3.109 \(\int \frac{\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx\)
Optimal. Leaf size=192 \[ \frac{7 A+i B}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{a^{5/2} d}+\frac{A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{3 A+i B}{6 a d (a+i a \tan (c+d x))^{3/2}} \]
[Out]
(-2*A*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(a^(5/2)*d) + ((A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]
/(Sqrt[2]*Sqrt[a])])/(4*Sqrt[2]*a^(5/2)*d) + (A + I*B)/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + (3*A + I*B)/(6*a*d
*(a + I*a*Tan[c + d*x])^(3/2)) + (7*A + I*B)/(4*a^2*d*Sqrt[a + I*a*Tan[c + d*x]])
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Rubi [A] time = 0.676441, antiderivative size = 192, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 7, integrand size = 34, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.206, Rules used =
{3596, 3600, 3480, 206, 3599, 63, 208} \[ \frac{7 A+i B}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{a^{5/2} d}+\frac{A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{3 A+i B}{6 a d (a+i a \tan (c+d x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[(Cot[c + d*x]*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^(5/2),x]
[Out]
(-2*A*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]/Sqrt[a]])/(a^(5/2)*d) + ((A - I*B)*ArcTanh[Sqrt[a + I*a*Tan[c + d*x]]
/(Sqrt[2]*Sqrt[a])])/(4*Sqrt[2]*a^(5/2)*d) + (A + I*B)/(5*d*(a + I*a*Tan[c + d*x])^(5/2)) + (3*A + I*B)/(6*a*d
*(a + I*a*Tan[c + d*x])^(3/2)) + (7*A + I*B)/(4*a^2*d*Sqrt[a + I*a*Tan[c + d*x]])
Rule 3596
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*A + b*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^(n + 1))/(2
*f*m*(b*c - a*d)), x] + Dist[1/(2*a*m*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Si
mp[A*(b*c*m - a*d*(2*m + n + 1)) + B*(a*c*m - b*d*(n + 1)) + d*(A*b - a*B)*(m + n + 1)*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] && LtQ[m, 0] && !GtQ[n,
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]
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 \frac{\cot (c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx &=\frac{A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{\int \frac{\cot (c+d x) \left (5 a A-\frac{5}{2} a (i A-B) \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=\frac{A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{3 A+i B}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac{\int \frac{\cot (c+d x) \left (15 a^2 A-\frac{15}{4} a^2 (3 i A-B) \tan (c+d x)\right )}{\sqrt{a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=\frac{A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{3 A+i B}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac{7 A+i B}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{\int \cot (c+d x) \sqrt{a+i a \tan (c+d x)} \left (15 a^3 A-\frac{15}{8} a^3 (7 i A-B) \tan (c+d x)\right ) \, dx}{15 a^6}\\ &=\frac{A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{3 A+i B}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac{7 A+i B}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{A \int \cot (c+d x) (a-i a \tan (c+d x)) \sqrt{a+i a \tan (c+d x)} \, dx}{a^4}+\frac{(i A+B) \int \sqrt{a+i a \tan (c+d x)} \, dx}{8 a^3}\\ &=\frac{A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{3 A+i B}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac{7 A+i B}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}+\frac{A \operatorname{Subst}\left (\int \frac{1}{x \sqrt{a+i a x}} \, dx,x,\tan (c+d x)\right )}{a^2 d}+\frac{(A-i B) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{4 a^2 d}\\ &=\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}+\frac{A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{3 A+i B}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac{7 A+i B}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}-\frac{(2 i A) \operatorname{Subst}\left (\int \frac{1}{i-\frac{i x^2}{a}} \, dx,x,\sqrt{a+i a \tan (c+d x)}\right )}{a^3 d}\\ &=-\frac{2 A \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{a}}\right )}{a^{5/2} d}+\frac{(A-i B) \tanh ^{-1}\left (\frac{\sqrt{a+i a \tan (c+d x)}}{\sqrt{2} \sqrt{a}}\right )}{4 \sqrt{2} a^{5/2} d}+\frac{A+i B}{5 d (a+i a \tan (c+d x))^{5/2}}+\frac{3 A+i B}{6 a d (a+i a \tan (c+d x))^{3/2}}+\frac{7 A+i B}{4 a^2 d \sqrt{a+i a \tan (c+d x)}}\\ \end{align*}
Mathematica [A] time = 4.17522, size = 233, normalized size = 1.21 \[ \frac{e^{-6 i (c+d x)} \left (1+e^{2 i (c+d x)}\right )^{3/2} \sec ^2(c+d x) \left (\sqrt{1+e^{2 i (c+d x)}} \left (3 A \left (7 e^{2 i (c+d x)}+41 e^{4 i (c+d x)}+1\right )+i B \left (11 e^{2 i (c+d x)}+23 e^{4 i (c+d x)}+3\right )\right )+15 (A-i B) e^{5 i (c+d x)} \sinh ^{-1}\left (e^{i (c+d x)}\right )-120 \sqrt{2} A e^{5 i (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{2} e^{i (c+d x)}}{\sqrt{1+e^{2 i (c+d x)}}}\right )\right )}{240 a^2 d \sqrt{a+i a \tan (c+d x)}} \]
Antiderivative was successfully verified.
