3.802 \(\int \frac {a+b \tan (c+d x)}{\sqrt {\cot (c+d x)}} \, dx\)

Optimal. Leaf size=166 \[ \frac {(a+b) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {(a+b) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {(a-b) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {(a-b) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d}+\frac {2 b}{d \sqrt {\cot (c+d x)}} \]

[Out]

-1/2*(a-b)*arctan(-1+2^(1/2)*cot(d*x+c)^(1/2))/d*2^(1/2)-1/2*(a-b)*arctan(1+2^(1/2)*cot(d*x+c)^(1/2))/d*2^(1/2
)+1/4*(a+b)*ln(1+cot(d*x+c)-2^(1/2)*cot(d*x+c)^(1/2))/d*2^(1/2)-1/4*(a+b)*ln(1+cot(d*x+c)+2^(1/2)*cot(d*x+c)^(
1/2))/d*2^(1/2)+2*b/d/cot(d*x+c)^(1/2)

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Rubi [A]  time = 0.13, antiderivative size = 166, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 9, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.429, Rules used = {3673, 3529, 3534, 1168, 1162, 617, 204, 1165, 628} \[ \frac {(a+b) \log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}-\frac {(a+b) \log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{2 \sqrt {2} d}+\frac {(a-b) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {(a-b) \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )}{\sqrt {2} d}+\frac {2 b}{d \sqrt {\cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + b*Tan[c + d*x])/Sqrt[Cot[c + d*x]],x]

[Out]

((a - b)*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]])/(Sqrt[2]*d) - ((a - b)*ArcTan[1 + Sqrt[2]*Sqrt[Cot[c + d*x]]]
)/(Sqrt[2]*d) + (2*b)/(d*Sqrt[Cot[c + d*x]]) + ((a + b)*Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2
*Sqrt[2]*d) - ((a + b)*Log[1 + Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]])/(2*Sqrt[2]*d)

Rule 204

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> -Simp[ArcTan[(Rt[-b, 2]*x)/Rt[-a, 2]]/(Rt[-a, 2]*Rt[-b, 2]), x] /
; FreeQ[{a, b}, x] && PosQ[a/b] && (LtQ[a, 0] || LtQ[b, 0])

Rule 617

Int[((a_) + (b_.)*(x_) + (c_.)*(x_)^2)^(-1), x_Symbol] :> With[{q = 1 - 4*Simplify[(a*c)/b^2]}, Dist[-2/b, Sub
st[Int[1/(q - x^2), x], x, 1 + (2*c*x)/b], x] /; RationalQ[q] && (EqQ[q^2, 1] ||  !RationalQ[b^2 - 4*a*c])] /;
FreeQ[{a, b, c}, x] && NeQ[b^2 - 4*a*c, 0]

Rule 628

Int[((d_) + (e_.)*(x_))/((a_.) + (b_.)*(x_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(d*Log[RemoveContent[a + b*x +
c*x^2, x]])/b, x] /; FreeQ[{a, b, c, d, e}, x] && EqQ[2*c*d - b*e, 0]

Rule 1162

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(2*d)/e, 2]}, Dist[e/(2*c), Int[1/S
imp[d/e + q*x + x^2, x], x], x] + Dist[e/(2*c), Int[1/Simp[d/e - q*x + x^2, x], x], x]] /; FreeQ[{a, c, d, e},
x] && EqQ[c*d^2 - a*e^2, 0] && PosQ[d*e]

Rule 1165

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[(-2*d)/e, 2]}, Dist[e/(2*c*q), Int[
(q - 2*x)/Simp[d/e + q*x - x^2, x], x], x] + Dist[e/(2*c*q), Int[(q + 2*x)/Simp[d/e - q*x - x^2, x], x], x]] /
; FreeQ[{a, c, d, e}, x] && EqQ[c*d^2 - a*e^2, 0] && NegQ[d*e]

Rule 1168

Int[((d_) + (e_.)*(x_)^2)/((a_) + (c_.)*(x_)^4), x_Symbol] :> With[{q = Rt[a*c, 2]}, Dist[(d*q + a*e)/(2*a*c),
Int[(q + c*x^2)/(a + c*x^4), x], x] + Dist[(d*q - a*e)/(2*a*c), Int[(q - c*x^2)/(a + c*x^4), x], x]] /; FreeQ
[{a, c, d, e}, x] && NeQ[c*d^2 + a*e^2, 0] && NeQ[c*d^2 - a*e^2, 0] && NegQ[-(a*c)]

