3.9.69 \(\int \frac {\sqrt {\cot (c+d x)}}{(a+b \tan (c+d x))^{3/2}} \, dx\) [869]
Optimal. Leaf size=199 \[ \frac {i \text {ArcTan}\left (\frac {\sqrt {i a-b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a-b)^{3/2} d}+\frac {i \tanh ^{-1}\left (\frac {\sqrt {i a+b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right ) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}}{(i a+b)^{3/2} d}+\frac {2 b^2}{a \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}} \]
[Out]
I*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)^(3/2
)/d+I*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)
^(3/2)/d+2*b^2/a/(a^2+b^2)/d/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2)
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Rubi [A]
time = 0.38, antiderivative size = 199, normalized size of antiderivative = 1.00, number of steps
used = 9, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {4326, 3650,
3697, 3696, 95, 209, 212} \begin {gather*} \frac {2 b^2}{a d \left (a^2+b^2\right ) \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {i \sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \text {ArcTan}\left (\frac {\sqrt {-b+i a} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{d (-b+i a)^{3/2}}+\frac {i \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)^{3/2}} \end {gather*}
Antiderivative was successfully verified.
[In]
Int[Sqrt[Cot[c + d*x]]/(a + b*Tan[c + d*x])^(3/2),x]
[Out]
(I*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)^(3/2)*d) + (I*ArcTanh[(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)^(3/2)*d) + (2*b^2)/(a*(a^2 + b^2)*d*Sqrt[Cot[c + d*x]]*Sqrt[a + b*Tan[c + d*
x]])
Rule 95
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 209
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 212
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[-b, 2]))*ArcTanh[Rt[-b, 2]*(x/Rt[a, 2])], x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 3650
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Si
mp[b^2*(a + b*Tan[e + f*x])^(m + 1)*((c + d*Tan[e + f*x])^(n + 1)/(f*(m + 1)*(a^2 + b^2)*(b*c - a*d))), x] + D
ist[1/((m + 1)*(a^2 + b^2)*(b*c - a*d)), Int[(a + b*Tan[e + f*x])^(m + 1)*(c + d*Tan[e + f*x])^n*Simp[a*(b*c -
a*d)*(m + 1) - b^2*d*(m + n + 2) - b*(b*c - a*d)*(m + 1)*Tan[e + f*x] - b^2*d*(m + n + 2)*Tan[e + f*x]^2, x],
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] && NeQ[c^2 + d^2, 0] && I
ntegerQ[2*m] && LtQ[m, -1] && (LtQ[n, 0] || IntegerQ[m]) && !(ILtQ[n, -1] && ( !IntegerQ[m] || (EqQ[c, 0] &&
NeQ[a, 0])))
Rule 3696
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 3697
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 4326
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 {\sqrt {\cot (c+d x)}}{(a+b \tan (c+d x))^{3/2}} \, dx &=\left (\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {1}{\sqrt {\tan (c+d x)} (a+b \tan (c+d x))^{3/2}} \, dx\\ &=\frac {2 b^2}{a \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {\left (2 \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \int \frac {\frac {a^2}{2}-\frac {1}{2} a b \tan (c+d x)}{\sqrt {\tan (c+d x)} \sqrt {a+b \tan (c+d x)}} \, dx}{a \left (a^2+b^2\right )}\\ &=\frac {2 b^2}{a \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {\left ((a-i b) \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 )}+\frac {\left ((a+i b) \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 )}\\ &=\frac {2 b^2}{a \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {\left ((a-i b) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {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 ) d}+\frac {\left ((a+i b) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {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 ) d}\\ &=\frac {2 b^2}{a \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}+\frac {\left ((a-i b) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {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 ) d}+\frac {\left ((a+i b) \sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)}\right ) \text {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 ) d}\\ &=\frac {i \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)^{3/2} d}+\frac {i \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)^{3/2} d}+\frac {2 b^2}{a \left (a^2+b^2\right ) d \sqrt {\cot (c+d x)} \sqrt {a+b \tan (c+d x)}}\\ \end {align*}
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Mathematica [A]
time = 1.04, size = 203, normalized size = 1.02 \begin {gather*} -\frac {\sqrt {\cot (c+d x)} \sqrt {\tan (c+d x)} \left (\frac {(-1)^{3/4} (a+i b) \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {-a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {-a+i b}}+\frac {\sqrt [4]{-1} (i a+b) \text {ArcTan}\left (\frac {\sqrt [4]{-1} \sqrt {a+i b} \sqrt {\tan (c+d x)}}{\sqrt {a+b \tan (c+d x)}}\right )}{\sqrt {a+i b}}-\frac {2 b^2 \sqrt {\tan (c+d x)}}{a \sqrt {a+b \tan (c+d x)}}\right )}{\left (a^2+b^2\right ) d} \end {gather*}
Antiderivative was successfully verified.
