^(5/2)*A*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d + (4*Sqrt[2]*a^(5/2)*(A - I*B)*ArcTanh[Sqrt[a + I
*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (2*a^2*(A - (2*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/d + (((2*I)/3)*a*B*(
a + I*a*Tan[c + d*x])^(3/2))/d
________________________________________________________________________________________
Rubi [A] time = 0.534141, antiderivative size = 147, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 7, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.206, Rules used =
{3594, 3600, 3480, 206, 3599, 63, 208} \[ -\frac{2 a^2 (A-2 i B) \sqrt{a+i a \tan (c+d x)}}{d}+\frac{4 \sqrt{2} a^{5/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{2 a^{5/2} A \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{d}+\frac{2 i a B (a+i a \tan (c+d x))^{3/2}}{3 d} \]
Antiderivative was successfully verified.
[In]
Int[Cot[c + d*x]*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]
[Out]
(-2*a^(5/2)*A*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/d + (4*Sqrt[2]*a^(5/2)*(A - I*B)*ArcTanh[Sqrt[a + I
*a*Tan[c + d*x]]/(Sqrt[2]*Sqrt[a])])/d - (2*a^2*(A - (2*I)*B)*Sqrt[a + I*a*Tan[c + d*x]])/d + (((2*I)/3)*a*B*(
a + I*a*Tan[c + d*x])^(3/2))/d
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]
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]
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 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 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 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 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 (c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \, dx &=\frac{2 i a B (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac{2}{3} \int \cot (c+d x) (a+i a \tan (c+d x))^{3/2} \left (\frac{3 a A}{2}+\frac{3}{2} a (i A+2 B) \tan (c+d x)\right ) \, dx\\ &=-\frac{2 a^2 (A-2 i B) \sqrt{a+i a \tan (c+d x)}}{d}+\frac{2 i a B (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac{4}{3} \int \cot (c+d x) \sqrt{a+i a \tan (c+d x)} \left (\frac{3 a^2 A}{4}+\frac{3}{4} a^2 (3 i A+4 B) \tan (c+d x)\right ) \, dx\\ &=-\frac{2 a^2 (A-2 i B) \sqrt{a+i a \tan (c+d x)}}{d}+\frac{2 i a B (a+i a \tan (c+d x))^{3/2}}{3 d}+(a A) \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)} \, dx+\left (4 a^2 (i A+B)\right ) \int \sqrt{a+i a \tan (c+d x)} \, dx\\ &=-\frac{2 a^2 (A-2 i B) \sqrt{a+i a \tan (c+d x)}}{d}+\frac{2 i a B (a+i a \tan (c+d x))^{3/2}}{3 d}+\frac{\left (a^3 A\right ) \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{d}+\frac{\left (8 a^3 (A-i B)\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{d}\\ &=\frac{4 \sqrt{2} a^{5/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{2 a^2 (A-2 i B) \sqrt{a+i a \tan (c+d x)}}{d}+\frac{2 i a B (a+i a \tan (c+d x))^{3/2}}{3 d}-\frac{\left (2 i a^2 A\right ) \operatorname{Subst}\left (\int \frac{1}{i-\frac{i x^2}{a}} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{d}\\ &=-\frac{2 a^{5/2} A \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{d}+\frac{4 \sqrt{2} a^{5/2} (A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{d}-\frac{2 a^2 (A-2 i B) \sqrt{a+i a \tan (c+d x)}}{d}+\frac{2 i a B (a+i a \tan (c+d x))^{3/2}}{3 d}\\ \end{align*}
Mathematica [B] time = 7.93852, size = 429, normalized size = 2.92 \[ \frac{\cos ^3(c+d x) (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \left ((3 A-8 i B) \left (-\frac{2}{3} \cos (2 c)+\frac{2}{3} i \sin (2 c)\right )+\sec (c+d x) \left (-\frac{2}{3} B \sin (3 c+d x)-\frac{2}{3} i B \cos (3 c+d x)\right )\right )}{d (\cos (d x)+i \sin (d x))^2 (A \cos (c+d x)+B \sin (c+d x))}+\frac{e^{-2 i c} \sqrt{e^{i d x}} (a+i a \tan (c+d x))^{5/2} (A+B \tan (c+d x)) \left (8 (A-i B) \sinh ^{-1}\left (e^{i (c+d x)}\right )+\sqrt{2} A \left (\log \left (1-e^{i (c+d x)}\right )-\log \left (1+e^{i (c+d x)}\right )+\log \left (\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}-e^{i (c+d x)}+1\right )-\log \left (\sqrt{2} \sqrt{1+e^{2 i (c+d x)}}+e^{i (c+d x)}+1\right )\right )\right )}{\sqrt{2} d \sqrt{\frac{e^{i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt{1+e^{2 i (c+d x)}} \sec ^{\frac{7}{2}}(c+d x) (\cos (d x)+i \sin (d x))^{5/2} (A \cos (c+d x)+B \sin (c+d x))} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Cot[c + d*x]*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]),x]
[Out]
(Sqrt[E^(I*d*x)]*(8*(A - I*B)*ArcSinh[E^(I*(c + d*x))] + Sqrt[2]*A*(Log[1 - E^(I*(c + d*x))] - Log[1 + E^(I*(c
+ d*x))] + Log[1 - E^(I*(c + d*x)) + Sqrt[2]*Sqrt[1 + E^((2*I)*(c + d*x))]] - Log[1 + E^(I*(c + d*x)) + Sqrt[
2]*Sqrt[1 + E^((2*I)*(c + d*x))]]))*(a + I*a*Tan[c + d*x])^(5/2)*(A + B*Tan[c + d*x]))/(Sqrt[2]*d*E^((2*I)*c)*