Rule 3529

Int[((a_.) + (b_.)*tan[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*tan[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[((
b*c - a*d)*(a + b*Tan[e + f*x])^(m + 1))/(f*(m + 1)*(a^2 + b^2)), x] + Dist[1/(a^2 + b^2), Int[(a + b*Tan[e +
f*x])^(m + 1)*Simp[a*c + b*d - (b*c - a*d)*Tan[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c
- a*d, 0] && NeQ[a^2 + b^2, 0] && LtQ[m, -1]

Rule 3534

Int[((c_) + (d_.)*tan[(e_.) + (f_.)*(x_)])/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[2/f, Subst[I
nt[(b*c + d*x^2)/(b^2 + x^4), x], x, Sqrt[b*Tan[e + f*x]]], x] /; FreeQ[{b, c, d, e, f}, x] && NeQ[c^2 - d^2,
0] && NeQ[c^2 + d^2, 0]

Rule 3673

Int[(cot[(e_.) + (f_.)*(x_)]*(d_.))^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^(n_.))^(p_.), x_Symbol] :> Dist
[d^(n*p), Int[(d*Cot[e + f*x])^(m - n*p)*(b + a*Cot[e + f*x]^n)^p, x], x] /; FreeQ[{a, b, d, e, f, m, n, p}, x
] &&  !IntegerQ[m] && IntegersQ[n, p]

Rubi steps

\begin {align*} \int \frac {a+b \tan (c+d x)}{\sqrt {\cot (c+d x)}} \, dx &=\int \frac {b+a \cot (c+d x)}{\cot ^{\frac {3}{2}}(c+d x)} \, dx\\ &=\frac {2 b}{d \sqrt {\cot (c+d x)}}+\int \frac {a-b \cot (c+d x)}{\sqrt {\cot (c+d x)}} \, dx\\ &=\frac {2 b}{d \sqrt {\cot (c+d x)}}+\frac {2 \operatorname {Subst}\left (\int \frac {-a+b x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}\\ &=\frac {2 b}{d \sqrt {\cot (c+d x)}}-\frac {(a-b) \operatorname {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}-\frac {(a+b) \operatorname {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\cot (c+d x)}\right )}{d}\\ &=\frac {2 b}{d \sqrt {\cot (c+d x)}}-\frac {(a-b) \operatorname {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 d}-\frac {(a-b) \operatorname {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 d}+\frac {(a+b) \operatorname {Subst}\left (\int \frac {\sqrt {2}+2 x}{-1-\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \sqrt {2} d}+\frac {(a+b) \operatorname {Subst}\left (\int \frac {\sqrt {2}-2 x}{-1+\sqrt {2} x-x^2} \, dx,x,\sqrt {\cot (c+d x)}\right )}{2 \sqrt {2} d}\\ &=\frac {2 b}{d \sqrt {\cot (c+d x)}}+\frac {(a+b) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}-\frac {(a+b) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}-\frac {(a-b) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {(a-b) \operatorname {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}\\ &=\frac {(a-b) \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}-\frac {(a-b) \tan ^{-1}\left (1+\sqrt {2} \sqrt {\cot (c+d x)}\right )}{\sqrt {2} d}+\frac {2 b}{d \sqrt {\cot (c+d x)}}+\frac {(a+b) \log \left (1-\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}-\frac {(a+b) \log \left (1+\sqrt {2} \sqrt {\cot (c+d x)}+\cot (c+d x)\right )}{2 \sqrt {2} d}\\ \end {align*}

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Mathematica [C]  time = 0.37, size = 194, normalized size = 1.17 \[ \frac {\sqrt {\tan (c+d x)} \sqrt {\cot (c+d x)} \left (8 a \tan ^{\frac {3}{2}}(c+d x) \, _2F_1\left (\frac {3}{4},1;\frac {7}{4};-\tan ^2(c+d x)\right )+3 b \left (2 \left (\sqrt {2} \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-\sqrt {2} \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )\right )+8 \sqrt {\tan (c+d x)}+\sqrt {2} \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )-\sqrt {2} \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )\right )\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Tan[c + d*x])/Sqrt[Cot[c + d*x]],x]

[Out]

(Sqrt[Cot[c + d*x]]*Sqrt[Tan[c + d*x]]*(3*b*(2*(Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] - Sqrt[2]*ArcTa
n[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]]) + Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] - Sqrt[2]*Log[
1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] + 8*Sqrt[Tan[c + d*x]]) + 8*a*Hypergeometric2F1[3/4, 1, 7/4, -T
an[c + d*x]^2]*Tan[c + d*x]^(3/2)))/(12*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((a+b*tan(d*x+c))/cot(d*x+c)^(1/2),x, algorithm="fricas")

