3.864 \(\int \frac{1}{\sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx\)
Optimal. Leaf size=155 \[ \frac{i \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{-b+i a}}-\frac{i \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{b+i a}} \]
[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]])/
(Sqrt[I*a - b]*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]])/(Sqrt[I*a + b]*d)
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Rubi [A] time = 0.194775, antiderivative size = 155, normalized size of antiderivative = 1.,
number of steps used = 8, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used =
{4241, 3575, 910, 93, 205, 208} \[ \frac{i \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{-b+i a}}-\frac{i \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{b+i a}} \]
Antiderivative was successfully verified.
[In]
Int[1/(Sqrt[Cot[c + d*x]]*Sqrt[a + b*Tan[c + d*x]]),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]])/
(Sqrt[I*a - b]*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]])/(Sqrt[I*a + b]*d)
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]
Rule 3575
Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Wit
h[{ff = FreeFactors[Tan[e + f*x], x]}, Dist[ff/f, Subst[Int[((a + b*ff*x)^m*(c + d*ff*x)^n)/(1 + ff^2*x^2), x]
, x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 + b^2, 0] &&
NeQ[c^2 + d^2, 0]
Rule 910
Int[((d_.) + (e_.)*(x_))^(m_)/(Sqrt[(f_.) + (g_.)*(x_)]*((a_.) + (c_.)*(x_)^2)), x_Symbol] :> Int[ExpandIntegr
and[1/(Sqrt[d + e*x]*Sqrt[f + g*x]), (d + e*x)^(m + 1/2)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] &
& NeQ[c*d^2 + a*e^2, 0] && IGtQ[m + 1/2, 0]
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 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 208
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]
Rubi steps
\begin{align*} \int \frac{1}{\sqrt{\cot (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx &=\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}} \, dx\\ &=\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{x}}{\sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \left (-\frac{1}{2 (i-x) \sqrt{x} \sqrt{a+b x}}+\frac{1}{2 \sqrt{x} (i+x) \sqrt{a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{(i-x) \sqrt{x} \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}+\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} (i+x) \sqrt{a+b x}} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{i-(-a+i b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d}-\frac{\left (\sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{i-(a+i b) x^2} \, dx,x,\frac{\sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\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)}}{\sqrt{i a-b} 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)}}{\sqrt{i a+b} d}\\ \end{align*}
Mathematica [A] time = 0.140938, size = 143, normalized size = 0.92 \[ \frac{\sqrt [4]{-1} \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \left (\frac{\tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{-a-i b}}-\frac{\tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{a-i b}}\right )}{d} \]
Antiderivative was successfully verified.
[In]
Integrate[1/(Sqrt[Cot[c + d*x]]*Sqrt[a + b*Tan[c + d*x]]),x]
[Out]
((-1)^(1/4)*(ArcTanh[((-1)^(1/4)*Sqrt[-a - I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]/Sqrt[-a - I*b] -
ArcTanh[((-1)^(1/4)*Sqrt[a - I*b]*Sqrt[Tan[c + d*x]])/Sqrt[a + b*Tan[c + d*x]]]/Sqrt[a - I*b])*Sqrt[Cot[c + d
*x]]*Sqrt[Tan[c + d*x]])/d
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Maple [C] time = 0.39, size = 1630, normalized size = 10.5 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(1/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2),x)
[Out]
-1/d/(I*a-(a^2+b^2)^(1/2)+b)/(I*a+(a^2+b^2)^(1/2)-b)*(I*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*(a^2+b^2)^(1/2)-I*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*b-I*EllipticPi((((a^2+b^2)^(1/2)*sin(d*x+c)-b*si
n(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*(a^2+b^2)^(1/2)+I*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*b-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))*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+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))*b^2-2*EllipticPi((((a^2+b^2)^(1/2)*s
in(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))*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+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))*b^2)*cos(d*x+c)*(a*(cos(d*x+c)-1)
/(-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^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)*2^(1/2)*(1/cos(d*x+c)*(a*cos(d*x+c)+b*sin(d*x+c)))^(1/2)*sin(d*x+c)/(a*cos(d*x+c)+b*sin(d*x+c))/(cos(
d*x+c)-1)/(cos(d*x+c)/sin(d*x+c))^(1/2)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \tan \left (d x + c\right ) + a} \sqrt{\cot \left (d x + c\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
integrate(1/(sqrt(b*tan(d*x + c) + a)*sqrt(cot(d*x + c))), x)
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Fricas [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
Timed out
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{a + b \tan{\left (c + d x \right )}} \sqrt{\cot{\left (c + d x \right )}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/cot(d*x+c)**(1/2)/(a+b*tan(d*x+c))**(1/2),x)
[Out]
Integral(1/(sqrt(a + b*tan(c + d*x))*sqrt(cot(c + d*x))), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\sqrt{b \tan \left (d x + c\right ) + a} \sqrt{\cot \left (d x + c\right )}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(1/cot(d*x+c)^(1/2)/(a+b*tan(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate(1/(sqrt(b*tan(d*x + c) + a)*sqrt(cot(d*x + c))), x)