[In]
Integrate[Sqrt[Cot[c + d*x]]/(a + b*Tan[c + d*x])^(3/2),x]
[Out]
-((Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(((-1)^(3/4)*(a + I*b)*ArcTan[((-1)^(1/4)*Sqrt[-a + I*b]*Sqrt[Tan[c +
d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[-a + I*b] + ((-1)^(1/4)*(I*a + b)*ArcTan[((-1)^(1/4)*Sqrt[a + I*b]*Sqr
t[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]])/Sqrt[a + I*b] - (2*b^2*Sqrt[Tan[c + d*x]])/(a*Sqrt[a + b*Tan[c + d
*x]])))/((a^2 + b^2)*d))
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Maple [C] Result contains higher order function than in optimal. Order 4 vs. order
3.
time = 38.35, size = 4830, normalized size = 24.27
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Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(3/2),x,method=_RETURNVERBOSE)
[Out]
1/d*(cos(d*x+c)/sin(d*x+c))^(1/2)*((a*cos(d*x+c)+b*sin(d*x+c))/cos(d*x+c))^(1/2)*sin(d*x+c)*(-I*(((a^2+b^2)^(1
/2)*sin(d*x+c)-b*sin(d*x+c)-a*cos(d*x+c)+a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2)*EllipticPi((((a^2+b^2)^(1/2
)*sin(d*x+c)-b*sin(d*x+c)-a*cos(d*x+c)+a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2),(-b+(a^2+b^2)^(1/2))/(I*a+(a^
2+b^2)^(1/2)-b),1/2*2^(1/2)*((-b+(a^2+b^2)^(1/2))/(a^2+b^2)^(1/2))^(1/2))*(((a^2+b^2)^(1/2)*sin(d*x+c)+b*sin(d
*x+c)+a*cos(d*x+c)-a)/(a^2+b^2)^(1/2)/sin(d*x+c))^(1/2)*(a*(-1+cos(d*x+c))/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1
/2)*sin(d*x+c)*a^4+3*I*(a^2+b^2)^(1/2)*(((a^2+b^2)^(1/2)*sin(d*x+c)-b*sin(d*x+c)-a*cos(d*x+c)+a)/(-b+(a^2+b^2)
^(1/2))/sin(d*x+c))^(1/2)*EllipticPi((((a^2+b^2)^(1/2)*sin(d*x+c)-b*sin(d*x+c)-a*cos(d*x+c)+a)/(-b+(a^2+b^2)^(
1/2))/sin(d*x+c))^(1/2),(-b+(a^2+b^2)^(1/2))/(I*a+(a^2+b^2)^(1/2)-b),1/2*2^(1/2)*((-b+(a^2+b^2)^(1/2))/(a^2+b^
2)^(1/2))^(1/2))*(((a^2+b^2)^(1/2)*sin(d*x+c)+b*sin(d*x+c)+a*cos(d*x+c)-a)/(a^2+b^2)^(1/2)/sin(d*x+c))^(1/2)*(
a*(-1+cos(d*x+c))/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2)*sin(d*x+c)*a^2*b-3*I*(((a^2+b^2)^(1/2)*sin(d*x+c)-b*s
in(d*x+c)-a*cos(d*x+c)+a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2)*EllipticPi((((a^2+b^2)^(1/2)*sin(d*x+c)-b*sin
(d*x+c)-a*cos(d*x+c)+a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2),(-b+(a^2+b^2)^(1/2))/(I*a+(a^2+b^2)^(1/2)-b),1/
2*2^(1/2)*((-b+(a^2+b^2)^(1/2))/(a^2+b^2)^(1/2))^(1/2))*(((a^2+b^2)^(1/2)*sin(d*x+c)+b*sin(d*x+c)+a*cos(d*x+c)
-a)/(a^2+b^2)^(1/2)/sin(d*x+c))^(1/2)*(a*(-1+cos(d*x+c))/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2)*sin(d*x+c)*a^2
*b^2+3*I*(((a^2+b^2)^(1/2)*sin(d*x+c)-b*sin(d*x+c)-a*cos(d*x+c)+a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2)*(((a
