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

Optimal. Leaf size=163 \[ \frac {(A+i B) \sqrt {\cot (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {(3 A+i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{a d}+\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d} \]

[Out]

(1/2+1/2*I)*(A-I*B)*arctanh((1+I)*a^(1/2)*tan(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+
c)^(1/2)/d/a^(1/2)+(A+I*B)*cot(d*x+c)^(1/2)/d/(a+I*a*tan(d*x+c))^(1/2)-(3*A+I*B)*cot(d*x+c)^(1/2)*(a+I*a*tan(d
*x+c))^(1/2)/a/d

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Rubi [A]  time = 0.50, antiderivative size = 163, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4241, 3596, 3598, 12, 3544, 205} \[ \frac {(A+i B) \sqrt {\cot (c+d x)}}{d \sqrt {a+i a \tan (c+d x)}}-\frac {(3 A+i B) \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}{a d}+\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{\sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((1/2 + I/2)*(A - I*B)*ArcTanh[((1 + I)*Sqrt[a]*Sqrt[Tan[c + d*x]])/Sqrt[a + I*a*Tan[c + d*x]]]*Sqrt[Cot[c + d
*x]]*Sqrt[Tan[c + d*x]])/(Sqrt[a]*d) + ((A + I*B)*Sqrt[Cot[c + d*x]])/(d*Sqrt[a + I*a*Tan[c + d*x]]) - ((3*A +
I*B)*Sqrt[Cot[c + d*x]]*Sqrt[a + I*a*Tan[c + d*x]])/(a*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 205

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[a/b, 2]*ArcTan[x/Rt[a/b, 2]])/a, x] /; FreeQ[{a, b}, x]
&& PosQ[a/b]

Rule 3544

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

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 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 4241

Int[(cot[(a_.) + (b_.)*(x_)]*(c_.))^(m_.)*(u_), x_Symbol] :> Dist[(c*Cot[a + b*x])^m*(c*Tan[a + b*x])^m, Int[A
ctivateTrig[u]/(c*Tan[a + b*x])^m, x], x] /; FreeQ[{a, b, c, m}, x] &&  !IntegerQ[m] && KnownTangentIntegrandQ
[u, x]

Rubi steps

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

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Mathematica [A]  time = 3.52, size = 165, normalized size = 1.01 \[ \frac {e^{-2 i (c+d x)} \sqrt {\frac {a e^{2 i (c+d x)}}{1+e^{2 i (c+d x)}}} \sqrt {\cot (c+d x)} \left ((A-i B) e^{i (c+d x)} \sqrt {-1+e^{2 i (c+d x)}} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )-5 A e^{2 i (c+d x)}+A-i B \left (-1+e^{2 i (c+d x)}\right )\right )}{\sqrt {2} a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[(a*E^((2*I)*(c + d*x)))/(1 + E^((2*I)*(c + d*x)))]*(A - 5*A*E^((2*I)*(c + d*x)) - I*B*(-1 + E^((2*I)*(c
+ d*x))) + (A - I*B)*E^(I*(c + d*x))*Sqrt[-1 + E^((2*I)*(c + d*x))]*ArcTanh[E^(I*(c + d*x))/Sqrt[-1 + E^((2*I)
*(c + d*x))]])*Sqrt[Cot[c + d*x]])/(Sqrt[2]*a*d*E^((2*I)*(c + d*x)))

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fricas [B]  time = 0.48, size = 422, normalized size = 2.59 \[ \frac {{\left (a d \sqrt {\frac {2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {{\left (\sqrt {2} {\left (2 i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} - 2 i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}}{a d^{2}}} - 4 \, {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - a d \sqrt {\frac {2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}}{a d^{2}}} e^{\left (i \, d x + i \, c\right )} \log \left (\frac {{\left (\sqrt {2} {\left (-2 i \, a d e^{\left (2 i \, d x + 2 i \, c\right )} + 2 i \, a d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {2 i \, A^{2} + 4 \, A B - 2 i \, B^{2}}{a d^{2}}} - 4 \, {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 2 \, \sqrt {2} {\left ({\left (5 \, A + i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - A - i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-i \, d x - i \, c\right )}}{4 \, a d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/4*(a*d*sqrt((2*I*A^2 + 4*A*B - 2*I*B^2)/(a*d^2))*e^(I*d*x + I*c)*log((sqrt(2)*(2*I*a*d*e^(2*I*d*x + 2*I*c) -
2*I*a*d)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt((
2*I*A^2 + 4*A*B - 2*I*B^2)/(a*d^2)) - 4*(A - I*B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) - a*d*sqrt((2
*I*A^2 + 4*A*B - 2*I*B^2)/(a*d^2))*e^(I*d*x + I*c)*log((sqrt(2)*(-2*I*a*d*e^(2*I*d*x + 2*I*c) + 2*I*a*d)*sqrt(
a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2*I*c) - 1))*sqrt((2*I*A^2 + 4*A*B
- 2*I*B^2)/(a*d^2)) - 4*(A - I*B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) - 2*sqrt(2)*((5*A + I*B)*e^(
2*I*d*x + 2*I*c) - A - I*B)*sqrt(a/(e^(2*I*d*x + 2*I*c) + 1))*sqrt((I*e^(2*I*d*x + 2*I*c) + I)/(e^(2*I*d*x + 2
*I*c) - 1)))*e^(-I*d*x - I*c)/(a*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (B \tan \left (d x + c\right ) + A\right )} \cot \left (d x + c\right )^{\frac {3}{2}}}{\sqrt {i \, a \tan \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

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maple [B]  time = 4.22, size = 482, normalized size = 2.96 \[ \frac {\left (-\frac {1}{2}-\frac {i}{2}\right ) \left (i A \sqrt {2}\, \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right )-i B \sqrt {2}\, \cos \left (d x +c \right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right )+A \sqrt {2}\, \cos \left (d x +c \right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right )-i B \sqrt {2}\, \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}+B \sqrt {2}\, \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sin \left (d x +c \right )+A \sqrt {2}\, \arctan \left (\left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {2}\right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}-2 i A \cos \left (d x +c \right )+3 i A \sin \left (d x +c \right )+i B \sin \left (d x +c \right )+2 A \cos \left (d x +c \right )+3 A \sin \left (d x +c \right )-B \sin \left (d x +c \right )\right ) \left (\frac {\cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )^{\frac {3}{2}} \sqrt {\frac {a \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )}{d \left (i \sin \left (d x +c \right )+\cos \left (d x +c \right )\right ) \cos \left (d x +c \right ) a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: RuntimeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cot(d*x+c)^(3/2)*(A+B*tan(d*x+c))/(a+I*a*tan(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Exception raised: RuntimeError >> ECL says: THROW: The catch RAT-ERR is undefined.

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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