3.878 \(\int \frac {1}{\cot ^{\frac {3}{2}}(c+d x) (a+b \tan (c+d x))^{5/2}} \, dx\)
Optimal. Leaf size=239 \[ \frac {2 a}{3 d \left (a^2+b^2\right ) \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {4 \left (a^2-2 b^2\right )}{3 d \left (a^2+b^2\right )^2 \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}+\frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}} \]
[Out]
arctan((I*a-b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/(I*a-b)^(5/2)/
d+arctanh((I*a+b)^(1/2)*tan(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2))*cot(d*x+c)^(1/2)*tan(d*x+c)^(1/2)/(I*a+b)^(5/
2)/d+4/3*(a^2-2*b^2)/(a^2+b^2)^2/d/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2)+2/3*a/(a^2+b^2)/d/cot(d*x+c)^(1/2)/
(a+b*tan(d*x+c))^(3/2)
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Rubi [A] time = 0.75, antiderivative size = 239, normalized size of antiderivative = 1.00,
number of steps used = 10, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used
= {4241, 3567, 3649, 3616, 3615, 93, 203, 206} \[ \frac {2 a}{3 d \left (a^2+b^2\right ) \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {4 \left (a^2-2 b^2\right )}{3 d \left (a^2+b^2\right )^2 \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tan ^{-1}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{5/2}}+\frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \tanh ^{-1}\left (\frac {\sqrt {b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (b+i a)^{5/2}} \]
Antiderivative was successfully verified.
[In]
Int[1/(Cot[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^(5/2)),x]
[Out]
(ArcTan[(Sqrt[I*a - b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/((
I*a - b)^(5/2)*d) + (ArcTanh[(Sqrt[I*a + b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]*Sqrt[Cot[c + d*x]]*S
qrt[Tan[c + d*x]])/((I*a + b)^(5/2)*d) + (2*a)/(3*(a^2 + b^2)*d*Sqrt[Cot[c + d*x]]*(a + b*Tan[c + d*x])^(3/2))
+ (4*(a^2 - 2*b^2))/(3*(a^2 + b^2)^2*d*Sqrt[Cot[c + d*x]]*Sqrt[a + b*Tan[c + d*x]])
Rule 93
Int[(((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_))/((e_.) + (f_.)*(x_)), x_Symbol] :> With[{q = Denomin
ator[m]}, Dist[q, Subst[Int[x^(q*(m + 1) - 1)/(b*e - a*f - (d*e - c*f)*x^q), x], x, (a + b*x)^(1/q)/(c + d*x)^
(1/q)], x]] /; FreeQ[{a, b, c, d, e, f}, x] && EqQ[m + n + 1, 0] && RationalQ[n] && LtQ[-1, m, 0] && SimplerQ[
a + b*x, c + d*x]
Rule 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 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 3567
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[((b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[
1/((m + 1)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n - 2)*Simp[a*c^2*(m + 1) + a*
d^2*(n - 1) + b*c*d*(m - n + 2) - (b*c^2 - 2*a*c*d - b*d^2)*(m + 1)*Tan[e + f*x] - d*(b*c - a*d)*(m + n)*Tan[e
+ f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d
^2, 0] && LtQ[m, -1] && LtQ[1, n, 2] && IntegerQ[2*m]
Rule 3615
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[A^2/f, Subst[Int[((a + b*x)^m*(c + d*x)^n)/(A - B*x), 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] && NeQ[a^2 + b^2, 0] && EqQ[A^2 +
B^2, 0]
Rule 3616
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*tan[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*tan[(e
_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(A + I*B)/2, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 -
I*Tan[e + f*x]), x], x] + Dist[(A - I*B)/2, Int[(a + b*Tan[e + f*x])^m*(c + d*Tan[e + f*x])^n*(1 + I*Tan[e +
f*x]), x], x] /; FreeQ[{a, b, c, d, e, f, A, B, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[A^2
+ B^2, 0]
Rule 3649
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_)*((A_.) + (B_.)*t
an[(e_.) + (f_.)*(x_)] + (C_.)*tan[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[((A*b^2 - a*(b*B - a*C))*(a + b*T
an[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^(n + 1))/(f*(m + 1)*(b*c - a*d)*(a^2 + b^2)), x] + Dist[1/((m + 1)*(
b*c - a*d)*(a^2 + b^2)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[A*(a*(b*c - a*d)*(m + 1)
- b^2*d*(m + n + 2)) + (b*B - a*C)*(b*c*(m + 1) + a*d*(n + 1)) - (m + 1)*(b*c - a*d)*(A*b - a*B - b*C)*Tan[e
+ f*x] - d*(A*b^2 - a*(b*B - a*C))*(m + n + 2)*Tan[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, A, B, C,
n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && LtQ[m, -1] && !(ILtQ[n, -1] && ( !I
ntegerQ[m] || (EqQ[c, 0] && NeQ[a, 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 {1}{\cot ^{\frac {3}{2}}(c+d x) (a+b \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))^{5/2}} \, dx\\ &=\frac {2 a}{3 \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}-\frac {\left (2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\frac {a}{2}-\frac {3}{2} b \tan (c+d x)-a \tan ^2(c+d x)}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx}{3 \left (a^2+b^2\right )}\\ &=\frac {2 a}{3 \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {4 \left (a^2-2 b^2\right )}{3 \left (a^2+b^2\right )^2 d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {\left (4 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\frac {3}{4} a \left (a^2-b^2\right )-\frac {3}{2} a^2 b \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{3 a \left (a^2+b^2\right )^2}\\ &=\frac {2 a}{3 \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {4 \left (a^2-2 b^2\right )}{3 \left (a^2+b^2\right )^2 d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {\left ((a-i b)^2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {1-i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 \left (a^2+b^2\right )^2}-\frac {\left ((a+i b)^2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {1+i \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{2 \left (a^2+b^2\right )^2}\\ &=\frac {2 a}{3 \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {4 \left (a^2-2 b^2\right )}{3 \left (a^2+b^2\right )^2 d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {\left ((a-i b)^2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{(1+i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 \left (a^2+b^2\right )^2 