^2+b^2)^(1/2)*sin(d*x+c)+b*sin(d*x+c)+a*cos(d*x+c)-a)/(a^2+b^2)^(1/2)/sin(d*x+c))^(1/2)*(a*(-1+cos(d*x+c))/(-b
+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2)*EllipticPi((((a^2+b^2)^(1/2)*sin(d*x+c)-b*sin(d*x+c)-a*cos(d*x+c)+a)/(-b+(
a^2+b^2)^(1/2))/sin(d*x+c))^(1/2),(-b+(a^2+b^2)^(1/2))/(-I*a+(a^2+b^2)^(1/2)-b),1/2*2^(1/2)*((-b+(a^2+b^2)^(1/
2))/(a^2+b^2)^(1/2))^(1/2))*sin(d*x+c)*a^2*b^2+I*(((a^2+b^2)^(1/2)*sin(d*x+c)-b*sin(d*x+c)-a*cos(d*x+c)+a)/(-b
+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2)*(((a^2+b^2)^(1/2)*sin(d*x+c)+b*sin(d*x+c)+a*cos(d*x+c)-a)/(a^2+b^2)^(1/2)/
sin(d*x+c))^(1/2)*(a*(-1+cos(d*x+c))/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2)*EllipticPi((((a^2+b^2)^(1/2)*sin(d
*x+c)-b*sin(d*x+c)-a*cos(d*x+c)+a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2),(-b+(a^2+b^2)^(1/2))/(-I*a+(a^2+b^2)
^(1/2)-b),1/2*2^(1/2)*((-b+(a^2+b^2)^(1/2))/(a^2+b^2)^(1/2))^(1/2))*sin(d*x+c)*a^4-3*I*(a^2+b^2)^(1/2)*(((a^2+
b^2)^(1/2)*sin(d*x+c)-b*sin(d*x+c)-a*cos(d*x+c)+a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2)*(((a^2+b^2)^(1/2)*si
n(d*x+c)+b*sin(d*x+c)+a*cos(d*x+c)-a)/(a^2+b^2)^(1/2)/sin(d*x+c))^(1/2)*(a*(-1+cos(d*x+c))/(-b+(a^2+b^2)^(1/2)
)/sin(d*x+c))^(1/2)*EllipticPi((((a^2+b^2)^(1/2)*sin(d*x+c)-b*sin(d*x+c)-a*cos(d*x+c)+a)/(-b+(a^2+b^2)^(1/2))/
sin(d*x+c))^(1/2),(-b+(a^2+b^2)^(1/2))/(-I*a+(a^2+b^2)^(1/2)-b),1/2*2^(1/2)*((-b+(a^2+b^2)^(1/2))/(a^2+b^2)^(1
/2))^(1/2))*sin(d*x+c)*a^2*b-(a^2+b^2)^(1/2)*EllipticPi((((a^2+b^2)^(1/2)*sin(d*x+c)-b*sin(d*x+c)-a*cos(d*x+c)
+a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2),(-b+(a^2+b^2)^(1/2))/(I*a+(a^2+b^2)^(1/2)-b),1/2*2^(1/2)*((-b+(a^2+
b^2)^(1/2))/(a^2+b^2)^(1/2))^(1/2))*(((a^2+b^2)^(1/2)*sin(d*x+c)+b*sin(d*x+c)+a*cos(d*x+c)-a)/(a^2+b^2)^(1/2)/
sin(d*x+c))^(1/2)*(a*(-1+cos(d*x+c))/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2)*sin(d*x+c)*(((a^2+b^2)^(1/2)*sin(d
*x+c)-b*sin(d*x+c)-a*cos(d*x+c)+a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2)*a^3+2*(a^2+b^2)^(1/2)*(((a^2+b^2)^(1
/2)*sin(d*x+c)-b*sin(d*x+c)-a*cos(d*x+c)+a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2)*EllipticPi((((a^2+b^2)^(1/2
)*sin(d*x+c)-b*sin(d*x+c)-a*cos(d*x+c)+a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2),(-b+(a^2+b^2)^(1/2))/(I*a+(a^
2+b^2)^(1/2)-b),1/2*2^(1/2)*((-b+(a^2+b^2)^(1/2))/(a^2+b^2)^(1/2))^(1/2))*(((a^2+b^2)^(1/2)*sin(d*x+c)+b*sin(d
*x+c)+a*cos(d*x+c)-a)/(a^2+b^2)^(1/2)/sin(d*x+c))^(1/2)*(a*(-1+cos(d*x+c))/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1
