3.565 \(\int \frac {A+B \tan (c+d x)}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=214 \[ \frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (B+i A) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {13 A-37 i B}{60 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {-B+i A}{5 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {A+11 i B}{30 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}} \]

[Out]

(1/8+1/8*I)*(I*A+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)/a^(5/2)/d+1/60*(13*A-37*I*B)/a^2/d/cot(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(1/2)+1/5*(I*A-B)/d/cot(d*x+c)
^(3/2)/(a+I*a*tan(d*x+c))^(5/2)+1/30*(A+11*I*B)/a/d/cot(d*x+c)^(1/2)/(a+I*a*tan(d*x+c))^(3/2)

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Rubi [A]  time = 0.76, antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 38, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {4241, 3595, 3596, 12, 3544, 205} \[ \frac {13 A-37 i B}{60 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (B+i A) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {-B+i A}{5 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {A+11 i B}{30 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[(A + B*Tan[c + d*x])/(Cot[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)),x]

[Out]

((1/8 + I/8)*(I*A + 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]])/(a^(5/2)*d) + (I*A - B)/(5*d*Cot[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)) + (A +
(11*I)*B)/(30*a*d*Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)) + (13*A - (37*I)*B)/(60*a^2*d*Sqrt[Cot[c +
d*x]]*Sqrt[a + I*a*Tan[c + d*x]])

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 3595

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*b - a*B)*(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n)/(2*a*f*
m), x] + Dist[1/(2*a^2*m), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1)*Simp[A*(a*c*m + b*d*n
) - B*(b*c*m + a*d*n) - d*(b*B*(m - n) - a*A*(m + n))*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f, A,
B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 + b^2, 0] && LtQ[m, 0] && GtQ[n, 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 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 {A+B \tan (c+d x)}{\cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}} \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\tan ^{\frac {3}{2}}(c+d x) (A+B \tan (c+d x))}{(a+i a \tan (c+d x))^{5/2}} \, dx\\ &=\frac {i A-B}{5 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}-\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\sqrt {\tan (c+d x)} \left (\frac {3}{2} a (i A-B)-a (A-4 i B) \tan (c+d x)\right )}{(a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=\frac {i A-B}{5 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {A+11 i B}{30 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {-\frac {1}{4} a^2 (A+11 i B)-\frac {1}{2} a^2 (7 i A+13 B) \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+i a \tan (c+d x)}} \, dx}{15 a^4}\\ &=\frac {i A-B}{5 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {A+11 i B}{30 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {13 A-37 i B}{60 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int -\frac {15 a^3 (A-i B) \sqrt {a+i a \tan (c+d x)}}{8 \sqrt {\tan (c+d x)}} \, dx}{15 a^6}\\ &=\frac {i A-B}{5 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {A+11 i B}{30 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {13 A-37 i B}{60 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left ((A-i 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}{8 a^3}\\ &=\frac {i A-B}{5 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {A+11 i B}{30 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {13 A-37 i B}{60 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (i (A-i 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 )}{4 a d}\\ &=\frac {\left (\frac {1}{8}+\frac {i}{8}\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)}}{a^{5/2} d}+\frac {i A-B}{5 d \cot ^{\frac {3}{2}}(c+d x) (a+i a \tan (c+d x))^{5/2}}+\frac {A+11 i B}{30 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {13 A-37 i B}{60 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}\\ \end {align*}

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Mathematica [A]  time = 7.96, size = 169, normalized size = 0.79 \[ \frac {\cot ^{\frac {3}{2}}(c+d x) \sec ^2(c+d x) \left (40 (B+i A) \sin (2 (c+d x))+4 (7 A-13 i B) \cos (2 (c+d x))-\frac {30 (A-i B) e^{3 i (c+d x)} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )}{\sqrt {-1+e^{2 i (c+d x)}}}+2 A+22 i B\right )}{120 a^2 d (\cot (c+d x)+i)^2 \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(A + B*Tan[c + d*x])/(Cot[c + d*x]^(3/2)*(a + I*a*Tan[c + d*x])^(5/2)),x]

[Out]

(Cot[c + d*x]^(3/2)*Sec[c + d*x]^2*(2*A + (22*I)*B - (30*(A - I*B)*E^((3*I)*(c + d*x))*ArcTanh[E^(I*(c + d*x))
/Sqrt[-1 + E^((2*I)*(c + d*x))]])/Sqrt[-1 + E^((2*I)*(c + d*x))] + 4*(7*A - (13*I)*B)*Cos[2*(c + d*x)] + 40*(I
*A + B)*Sin[2*(c + d*x)]))/(120*a^2*d*(I + Cot[c + d*x])^2*Sqrt[a + I*a*Tan[c + d*x]])

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fricas [B]  time = 0.61, size = 480, normalized size = 2.24 \[ -\frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (-\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{5} d^{2}}} + {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {4 \, {\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {-i \, A^{2} - 2 \, A B + i \, B^{2}}{a^{5} d^{2}}} - {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{i \, A + B}\right ) - \sqrt {2} {\left ({\left (-17 i \, A - 23 \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + {\left (16 i \, A + 34 \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} + {\left (4 i \, A - 14 \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 i \, A + 3 \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