d}-\frac {\left ((a+i b)^2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{(1-i x) \sqrt {x} \sqrt {a+b x}} \, dx,x,\tan (c+d x)\right )}{2 \left (a^2+b^2\right )^2 d}\\ &=\frac {2 a}{3 \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {4 \left (a^2-2 b^2\right )}{3 \left (a^2+b^2\right )^2 d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}-\frac {\left ((a-i b)^2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{1-(-i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\left (a^2+b^2\right )^2 d}-\frac {\left ((a+i b)^2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{1-(i a+b) x^2} \, dx,x,\frac {\sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\left (a^2+b^2\right )^2 d}\\ &=\frac {\tan ^{-1}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a-b)^{5/2} d}+\frac {\tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a+b)^{5/2} d}+\frac {2 a}{3 \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} (a+b \tan (c+d x))^{3/2}}+\frac {4 \left (a^2-2 b^2\right )}{3 \left (a^2+b^2\right )^2 d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}\\ \end {align*}
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Mathematica [A] time = 4.19, size = 218, normalized size = 0.91 \[ \frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (\frac {2 \sqrt {\tan (c+d x)} \left (2 b \left (a^2-2 b^2\right ) \tan (c+d x)+3 a \left (a^2-b^2\right )\right )}{\left (a^2+b^2\right )^2 (a+b \tan (c+d x))^{3/2}}+\frac {3 (-1)^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(-a+i b)^{5/2}}+\frac {3 (-1)^{3/4} \tan ^{-1}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{(a+i b)^{5/2}}\right )}{3 d} \]
Antiderivative was successfully verified.
[In]
Integrate[1/(Cot[c + d*x]^(3/2)*(a + b*Tan[c + d*x])^(5/2)),x]
[Out]
(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*((3*(-1)^(3/4)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c + d*x]])/Sq
rt[a + b*Tan[c + d*x]]])/(-a + I*b)^(5/2) + (3*(-1)^(3/4)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqrt[Tan[c + d*x]])
/Sqrt[a + b*Tan[c + d*x]]])/(a + I*b)^(5/2) + (2*Sqrt[Tan[c + d*x]]*(3*a*(a^2 - b^2) + 2*b*(a^2 - 2*b^2)*Tan[c
+ d*x]))/((a^2 + b^2)^2*(a + b*Tan[c + d*x])^(3/2))))/(3*d)
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fricas [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(1/cot(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="fricas")
[Out]
Timed out
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giac [F(-2)] time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/cot(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="giac")
[Out]
Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Warning, need to choose a branch for the root of
a polynomial with parameters. This might be wrong.The choice was done assuming [d]=[-59,63]sym2poly/r2sym(con
st gen & e,const index_m & i,const vecteur & l) Error: Bad Argument ValueWarning, need to choose a branch for
the root of a polynomial with parameters. This might be wrong.The choice was done assuming [d]=[74,98]sym2poly
/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Valuesym2poly/r2sym(const gen &
e,const index_m & i,const vecteur & l) Error: Bad Argument Valuesym2poly/r2sym(const gen & e,const index_m & i
,const vecteur & l) Error: Bad Argument Valuesym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l)
Error: Bad Argument Valuesym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argumen
t Valuesym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Valuesym2poly/r2s
ym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Valuesym2poly/r2sym(const gen & e,co
nst index_m & i,const vecteur & l) Error: Bad Argument Valuesym2poly/r2sym(const gen & e,const index_m & i,con
st vecteur & l) Error: Bad Argument Valuesym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Err
or: Bad Argument Valuesym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Va
luesym2poly/r2sym(const gen & e,const index_m & i,const vecteur & l) Error: Bad Argument Valuesym2poly/r2sym(c
onst gen & e,const index_m & i,const vecteur & l) Error: Bad Argument ValueWarning, integration of abs or sign
assumes constant sign by intervals (correct if the argument is real):Check [abs(sin(d*t_nostep+c))]Unable to
check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unab
le to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/
2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_n
ostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*
pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2
)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_nostep/2)>(-4*pi/t_nostep/2)Unable to check sign: (4*pi/t_no
step/2)>(-4*pi/t_nostep/2)Discontinuities at zeroes of sin(d*t_nostep+c) were not checkedWarning, integration
of abs or sign assumes constant sign by intervals (correct if the argument is real):Check [abs(t_nostep),abs(t
_nostep^2-1)]Discontinuities at zeroes of t_nostep^2-1 were not checkedEvaluation time: 41.96Done
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maple [C] time = 2.16, size = 20598, normalized size = 86.18 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/cot(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(5/2),x)
[Out]
result too large to display
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {1}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac {5}{2}} \cot \left (d x + c\right )^{\frac {3}{2}}}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/cot(d*x+c)^(3/2)/(a+b*tan(d*x+c))^(5/2),x, algorithm="maxima")
[Out]
integrate(1/((b*tan(d*x + c) + a)^(5/2)*cot(d*x + c)^(3/2)), x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\mathrm {cot}\left (c+d\,x\right )}^{3/2}\,{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{5/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/(cot(c + d*x)^(3/2)*(a + b*tan(c + d*x))^(5/2)),x)
[Out]
int(1/(cot(c + d*x)^(3/2)*(a + b*tan(c + d*x))^(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(1/cot(d*x+c)**(3/2)/(a+b*tan(d*x+c))**(5/2),x)
[Out]
Timed out
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