Sqrt[E^(I*(c + d*x))/(1 + E^((2*I)*(c + d*x)))]*Sqrt[1 + E^((2*I)*(c + d*x))]*Sec[c + d*x]^(7/2)*(Cos[d*x] + I
*Sin[d*x])^(5/2)*(A*Cos[c + d*x] + B*Sin[c + d*x])) + (Cos[c + d*x]^3*((3*A - (8*I)*B)*((-2*Cos[2*c])/3 + ((2*
I)/3)*Sin[2*c]) + Sec[c + d*x]*(((-2*I)/3)*B*Cos[3*c + d*x] - (2*B*Sin[3*c + d*x])/3))*(a + I*a*Tan[c + d*x])^
(5/2)*(A + B*Tan[c + d*x]))/(d*(Cos[d*x] + I*Sin[d*x])^2*(A*Cos[c + d*x] + B*Sin[c + d*x]))
________________________________________________________________________________________
Maple [B] time = 0.431, size = 965, normalized size = 6.6 \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)*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x)
[Out]
1/6/d*a^2*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(3*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))^(3/2)*sin(d*x+c)+12*I*B*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))^(3/2)*sin(d*x+c)+12*I*A*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))^(3/2)*sin(d*x+c)-28*I*B*cos(d*x+c)-
12*A*cos(d*x+c)*sin(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*arctanh(1/2*2^(1/2)*(-2*cos(d*x+c)/(co
s(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))+3*I*A*cos(d*x+c)*sin(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*arc
tan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+12*B*cos(d*x+c)*sin(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^
(3/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-12*I*A*cos(d*x+c)*sin(d*x+c)-3*A*cos(d*x+c)*sin
(d*x+c)*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*
x+c)+1))^(3/2)-12*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))^(3/2)*sin(d*x+c)+12*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))^(3/2)*sin(d*x+c)-3*A*ln(-(-(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c
)+cos(d*x+c)-1)/sin(d*x+c))*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*sin(d*x+c)-4*I*B+12*I*A*cos(d*x+c)*sin(d*x+c)
*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*arctan(1/2*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-12*A*co
s(d*x+c)^2+12*I*B*cos(d*x+c)*sin(d*x+c)*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(3/2)*arctanh(1/2*2^(1/2)*(-2*c
os(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)/cos(d*x+c))-32*B*cos(d*x+c)*sin(d*x+c)+12*A*cos(d*x+c)+32*I*B*cos(d
*x+c)^2+4*B*sin(d*x+c))/(I*sin(d*x+c)+cos(d*x+c)-1)/cos(d*x+c)
________________________________________________________________________________________
Maxima [F(-2)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="maxima")
[Out]
Exception raised: ValueError
________________________________________________________________________________________
Fricas [B] time = 1.884, size = 1652, normalized size = 11.24 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="fricas")
[Out]
-1/6*(4*sqrt(2)*((3*A - 8*I*B)*a^2*e^(2*I*d*x + 2*I*c) + 3*(A - 2*I*B)*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*
e^(I*d*x + I*c) + 6*sqrt(A^2*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*(A*a^2*e^(2*I*d*x + 2*I*c) + A*
a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + 2*sqrt(A^2*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*
x - 2*I*c)/(A*a^2)) - 6*sqrt(A^2*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*(A*a^2*e^(2*I*d*x + 2*I*c)
+ A*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - 2*sqrt(A^2*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*
I*d*x - 2*I*c)/(A*a^2)) - 3*sqrt((32*A^2 - 64*I*A*B - 32*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2
)*((4*I*A + 4*B)*a^2*e^(2*I*d*x + 2*I*c) + (4*I*A + 4*B)*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c
) + I*sqrt((32*A^2 - 64*I*A*B - 32*B^2)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B)*a^
2)) + 3*sqrt((32*A^2 - 64*I*A*B - 32*B^2)*a^5/d^2)*(d*e^(2*I*d*x + 2*I*c) + d)*log((sqrt(2)*((4*I*A + 4*B)*a^2
*e^(2*I*d*x + 2*I*c) + (4*I*A + 4*B)*a^2)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) - I*sqrt((32*A^2 -
64*I*A*B - 32*B^2)*a^5/d^2)*d*e^(2*I*d*x + 2*I*c))*e^(-2*I*d*x - 2*I*c)/((4*I*A + 4*B)*a^2)))/(d*e^(2*I*d*x +
2*I*c) + d)
________________________________________________________________________________________
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)*(a+I*a*tan(d*x+c))**(5/2)*(A+B*tan(d*x+c)),x)
[Out]
Timed out
________________________________________________________________________________________
Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (B \tan \left (d x + c\right ) + A\right )}{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}} \cot \left (d x + c\right )\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)*(a+I*a*tan(d*x+c))^(5/2)*(A+B*tan(d*x+c)),x, algorithm="giac")
[Out]
integrate((B*tan(d*x + c) + A)*(I*a*tan(d*x + c) + a)^(5/2)*cot(d*x + c), x)