-1/120*(15*sqrt(1/2)*a^3*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(-4*(sqrt(2)*sqrt(1
/2)*(a^3*d*e^(2*I*d*x + 2*I*c) - a^3*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((-I*A^2 - 2*A*B + I*B^2)/(a^5*d^2)) + (A - I*B)*a*e^(I*d*x + I*c))*e^(-I*d*x - I*
c)/(I*A + B)) - 15*sqrt(1/2)*a^3*d*sqrt((-I*A^2 - 2*A*B + I*B^2)/(a^5*d^2))*e^(5*I*d*x + 5*I*c)*log(4*(sqrt(2)
*sqrt(1/2)*(a^3*d*e^(2*I*d*x + 2*I*c) - a^3*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((-I*A^2 - 2*A*B + I*B^2)/(a^5*d^2)) - (A - I*B)*a*e^(I*d*x + I*c))*e^(-I*d
*x - I*c)/(I*A + B)) - sqrt(2)*((-17*I*A - 23*B)*e^(6*I*d*x + 6*I*c) + (16*I*A + 34*B)*e^(4*I*d*x + 4*I*c) + (
4*I*A - 14*B)*e^(2*I*d*x + 2*I*c) - 3*I*A + 3*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^(-5*I*d*x - 5*I*c)/(a^3*d)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cot \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x, algorithm="giac")

[Out]

integrate((B*tan(d*x + c) + A)/((I*a*tan(d*x + c) + a)^(5/2)*cot(d*x + c)^(3/2)), x)

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maple [B]  time = 4.05, size = 1212, normalized size = 5.66 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A+B*tan(d*x+c))/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),x)

[Out]

(-1/120-1/120*I)/d*(-60*I*A*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*sin(d*x+c)*2^(1/2)*
cos(d*x+c)^2+13*A*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-15*A*2^(1/2)*arctan((1/2+1/2*I)*((-1+cos(d*x+c
))/sin(d*x+c))^(1/2)*2^(1/2))-37*B*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+45*A*cos(d*x+c)*2^(1/2)*arcta
n((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))+15*B*sin(d*x+c)*2^(1/2)*arctan((1/2+1/2*I)*((-1+cos(
d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))+30*I*A*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*sin(d
*x+c)*cos(d*x+c)*2^(1/2)-60*A*cos(d*x+c)^3*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*2^(1
/2)-28*A*cos(d*x+c)^2*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+52*B*cos(d*x+c)^2*sin(d*x+c)*((-1+cos(d*x+
c))/sin(d*x+c))^(1/2)+40*I*A*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3-40*I*B*((-1+cos(d*x+c))/sin(d*x+c
))^(1/2)*cos(d*x+c)^3-13*I*A*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-40*I*A*cos(d*x+c)*((-1+cos(d*x+c))/
sin(d*x+c))^(1/2)-37*I*B*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)+40*I*B*cos(d*x+c)*((-1+cos(d*x+c))/sin(
d*x+c))^(1/2)+15*I*B*2^(1/2)*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))+30*A*arctan((1/2+1
/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*cos(d*x+c)^2*2^(1/2)+40*A*cos(d*x+c)^3*((-1+cos(d*x+c))/sin(
d*x+c))^(1/2)-40*A*cos(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+40*B*cos(d*x+c)^3*((-1+cos(d*x+c))/sin(d*x+c)
)^(1/2)-40*B*cos(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)+60*I*B*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+
c))^(1/2)*2^(1/2))*2^(1/2)*cos(d*x+c)^3+28*I*A*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2+15*I
*A*sin(d*x+c)*2^(1/2)*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))+52*I*B*sin(d*x+c)*((-1+co
s(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^2-30*I*B*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))
*2^(1/2)*cos(d*x+c)^2-45*I*B*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*cos(d*x+c)*2^(1/2)
-60*B*cos(d*x+c)^2*sin(d*x+c)*arctan((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*2^(1/2)+30*B*arct
an((1/2+1/2*I)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*2^(1/2))*cos(d*x+c)*sin(d*x+c)*2^(1/2))*cos(d*x+c)^2*(a*(I*s
in(d*x+c)+cos(d*x+c))/cos(d*x+c))^(1/2)/(4*I*cos(d*x+c)^2*sin(d*x+c)+4*cos(d*x+c)^3-I*sin(d*x+c)-3*cos(d*x+c))
/((-1+cos(d*x+c))/sin(d*x+c))^(1/2)/sin(d*x+c)^2/(cos(d*x+c)/sin(d*x+c))^(3/2)/a^3

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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((A+B*tan(d*x+c))/cot(d*x+c)^(3/2)/(a+I*a*tan(d*x+c))^(5/2),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 = -0.00 \[ \int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{{\mathrm {cot}\left (c+d\,x\right )}^{3/2}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((A + B*tan(c + d*x))/(cot(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(5/2)),x)

[Out]

int((A + B*tan(c + d*x))/(cot(c + d*x)^(3/2)*(a + a*tan(c + d*x)*1i)^(5/2)), 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((A+B*tan(d*x+c))/cot(d*x+c)**(3/2)/(a+I*a*tan(d*x+c))**(5/2),x)

[Out]

Timed out

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