3.235 \(\int \sqrt{e \cot (c+d x)} (a+a \sec (c+d x)) \, dx\)

Optimal. Leaf size=274 \[ \frac{a \sqrt{\sin (2 c+2 d x)} \sec (c+d x) \text{EllipticF}\left (c+d x-\frac{\pi }{4},2\right ) \sqrt{e \cot (c+d x)}}{d}-\frac{a \sqrt{\tan (c+d x)} \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right ) \sqrt{e \cot (c+d x)}}{\sqrt{2} d}+\frac{a \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right ) \sqrt{\tan (c+d x)} \sqrt{e \cot (c+d x)}}{\sqrt{2} d}-\frac{a \sqrt{\tan (c+d x)} \sqrt{e \cot (c+d x)} \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{a \sqrt{\tan (c+d x)} \sqrt{e \cot (c+d x)} \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d} \]

[Out]

(a*Sqrt[e*Cot[c + d*x]]*EllipticF[c - Pi/4 + d*x, 2]*Sec[c + d*x]*Sqrt[Sin[2*c + 2*d*x]])/d - (a*ArcTan[1 - Sq
rt[2]*Sqrt[Tan[c + d*x]]]*Sqrt[e*Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/(Sqrt[2]*d) + (a*ArcTan[1 + Sqrt[2]*Sqrt[Ta
n[c + d*x]]]*Sqrt[e*Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/(Sqrt[2]*d) - (a*Sqrt[e*Cot[c + d*x]]*Log[1 - Sqrt[2]*Sq
rt[Tan[c + d*x]] + Tan[c + d*x]]*Sqrt[Tan[c + d*x]])/(2*Sqrt[2]*d) + (a*Sqrt[e*Cot[c + d*x]]*Log[1 + Sqrt[2]*S
qrt[Tan[c + d*x]] + Tan[c + d*x]]*Sqrt[Tan[c + d*x]])/(2*Sqrt[2]*d)

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Rubi [A]  time = 0.201291, antiderivative size = 274, normalized size of antiderivative = 1., number of steps used = 16, number of rules used = 13, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.565, Rules used = {3900, 3884, 3476, 329, 211, 1165, 628, 1162, 617, 204, 2614, 2573, 2641} \[ -\frac{a \sqrt{\tan (c+d x)} \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right ) \sqrt{e \cot (c+d x)}}{\sqrt{2} d}+\frac{a \tan ^{-1}\left (\sqrt{2} \sqrt{\tan (c+d x)}+1\right ) \sqrt{\tan (c+d x)} \sqrt{e \cot (c+d x)}}{\sqrt{2} d}-\frac{a \sqrt{\tan (c+d x)} \sqrt{e \cot (c+d x)} \log \left (\tan (c+d x)-\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{a \sqrt{\tan (c+d x)} \sqrt{e \cot (c+d x)} \log \left (\tan (c+d x)+\sqrt{2} \sqrt{\tan (c+d x)}+1\right )}{2 \sqrt{2} d}+\frac{a \sqrt{\sin (2 c+2 d x)} \sec (c+d x) F\left (\left .c+d x-\frac{\pi }{4}\right |2\right ) \sqrt{e \cot (c+d x)}}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[e*Cot[c + d*x]]*(a + a*Sec[c + d*x]),x]

[Out]

(a*Sqrt[e*Cot[c + d*x]]*EllipticF[c - Pi/4 + d*x, 2]*Sec[c + d*x]*Sqrt[Sin[2*c + 2*d*x]])/d - (a*ArcTan[1 - Sq
rt[2]*Sqrt[Tan[c + d*x]]]*Sqrt[e*Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/(Sqrt[2]*d) + (a*ArcTan[1 + Sqrt[2]*Sqrt[Ta
n[c + d*x]]]*Sqrt[e*Cot[c + d*x]]*Sqrt[Tan[c + d*x]])/(Sqrt[2]*d) - (a*Sqrt[e*Cot[c + d*x]]*Log[1 - Sqrt[2]*Sq
rt[Tan[c + d*x]] + Tan[c + d*x]]*Sqrt[Tan[c + d*x]])/(2*Sqrt[2]*d) + (a*Sqrt[e*Cot[c + d*x]]*Log[1 + Sqrt[2]*S
qrt[Tan[c + d*x]] + Tan[c + d*x]]*Sqrt[Tan[c + d*x]])/(2*Sqrt[2]*d)

Rule 3900

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*((a_) + (b_.)*sec[(c_.) + (d_.)*(x_)])^(n_.), x_Symbol] :> Dist[(e*Co
t[c + d*x])^m*Tan[c + d*x]^m, Int[(a + b*Sec[c + d*x])^n/Tan[c + d*x]^m, x], x] /; FreeQ[{a, b, c, d, e, m, n}
, x] &&  !IntegerQ[m]