[Out]

Timed out

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {b \tan \left (d x + c\right ) + a}{\sqrt {\cot \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))/cot(d*x+c)^(1/2),x, algorithm="giac")

[Out]

integrate((b*tan(d*x + c) + a)/sqrt(cot(d*x + c)), x)

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maple [C]  time = 1.13, size = 1112, normalized size = 6.70 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+b*tan(d*x+c))/cot(d*x+c)^(1/2),x)

[Out]

1/2/d*(1+cos(d*x+c))^2*(-1+cos(d*x+c))*(I*sin(d*x+c)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1
/2+1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((1-cos
(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*a-I*sin(d*x+c)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1
/2+1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((1-cos
(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*b-I*sin(d*x+c)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1
/2-1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((1-cos
(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*a+I*sin(d*x+c)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1
/2-1/2*I,1/2*2^(1/2))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((1-cos
(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*b+sin(d*x+c)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c
)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d
*x+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*a+sin(d*x+c)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+
sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x
+c))^(1/2),1/2+1/2*I,1/2*2^(1/2))*b+sin(d*x+c)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+si
n(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c
))^(1/2),1/2-1/2*I,1/2*2^(1/2))*a+sin(d*x+c)*((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(
d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*EllipticPi(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))
^(1/2),1/2-1/2*I,1/2*2^(1/2))*b-2*sin(d*x+c)*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2
))*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2)*((1-cos(d*x+c)+sin(d*x+c))
/sin(d*x+c))^(1/2)*b+2*cos(d*x+c)*2^(1/2)*b-2*b*2^(1/2))/sin(d*x+c)^4/(cos(d*x+c)/sin(d*x+c))^(1/2)*2^(1/2)

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maxima [A]  time = 0.62, size = 139, normalized size = 0.84 \[ -\frac {2 \, \sqrt {2} {\left (a - b\right )} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} {\left (a - b\right )} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + \sqrt {2} {\left (a + b\right )} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - \sqrt {2} {\left (a + b\right )} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) - 8 \, b \sqrt {\tan \left (d x + c\right )}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))/cot(d*x+c)^(1/2),x, algorithm="maxima")

[Out]

-1/4*(2*sqrt(2)*(a - b)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + 2*sqrt(2)*(a - b)*arctan(-1/2*s
qrt(2)*(sqrt(2) - 2/sqrt(tan(d*x + c)))) + sqrt(2)*(a + b)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1
) - sqrt(2)*(a + b)*log(-sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) - 8*b*sqrt(tan(d*x + c)))/d

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mupad [B]  time = 5.73, size = 99, normalized size = 0.60 \[ \frac {2\,b}{d\,\sqrt {\mathrm {cot}\left (c+d\,x\right )}}+\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atan}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {cot}\left (c+d\,x\right )}\right )\,1{}\mathrm {i}}{d}+\frac {{\left (-1\right )}^{1/4}\,a\,\mathrm {atanh}\left ({\left (-1\right )}^{1/4}\,\sqrt {\mathrm {cot}\left (c+d\,x\right )}\right )\,1{}\mathrm {i}}{d}+\frac {{\left (-1\right )}^{1/4}\,b\,\mathrm {atan}\left (\frac {{\left (-1\right )}^{1/4}}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}}\right )\,1{}\mathrm {i}}{d}+\frac {{\left (-1\right )}^{1/4}\,b\,\mathrm {atanh}\left (\frac {{\left (-1\right )}^{1/4}}{\sqrt {\mathrm {cot}\left (c+d\,x\right )}}\right )\,1{}\mathrm {i}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(c + d*x))/cot(c + d*x)^(1/2),x)

[Out]

(2*b)/(d*cot(c + d*x)^(1/2)) + ((-1)^(1/4)*a*atan((-1)^(1/4)*cot(c + d*x)^(1/2))*1i)/d + ((-1)^(1/4)*a*atanh((
-1)^(1/4)*cot(c + d*x)^(1/2))*1i)/d + ((-1)^(1/4)*b*atan((-1)^(1/4)/cot(c + d*x)^(1/2))*1i)/d + ((-1)^(1/4)*b*
atanh((-1)^(1/4)/cot(c + d*x)^(1/2))*1i)/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {a + b \tan {\left (c + d x \right )}}{\sqrt {\cot {\left (c + d x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*tan(d*x+c))/cot(d*x+c)**(1/2),x)

[Out]

Integral((a + b*tan(c + d*x))/sqrt(cot(c + d*x)), x)

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