/2)*sin(d*x+c)*a*b^2-(a^2+b^2)^(1/2)*(((a^2+b^2)^(1/2)*sin(d*x+c)+b*sin(d*x+c)+a*cos(d*x+c)-a)/(a^2+b^2)^(1/2)
/sin(d*x+c))^(1/2)*(a*(-1+cos(d*x+c))/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2)*EllipticPi((((a^2+b^2)^(1/2)*sin(
d*x+c)-b*sin(d*x+c)-a*cos(d*x+c)+a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2),(-b+(a^2+b^2)^(1/2))/(-I*a+(a^2+b^2
)^(1/2)-b),1/2*2^(1/2)*((-b+(a^2+b^2)^(1/2))/(a^2+b^2)^(1/2))^(1/2))*sin(d*x+c)*(((a^2+b^2)^(1/2)*sin(d*x+c)-b
*sin(d*x+c)-a*cos(d*x+c)+a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2)*a^3+2*(a^2+b^2)^(1/2)*(((a^2+b^2)^(1/2)*sin
(d*x+c)-b*sin(d*x+c)-a*cos(d*x+c)+a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1/2)*(((a^2+b^2)^(1/2)*sin(d*x+c)+b*sin
(d*x+c)+a*cos(d*x+c)-a)/(a^2+b^2)^(1/2)/sin(d*x+c))^(1/2)*(a*(-1+cos(d*x+c))/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^
(1/2)*EllipticPi((((a^2+b^2)^(1/2)*sin(d*x+c)-b*sin(d*x+c)-a*cos(d*x+c)+a)/(-b+(a^2+b^2)^(1/2))/sin(d*x+c))^(1
/2),(-b+(a^2+b^2)^(1/2))/(-I*a+(a^2+b^2)^(1/2)-b),1/2*2^(1/2)*((-b+(a^2+b^2)^(1/2))/(a^2+b^2)^(1/2))^(1/2))*si
n(d*x+c)*a*b^2+2*(a^2+b^2)^(1/2)*(((a^2+b^2)^(1...
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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(3/2),x, algorithm="maxima")
[Out]
integrate(sqrt(cot(d*x + c))/(b*tan(d*x + c) + a)^(3/2), x)
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Fricas [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(3/2),x, algorithm="fricas")
[Out]
Timed out
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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sqrt {\cot {\left (c + d x \right )}}}{\left (a + b \tan {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)**(1/2)/(a+b*tan(d*x+c))**(3/2),x)
[Out]
Integral(sqrt(cot(c + d*x))/(a + b*tan(c + d*x))**(3/2), x)
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Giac [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(3/2),x, algorithm="giac")
[Out]
Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,sageVARx):;OUTP
UT:Warning, integration of abs or sign assumes constant sign by intervals (correct if the argument is real):Ch
eck [abs(si
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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {\sqrt {\mathrm {cot}\left (c+d\,x\right )}}{{\left (a+b\,\mathrm {tan}\left (c+d\,x\right )\right )}^{3/2}} \,d x \end {gather*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(c + d*x)^(1/2)/(a + b*tan(c + d*x))^(3/2),x)
[Out]
int(cot(c + d*x)^(1/2)/(a + b*tan(c + d*x))^(3/2), x)
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