Rule 3884

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Dist[a, Int[(e*
Cot[c + d*x])^m, x], x] + Dist[b, Int[(e*Cot[c + d*x])^m*Csc[c + d*x], x], x] /; FreeQ[{a, b, c, d, e, m}, x]

Rule 3476

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Dist[b/d, Subst[Int[x^n/(b^2 + x^2), x], x, b*Tan[c + d
*x]], x] /; FreeQ[{b, c, d, n}, x] &&  !IntegerQ[n]

Rule 329

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = Denominator[m]}, Dist[k/c, Subst[I
nt[x^(k*(m + 1) - 1)*(a + (b*x^(k*n))/c^n)^p, x], x, (c*x)^(1/k)], x]] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0]
&& FractionQ[m] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 211

Int[((a_) + (b_.)*(x_)^4)^(-1), x_Symbol] :> With[{r = Numerator[Rt[a/b, 2]], s = Denominator[Rt[a/b, 2]]}, Di
st[1/(2*r), Int[(r - s*x^2)/(a + b*x^4), x], x] + Dist[1/(2*r), Int[(r + s*x^2)/(a + b*x^4), x], x]] /; FreeQ[
{a, b}, x] && (GtQ[a/b, 0] || (PosQ[a/b] && AtomQ[SplitProduct[SumBaseQ, a]] && AtomQ[SplitProduct[SumBaseQ, b
]]))

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 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 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 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 2614

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[e + f*x]]/(Sqrt[Co
s[e + f*x]]*Sqrt[b*Tan[e + f*x]]), Int[1/(Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]]), x], x] /; FreeQ[{b, e, f}, x
]

Rule 2573

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \sqrt{e \cot (c+d x)} (a+a \sec (c+d x)) \, dx &=\left (\sqrt{e \cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{a+a \sec (c+d x)}{\sqrt{\tan (c+d x)}} \, dx\\ &=\left (a \sqrt{e \cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{1}{\sqrt{\tan (c+d x)}} \, dx+\left (a \sqrt{e \cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \int \frac{\sec (c+d x)}{\sqrt{\tan (c+d x)}} \, dx\\ &=\frac{\left (a \sqrt{e \cot (c+d x)} \sqrt{\sin (c+d x)}\right ) \int \frac{1}{\sqrt{\cos (c+d x)} \sqrt{\sin (c+d x)}} \, dx}{\sqrt{\cos (c+d x)}}+\frac{\left (a \sqrt{e \cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d}\\ &=\left (a \sqrt{e \cot (c+d x)} \sec (c+d x) \sqrt{\sin (2 c+2 d x)}\right ) \int \frac{1}{\sqrt{\sin (2 c+2 d x)}} \, dx+\frac{\left (2 a \sqrt{e \cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=\frac{a \sqrt{e \cot (c+d x)} F\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt{\sin (2 c+2 d x)}}{d}+\frac{\left (a \sqrt{e \cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1-x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}+\frac{\left (a \sqrt{e \cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1+x^2}{1+x^4} \, dx,x,\sqrt{\tan (c+d x)}\right )}{d}\\ &=\frac{a \sqrt{e \cot (c+d x)} F\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt{\sin (2 c+2 d x)}}{d}+\frac{\left (a \sqrt{e \cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1-\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 d}+\frac{\left (a \sqrt{e \cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{1+\sqrt{2} x+x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 d}-\frac{\left (a \sqrt{e \cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}+2 x}{-1-\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \sqrt{2} d}-\frac{\left (a \sqrt{e \cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{2}-2 x}{-1+\sqrt{2} x-x^2} \, dx,x,\sqrt{\tan (c+d x)}\right )}{2 \sqrt{2} d}\\ &=\frac{a \sqrt{e \cot (c+d x)} F\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt{\sin (2 c+2 d x)}}{d}-\frac{a \sqrt{e \cot (c+d x)} \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right ) \sqrt{\tan (c+d x)}}{2 \sqrt{2} d}+\frac{a \sqrt{e \cot (c+d x)} \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right ) \sqrt{\tan (c+d x)}}{2 \sqrt{2} d}+\frac{\left (a \sqrt{e \cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1-\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}-\frac{\left (a \sqrt{e \cot (c+d x)} \sqrt{\tan (c+d x)}\right ) \operatorname{Subst}\left (\int \frac{1}{-1-x^2} \, dx,x,1+\sqrt{2} \sqrt{\tan (c+d x)}\right )}{\sqrt{2} d}\\ &=\frac{a \sqrt{e \cot (c+d x)} F\left (\left .c-\frac{\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt{\sin (2 c+2 d x)}}{d}-\frac{a \tan ^{-1}\left (1-\sqrt{2} \sqrt{\tan (c+d x)}\right ) \sqrt{e \cot (c+d x)} \sqrt{\tan (c+d x)}}{\sqrt{2} d}+\frac{a \tan ^{-1}\left (1+\sqrt{2} \sqrt{\tan (c+d x)}\right ) \sqrt{e \cot (c+d x)} \sqrt{\tan (c+d x)}}{\sqrt{2} d}-\frac{a \sqrt{e \cot (c+d x)} \log \left (1-\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right ) \sqrt{\tan (c+d x)}}{2 \sqrt{2} d}+\frac{a \sqrt{e \cot (c+d x)} \log \left (1+\sqrt{2} \sqrt{\tan (c+d x)}+\tan (c+d x)\right ) \sqrt{\tan (c+d x)}}{2 \sqrt{2} d}\\ \end{align*}

