3.3.37 \(\int \frac {a+a \sec (c+d x)}{(e \cot (c+d x))^{3/2}} \, dx\) [237]

Optimal. Leaf size=320 \[ \frac {2 \cot (c+d x) (3 a+a \sec (c+d x))}{3 d (e \cot (c+d x))^{3/2}}-\frac {a \cot (c+d x) \csc (c+d x) F\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {\sin (2 c+2 d x)}}{3 d (e \cot (c+d x))^{3/2}}+\frac {a \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}-\frac {a \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}+\frac {a \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}-\frac {a \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)} \]

[Out]

2/3*cot(d*x+c)*(3*a+a*sec(d*x+c))/d/(e*cot(d*x+c))^(3/2)+1/3*a*cot(d*x+c)*csc(d*x+c)*(sin(c+1/4*Pi+d*x)^2)^(1/
2)/sin(c+1/4*Pi+d*x)*EllipticF(cos(c+1/4*Pi+d*x),2^(1/2))*sin(2*d*x+2*c)^(1/2)/d/(e*cot(d*x+c))^(3/2)-1/2*a*ar
ctan(-1+2^(1/2)*tan(d*x+c)^(1/2))/d/(e*cot(d*x+c))^(3/2)*2^(1/2)/tan(d*x+c)^(3/2)-1/2*a*arctan(1+2^(1/2)*tan(d
*x+c)^(1/2))/d/(e*cot(d*x+c))^(3/2)*2^(1/2)/tan(d*x+c)^(3/2)+1/4*a*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d
/(e*cot(d*x+c))^(3/2)*2^(1/2)/tan(d*x+c)^(3/2)-1/4*a*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d/(e*cot(d*x+c)
)^(3/2)*2^(1/2)/tan(d*x+c)^(3/2)

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Rubi [A]
time = 0.19, antiderivative size = 320, normalized size of antiderivative = 1.00, number of steps used = 17, number of rules used = 14, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.609, Rules used = {3985, 3966, 3969, 3557, 335, 217, 1179, 642, 1176, 631, 210, 2694, 2653, 2720} \begin {gather*} \frac {a \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d \tan ^{\frac {3}{2}}(c+d x) (e \cot (c+d x))^{3/2}}-\frac {a \text {ArcTan}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d \tan ^{\frac {3}{2}}(c+d x) (e \cot (c+d x))^{3/2}}+\frac {2 \cot (c+d x) (a \sec (c+d x)+3 a)}{3 d (e \cot (c+d x))^{3/2}}+\frac {a \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \tan ^{\frac {3}{2}}(c+d x) (e \cot (c+d x))^{3/2}}-\frac {a \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \tan ^{\frac {3}{2}}(c+d x) (e \cot (c+d x))^{3/2}}-\frac {a \sqrt {\sin (2 c+2 d x)} \cot (c+d x) \csc (c+d x) F\left (\left .c+d x-\frac {\pi }{4}\right |2\right )}{3 d (e \cot (c+d x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])/(e*Cot[c + d*x])^(3/2),x]

[Out]

(2*Cot[c + d*x]*(3*a + a*Sec[c + d*x]))/(3*d*(e*Cot[c + d*x])^(3/2)) - (a*Cot[c + d*x]*Csc[c + d*x]*EllipticF[
c - Pi/4 + d*x, 2]*Sqrt[Sin[2*c + 2*d*x]])/(3*d*(e*Cot[c + d*x])^(3/2)) + (a*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d
*x]]])/(Sqrt[2]*d*(e*Cot[c + d*x])^(3/2)*Tan[c + d*x]^(3/2)) - (a*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqr
t[2]*d*(e*Cot[c + d*x])^(3/2)*Tan[c + d*x]^(3/2)) + (a*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*
Sqrt[2]*d*(e*Cot[c + d*x])^(3/2)*Tan[c + d*x]^(3/2)) - (a*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/
(2*Sqrt[2]*d*(e*Cot[c + d*x])^(3/2)*Tan[c + d*x]^(3/2))

Rule 210

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

Rule 217

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 335

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 631

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 642

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 1176

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 1179

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 2653

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 2694

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 2720

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

Rule 3557

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 3966

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)), x_Symbol] :> Simp[(-e)*(e*Cot
[c + d*x])^(m - 1)*((a*m + b*(m - 1)*Csc[c + d*x])/(d*m*(m - 1))), x] - Dist[e^2/m, Int[(e*Cot[c + d*x])^(m -
2)*(a*m + b*(m - 1)*Csc[c + d*x]), x], x] /; FreeQ[{a, b, c, d, e}, x] && GtQ[m, 1]

Rule 3969

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 3985

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]

