3.653 \(\int \frac{\sqrt{\cot (c+d x)} (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx\)
Optimal. Leaf size=151 \[ \frac{B \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{-b+i a}}+\frac{B \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{b+i a}} \]
[Out]
(B*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) + (B*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 = 0.214753, antiderivative size = 151, normalized size of antiderivative = 1.,
number of steps used = 9, number of rules used = 7, integrand size = 38, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.184, Rules used =
{21, 4241, 3575, 912, 93, 205, 208} \[ \frac{B \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tan ^{-1}\left (\frac{\sqrt{-b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{-b+i a}}+\frac{B \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \tanh ^{-1}\left (\frac{\sqrt{b+i a} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{d \sqrt{b+i a}} \]
Antiderivative was successfully verified.
[In]
Int[(Sqrt[Cot[c + d*x]]*(a*B + b*B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^(3/2),x]
[Out]
(B*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) + (B*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 21
Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])
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 912
Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)^n, 1/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g, m, n}, x] && NeQ[c*d^2 + a*e^2,
0] && !IntegerQ[m] && !IntegerQ[n]
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{\sqrt{\cot (c+d x)} (a B+b B \tan (c+d x))}{(a+b \tan (c+d x))^{3/2}} \, dx &=B \int \frac{\sqrt{\cot (c+d x)}}{\sqrt{a+b \tan (c+d x)}} \, dx\\ &=\left (B \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{1}{\sqrt{\tan (c+d x)} \sqrt{a+b \tan (c+d x)}} \, dx\\ &=\frac{\left (B \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \sqrt{a+b x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\left (B \sqrt{\cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \left (\frac{i}{2 (i-x) \sqrt{x} \sqrt{a+b x}}+\frac{i}{2 \sqrt{x} (i+x) \sqrt{a+b x}}\right ) \, dx,x,\tan (c+d x)\right )}{d}\\ &=\frac{\left (i B \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 (i B \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 (i B \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 (i B \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{B \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{B \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 = 0.193956, size = 145, normalized size = 0.96 \[ \frac{(-1)^{3/4} B \sqrt{\tan (c+d x)} \sqrt{\cot (c+d x)} \left (-\frac{\tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{-a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{-a-i b}}-\frac{\tanh ^{-1}\left (\frac{\sqrt [4]{-1} \sqrt{a-i b} \sqrt{\tan (c+d x)}}{\sqrt{a+b \tan (c+d x)}}\right )}{\sqrt{a-i b}}\right )}{d} \]
Antiderivative was successfully verified.
[In]
Integrate[(Sqrt[Cot[c + d*x]]*(a*B + b*B*Tan[c + d*x]))/(a + b*Tan[c + d*x])^(3/2),x]
[Out]
((-1)^(3/4)*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]) - 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 = 0.593, size = 2054, normalized size = 13.6 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(cot(d*x+c)^(1/2)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),x)
[Out]
B/d/(-I*a+(a^2+b^2)^(1/2)-b)/(I*a+(a^2+b^2)^(1/2)-b)/a*2^(1/2)*(cos(d*x+c)/sin(d*x+c))^(1/2)*(1/cos(d*x+c)*(a*
cos(d*x+c)+b*sin(d*x+c)))^(1/2)*(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*(a^2+b^2)^(1/2)-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*(a^2+b^2)^(1/2)-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^3-2*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+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^3+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+2*EllipticF((((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),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)*a^2+4*EllipticF((((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),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)*b^2-Elli
pticPi((((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*(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*(a^2+b^2)^(1/2)-4*EllipticF((((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),1/2*2^(1/2)*((-b+(a^2+b^2)^(1/2))/(a^2+b^2)^(1/2))^(1/2))*a^2*b-4*EllipticF((((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),1/2*2^(1/2)*((-b+(a^2+b^2)^(1
/2))/(a^2+b^2)^(1/2))^(1/2))*b^3+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+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)*(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)*sin(d*x+c)^2/(cos(d*x+c)-1)/(a*cos(d*x+c)+b*sin(d*x+c))
________________________________________________________________________________________
Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B b \tan \left (d x + c\right ) + B a\right )} \sqrt{\cot \left (d x + c\right )}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(1/2)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),x, algorithm="maxima")
[Out]
integrate((B*b*tan(d*x + c) + B*a)*sqrt(cot(d*x + c))/(b*tan(d*x + c) + a)^(3/2), x)
________________________________________________________________________________________
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(cot(d*x+c)^(1/2)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),x, algorithm="fricas")
[Out]
Timed out
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} B \int \frac{\sqrt{\cot{\left (c + d x \right )}}}{\sqrt{a + b \tan{\left (c + d x \right )}}}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)**(1/2)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))**(3/2),x)
[Out]
B*Integral(sqrt(cot(c + d*x))/sqrt(a + b*tan(c + d*x)), x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (B b \tan \left (d x + c\right ) + B a\right )} \sqrt{\cot \left (d x + c\right )}}{{\left (b \tan \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(cot(d*x+c)^(1/2)*(a*B+b*B*tan(d*x+c))/(a+b*tan(d*x+c))^(3/2),x, algorithm="giac")
[Out]
integrate((B*b*tan(d*x + c) + B*a)*sqrt(cot(d*x + c))/(b*tan(d*x + c) + a)^(3/2), x)