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

Optimal. Leaf size=216 \[ -\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}-\frac {-13 B+3 i A}{60 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {7 B+3 i A}{30 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {-B+i A}{5 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}} \]

[Out]

(-1/8-1/8*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)/a^(5/2)/d+1/60*(-3*I*A+13*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
)^(1/2)/(a+I*a*tan(d*x+c))^(5/2)+1/30*(3*I*A+7*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 = 0.75, antiderivative size = 216, 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} \[ -\frac {-13 B+3 i A}{60 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\frac {\left (\frac {1}{8}+\frac {i}{8}\right ) (A-i B) \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {(1+i) \sqrt {a} \sqrt {\tan (c+d x)}}{\sqrt {a+i a \tan (c+d x)}}\right )}{a^{5/2} d}+\frac {7 B+3 i A}{30 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}+\frac {-B+i A}{5 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/8 - I/8)*(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]])/(a^(5/2)*d) + (I*A - B)/(5*d*Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^(5/2)) + ((3*
I)*A + 7*B)/(30*a*d*Sqrt[Cot[c + d*x]]*(a + I*a*Tan[c + d*x])^(3/2)) - ((3*I)*A - 13*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)}{\sqrt {\cot (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 {\sqrt {\tan (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 \sqrt {\cot (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 {\frac {1}{2} a (i A-B)-a (2 A-3 i B) \tan (c+d x)}{\sqrt {\tan (c+d x)} (a+i a \tan (c+d x))^{3/2}} \, dx}{5 a^2}\\ &=\frac {i A-B}{5 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {3 i A+7 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 (9 i A+B)-\frac {1}{2} a^2 (3 A-7 i 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 \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {3 i A+7 B}{30 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}-\frac {3 i A-13 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 (i A+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 \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {3 i A+7 B}{30 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}-\frac {3 i A-13 B}{60 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}-\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}{8 a^3}\\ &=\frac {i A-B}{5 d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {3 i A+7 B}{30 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}-\frac {3 i A-13 B}{60 a^2 d \sqrt {\cot (c+d x)} \sqrt {a+i a \tan (c+d x)}}+\frac {\left (i (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 )}{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 \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{5/2}}+\frac {3 i A+7 B}{30 a d \sqrt {\cot (c+d x)} (a+i a \tan (c+d x))^{3/2}}-\frac {3 i A-13 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 = 7.56, size = 168, normalized size = 0.78 \[ \frac {\cot ^{\frac {3}{2}}(c+d x) \sec ^2(c+d x) \left (2 (2 (7 B+3 i A) \cos (2 (c+d x))+9 i A+20 i B \sin (2 (c+d x))+B)-\frac {30 i (A-i B) e^{3 i (c+d x)} \tanh ^{-1}\left (\frac {e^{i (c+d x)}}{\sqrt {-1+e^{2 i (c+d x)}}}\right )}{\sqrt {-1+e^{2 i (c+d x)}}}\right )}{120 a^2 d (\cot (c+d x)+i)^2 \sqrt {a+i a \tan (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Cot[c + d*x]^(3/2)*Sec[c + d*x]^2*(((-30*I)*(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))] + 2*((9*I)*A + B + 2*((3*I)*A + 7*B)*Cos[2*(c + d*x)] + (
20*I)*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 = 0.47, size = 485, normalized size = 2.25 \[ -\frac {{\left (15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (8 i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} - 8 i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} - 8 \, {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, {\left (i \, A + B\right )}}\right ) - 15 \, \sqrt {\frac {1}{2}} a^{3} d \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} e^{\left (5 i \, d x + 5 i \, c\right )} \log \left (\frac {{\left (\sqrt {2} \sqrt {\frac {1}{2}} {\left (-8 i \, a^{3} d e^{\left (2 i \, d x + 2 i \, c\right )} + 8 i \, a^{3} d\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}} \sqrt {\frac {i \, A^{2} + 2 \, A B - i \, B^{2}}{a^{5} d^{2}}} - 8 \, {\left (A - i \, B\right )} a e^{\left (i \, d x + i \, c\right )}\right )} e^{\left (-i \, d x - i \, c\right )}}{2 \, {\left (i \, A + B\right )}}\right ) - \sqrt {2} {\left ({\left (3 \, A - 17 i \, B\right )} e^{\left (6 i \, d x + 6 i \, c\right )} + 2 \, {\left (3 \, A + 8 i \, B\right )} e^{\left (4 i \, d x + 4 i \, c\right )} - 2 \, {\left (3 \, A - 2 i \, B\right )} e^{\left (2 i \, d x + 2 i \, c\right )} - 3 \, A - 3 i \, B\right )} \sqrt {\frac {a}{e^{\left (2 i \, d x + 2 i \, c\right )} + 1}} \sqrt {\frac {i \, e^{\left (2 i \, d x + 2 i \, c\right )} + i}{e^{\left (2 i \, d x + 2 i \, c\right )} - 1}}\right )} e^{\left (-5 i \, d x - 5 i \, c\right )}}{120 \, a^{3} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/cot(d*x+c)^(1/2)/(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(1/2*(sqrt(2)*sqrt(1
/2)*(8*I*a^3*d*e^(2*I*d*x + 2*I*c) - 8*I*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)) - 8*(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(1/
2*(sqrt(2)*sqrt(1/2)*(-8*I*a^3*d*e^(2*I*d*x + 2*I*c) + 8*I*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)) - 8*(A - I*B)*a*e^(I
*d*x + I*c))*e^(-I*d*x - I*c)/(I*A + B)) - sqrt(2)*((3*A - 17*I*B)*e^(6*I*d*x + 6*I*c) + 2*(3*A + 8*I*B)*e^(4*
I*d*x + 4*I*c) - 2*(3*A - 2*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))*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 \[ \int \frac {B \tan \left (d x + c\right ) + A}{{\left (i \, a \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \sqrt {\cot \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((A+B*tan(d*x+c))/cot(d*x+c)^(1/2)/(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)*sqrt(cot(d*x + c))), x)

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maple [B]  time = 4.10, size = 1092, normalized size = 5.06 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(-1/120+1/120*I)/d*(60*I*A*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)+3*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))+13*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)*arctan(
(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*cos(d*x+c)*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)-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))+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)+40*I*B*cos(d*x+c)^3*((-1+cos(d
*x+c))/sin(d*x+c))^(1/2)-3*I*A*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-40*I*B*cos(d*x+c)*((-1+cos(d*x+c)
)/sin(d*x+c))^(1/2)+13*I*B*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*sin(d*x+c)+12*A*cos(d*x+c)^2*sin(d*x+c)*((-1+cos
(d*x+c))/sin(d*x+c))^(1/2)-28*B*cos(d*x+c)^2*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(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*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*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)-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))+45*I*B*cos(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)-60*I*B*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)-12*I*A*
cos(d*x+c)^2*sin(d*x+c)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)-28*I*B*cos(d*x+c)^2*sin(d*x+c)*((-1+cos(d*x+c))/sin
(d*x+c))^(1/2)+30*I*B*cos(d*x+c)^2*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
*arctan((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)*(a*(
I*sin(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))/sin(d*x+c)/((-1+cos(d*x+c))/sin(d*x+c))^(1/2)/(cos(d*x+c)/sin(d*x+c))^(1/2)/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)^(1/2)/(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 \[ \int \frac {A+B\,\mathrm {tan}\left (c+d\,x\right )}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}\,{\left (a+a\,\mathrm {tan}\left (c+d\,x\right )\,1{}\mathrm {i}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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