Mathematica [C]  time = 1.74716, size = 169, normalized size = 0.62 \[ \frac{a (\cos (c+d x)+1) \sec ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x) \sqrt{e \cot (c+d x)} \left (\sqrt{\sin (2 (c+d x))} \sqrt{\csc ^2(c+d x)} \left (\log \left (\sin (c+d x)+\sqrt{\sin (2 (c+d x))}+\cos (c+d x)\right )-\sin ^{-1}(\cos (c+d x)-\sin (c+d x))\right )+4 \sqrt [4]{-1} \sqrt{\cot (c+d x)} \text{EllipticF}\left (i \sinh ^{-1}\left (\sqrt [4]{-1} \sqrt{\cot (c+d x)}\right ),-1\right )\right )}{4 d \sqrt{\csc ^2(c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[e*Cot[c + d*x]]*(a + a*Sec[c + d*x]),x]

[Out]

(a*(1 + Cos[c + d*x])*Sqrt[e*Cot[c + d*x]]*Sec[(c + d*x)/2]^2*Sec[c + d*x]*(4*(-1)^(1/4)*Sqrt[Cot[c + d*x]]*El
lipticF[I*ArcSinh[(-1)^(1/4)*Sqrt[Cot[c + d*x]]], -1] + Sqrt[Csc[c + d*x]^2]*(-ArcSin[Cos[c + d*x] - Sin[c + d
*x]] + Log[Cos[c + d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]])*Sqrt[Sin[2*(c + d*x)]]))/(4*d*Sqrt[Csc[c + d
*x]^2])

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Maple [C]  time = 0.261, size = 289, normalized size = 1.1 \begin{align*} -{\frac{a\sqrt{2} \left ( -1+\cos \left ( dx+c \right ) \right ) \left ( \cos \left ( dx+c \right ) +1 \right ) ^{2}}{2\,d \left ( \sin \left ( dx+c \right ) \right ) ^{2}\cos \left ( dx+c \right ) }\sqrt{{\frac{e\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}}\sqrt{-{\frac{-1+\cos \left ( dx+c \right ) -\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}}\sqrt{{\frac{-1+\cos \left ( dx+c \right ) +\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}}\sqrt{{\frac{-1+\cos \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}} \left ( i{\it EllipticPi} \left ( \sqrt{-{\frac{-1+\cos \left ( dx+c \right ) -\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}},{\frac{1}{2}}-{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) -i{\it EllipticPi} \left ( \sqrt{-{\frac{-1+\cos \left ( dx+c \right ) -\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}},{\frac{1}{2}}+{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) +{\it EllipticPi} \left ( \sqrt{-{\frac{-1+\cos \left ( dx+c \right ) -\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}},{\frac{1}{2}}-{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) +{\it EllipticPi} \left ( \sqrt{-{\frac{-1+\cos \left ( dx+c \right ) -\sin \left ( dx+c \right ) }{\sin \left ( dx+c \right ) }}},{\frac{1}{2}}+{\frac{i}{2}},{\frac{\sqrt{2}}{2}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))*(e*cot(d*x+c))^(1/2),x)

[Out]

-1/2*a/d*2^(1/2)*(e*cos(d*x+c)/sin(d*x+c))^(1/2)*(-1+cos(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)*(I*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))-I*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))+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))
+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)))/sin(d*x+c)^2/cos(d*x+c)*(co
s(d*x+c)+1)^2

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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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((a+a*sec(d*x+c))*(e*cot(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*} a \left (\int \sqrt{e \cot{\left (c + d x \right )}}\, dx + \int \sqrt{e \cot{\left (c + d x \right )}} \sec{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))*(e*cot(d*x+c))**(1/2),x)

[Out]

a*(Integral(sqrt(e*cot(c + d*x)), x) + Integral(sqrt(e*cot(c + d*x))*sec(c + d*x), x))

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{e \cot \left (d x + c\right )}{\left (a \sec \left (d x + c\right ) + a\right )}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(e*cot(d*x + c))*(a*sec(d*x + c) + a), x)