[In]
Integrate[(Cot[c + d*x]*(A + B*Tan[c + d*x]))/(a + I*a*Tan[c + d*x])^(5/2),x]
[Out]
((1 + E^((2*I)*(c + d*x)))^(3/2)*(Sqrt[1 + E^((2*I)*(c + d*x))]*(I*B*(3 + 11*E^((2*I)*(c + d*x)) + 23*E^((4*I)
*(c + d*x))) + 3*A*(1 + 7*E^((2*I)*(c + d*x)) + 41*E^((4*I)*(c + d*x)))) + 15*(A - I*B)*E^((5*I)*(c + d*x))*Ar
cSinh[E^(I*(c + d*x))] - 120*Sqrt[2]*A*E^((5*I)*(c + d*x))*ArcTanh[(Sqrt[2]*E^(I*(c + d*x)))/Sqrt[1 + E^((2*I)
*(c + d*x))]])*Sec[c + d*x]^2)/(240*a^2*d*E^((6*I)*(c + d*x))*Sqrt[a + I*a*Tan[c + d*x]])
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Maple [B] time = 0.388, size = 1084, normalized size = 5.7 \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+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x)
[Out]
1/240/d/a^3*(a*(I*sin(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)*(-64*I*B*cos(d*x+c)^4-120*I*A*(-2*cos(d*x+c)/(cos(d
*x+c)+1))^(1/2)*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))+192*A*cos(d*x+
c)^6+192*B*cos(d*x+c)^5*sin(d*x+c)-120*I*A*cos(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*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))+20*I*B*cos(d*x+c)^2-192*I*A*cos(d*x+c)^3*sin(d*x+
c)-192*I*A*cos(d*x+c)^5*sin(d*x+c)+15*I*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*sin(d*x+c)*arctan(1/2*2^(1/2)*(
I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)+192*I*B*cos(d*x+c)^6-15*A*
(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*
cos(d*x+c)/(cos(d*x+c)+1))^(1/2))*2^(1/2)+96*A*cos(d*x+c)^4-420*I*A*cos(d*x+c)*sin(d*x+c)+15*B*(-2*cos(d*x+c)/
(cos(d*x+c)+1))^(1/2)*sin(d*x+c)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos
(d*x+c)+1))^(1/2))*2^(1/2)+32*B*cos(d*x+c)^3*sin(d*x+c)+120*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))*sin(d*x+c)+15*I*B*2^(1/2)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/
2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))-15*A*(-2*cos(d*x+c)/(co
s(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1
/2))*2^(1/2)-120*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)*arctan(1/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/
2))-120*A*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*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))*sin(d*x+c)+15*I*B*2^(1/2)*cos(d*x+c)*(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*arctan(1/2*2^(1/2)*
(I*cos(d*x+c)-I-sin(d*x+c))/sin(d*x+c)/(-2*cos(d*x+c)/(cos(d*x+c)+1))^(1/2))+300*A*cos(d*x+c)^2-120*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))+60*B*cos(d*x+c)*sin(d*x+c))
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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+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
Exception raised: ValueError
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Fricas [B] time = 2.96029, size = 1764, normalized size = 9.19 \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+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
1/120*(15*sqrt(1/2)*a^3*d*sqrt((A^2 - 2*I*A*B - B^2)/(a^5*d^2))*e^(6*I*d*x + 6*I*c)*log((2*I*sqrt(1/2)*a^3*d*s
qrt((A^2 - 2*I*A*B - B^2)/(a^5*d^2))*e^(2*I*d*x + 2*I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*s
qrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(4*I*A + 4*B)) - 15*sqrt(1/2)*a^3*d*sqrt((A
^2 - 2*I*A*B - B^2)/(a^5*d^2))*e^(6*I*d*x + 6*I*c)*log((-2*I*sqrt(1/2)*a^3*d*sqrt((A^2 - 2*I*A*B - B^2)/(a^5*d
^2))*e^(2*I*d*x + 2*I*c) + sqrt(2)*((I*A + B)*e^(2*I*d*x + 2*I*c) + I*A + B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))
*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(4*I*A + 4*B)) - 60*a^3*d*sqrt(A^2/(a^5*d^2))*e^(6*I*d*x + 6*I*c)*log(-88/5
07*(2*sqrt(2)*(A*e^(2*I*d*x + 2*I*c) + A)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c) + (3*a^3*d*e^(2*I*
d*x + 2*I*c) + a^3*d)*sqrt(A^2/(a^5*d^2)))/(A*e^(2*I*d*x + 2*I*c) - A)) + 60*a^3*d*sqrt(A^2/(a^5*d^2))*e^(6*I*
d*x + 6*I*c)*log(-88/507*(2*sqrt(2)*(A*e^(2*I*d*x + 2*I*c) + A)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I
*c) - (3*a^3*d*e^(2*I*d*x + 2*I*c) + a^3*d)*sqrt(A^2/(a^5*d^2)))/(A*e^(2*I*d*x + 2*I*c) - A)) + sqrt(2)*((123*
A + 23*I*B)*e^(6*I*d*x + 6*I*c) + 2*(72*A + 17*I*B)*e^(4*I*d*x + 4*I*c) + 2*(12*A + 7*I*B)*e^(2*I*d*x + 2*I*c)
+ 3*A + 3*I*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*e^(I*d*x + I*c))*e^(-6*I*d*x - 6*I*c)/(a^3*d)
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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+B*tan(d*x+c))/(a+I*a*tan(d*x+c))**(5/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B \tan \left (d x + c\right ) + A\right )} \cot \left (d x + c\right )}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")
[Out]
integrate((B*tan(d*x + c) + A)*cot(d*x + c)/(I*a*tan(d*x + c) + a)^(5/2), x)