Rubi steps

\begin {align*} \int \frac {a+a \sec (c+d x)}{(e \cot (c+d x))^{3/2}} \, dx &=\frac {\int (a+a \sec (c+d x)) \tan ^{\frac {3}{2}}(c+d x) \, dx}{(e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}\\ &=\frac {2 \cot (c+d x) (3 a+a \sec (c+d x))}{3 d (e \cot (c+d x))^{3/2}}-\frac {2 \int \frac {\frac {3 a}{2}+\frac {1}{2} a \sec (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{3 (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}\\ &=\frac {2 \cot (c+d x) (3 a+a \sec (c+d x))}{3 d (e \cot (c+d x))^{3/2}}-\frac {a \int \frac {\sec (c+d x)}{\sqrt {\tan (c+d x)}} \, dx}{3 (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}-\frac {a \int \frac {1}{\sqrt {\tan (c+d x)}} \, dx}{(e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}\\ &=\frac {2 \cot (c+d x) (3 a+a \sec (c+d x))}{3 d (e \cot (c+d x))^{3/2}}-\frac {\left (a \cos ^{\frac {3}{2}}(c+d x)\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)}} \, dx}{3 (e \cot (c+d x))^{3/2} \sin ^{\frac {3}{2}}(c+d x)}-\frac {a \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (1+x^2\right )} \, dx,x,\tan (c+d x)\right )}{d (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}\\ &=\frac {2 \cot (c+d x) (3 a+a \sec (c+d x))}{3 d (e \cot (c+d x))^{3/2}}-\frac {\left (a \cot (c+d x) \csc (c+d x) \sqrt {\sin (2 c+2 d x)}\right ) \int \frac {1}{\sqrt {\sin (2 c+2 d x)}} \, dx}{3 (e \cot (c+d x))^{3/2}}-\frac {(2 a) \text {Subst}\left (\int \frac {1}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}\\ &=\frac {2 \cot (c+d x) (3 a+a \sec (c+d x))}{3 d (e \cot (c+d x))^{3/2}}-\frac {a \cot (c+d x) \csc (c+d x) F\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {\sin (2 c+2 d x)}}{3 d (e \cot (c+d x))^{3/2}}-\frac {a \text {Subst}\left (\int \frac {1-x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}-\frac {a \text {Subst}\left (\int \frac {1+x^2}{1+x^4} \, dx,x,\sqrt {\tan (c+d x)}\right )}{d (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}\\ &=\frac {2 \cot (c+d x) (3 a+a \sec (c+d x))}{3 d (e \cot (c+d x))^{3/2}}-\frac {a \cot (c+d x) \csc (c+d x) F\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {\sin (2 c+2 d x)}}{3 d (e \cot (c+d x))^{3/2}}-\frac {a \text {Subst}\left (\int \frac {1}{1-\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}-\frac {a \text {Subst}\left (\int \frac {1}{1+\sqrt {2} x+x^2} \, dx,x,\sqrt {\tan (c+d x)}\right )}{2 d (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}+\frac {a \text {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 (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}+\frac {a \text {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 (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}\\ &=\frac {2 \cot (c+d x) (3 a+a \sec (c+d x))}{3 d (e \cot (c+d x))^{3/2}}-\frac {a \cot (c+d x) \csc (c+d x) F\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {\sin (2 c+2 d x)}}{3 d (e \cot (c+d x))^{3/2}}+\frac {a \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}-\frac {a \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}-\frac {a \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}+\frac {a \text {Subst}\left (\int \frac {1}{-1-x^2} \, dx,x,1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}\\ &=\frac {2 \cot (c+d x) (3 a+a \sec (c+d x))}{3 d (e \cot (c+d x))^{3/2}}-\frac {a \cot (c+d x) \csc (c+d x) F\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {\sin (2 c+2 d x)}}{3 d (e \cot (c+d x))^{3/2}}+\frac {a \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}-\frac {a \tan ^{-1}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}+\frac {a \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}-\frac {a \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )}{2 \sqrt {2} d (e \cot (c+d x))^{3/2} \tan ^{\frac {3}{2}}(c+d x)}\\ \end {align*}

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Mathematica [C] Result contains complex when optimal does not.
time = 12.66, size = 224, normalized size = 0.70 \begin {gather*} \frac {a (1+\cos (c+d x)) \cos (2 (c+d x)) \csc (c+d x) \sqrt {\csc ^2(c+d x)} \sec ^2\left (\frac {1}{2} (c+d x)\right ) \left (-4 \sqrt [4]{-1} \cot ^{\frac {3}{2}}(c+d x) F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} \sqrt {\cot (c+d x)}\right )\right |-1\right )+\sqrt {\csc ^2(c+d x)} \left (4+12 \cos (c+d x)+3 \text {ArcSin}(\cos (c+d x)-\sin (c+d x)) \cot (c+d x) \sqrt {\sin (2 (c+d x))}-3 \cot (c+d x) \log \left (\cos (c+d x)+\sin (c+d x)+\sqrt {\sin (2 (c+d x))}\right ) \sqrt {\sin (2 (c+d x))}\right )\right )}{12 d (e \cot (c+d x))^{3/2} \left (-1+\cot ^2(c+d x)\right )} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])/(e*Cot[c + d*x])^(3/2),x]

[Out]

(a*(1 + Cos[c + d*x])*Cos[2*(c + d*x)]*Csc[c + d*x]*Sqrt[Csc[c + d*x]^2]*Sec[(c + d*x)/2]^2*(-4*(-1)^(1/4)*Cot
[c + d*x]^(3/2)*EllipticF[I*ArcSinh[(-1)^(1/4)*Sqrt[Cot[c + d*x]]], -1] + Sqrt[Csc[c + d*x]^2]*(4 + 12*Cos[c +
d*x] + 3*ArcSin[Cos[c + d*x] - Sin[c + d*x]]*Cot[c + d*x]*Sqrt[Sin[2*(c + d*x)]] - 3*Cot[c + d*x]*Log[Cos[c +
d*x] + Sin[c + d*x] + Sqrt[Sin[2*(c + d*x)]]]*Sqrt[Sin[2*(c + d*x)]])))/(12*d*(e*Cot[c + d*x])^(3/2)*(-1 + Co
t[c + d*x]^2))

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Maple [C] Result contains complex when optimal does not.
time = 0.26, size = 698, normalized size = 2.18

method	result	size
default	\(-\frac {a \left (-1+\cos \left (d x +c \right )\right ) \left (-3 i \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {-1+\cos \left (d x +c \right )-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticPi \left (\sqrt {-\frac {-1+\cos \left (d x +c \right )-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+3 i \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {-1+\cos \left (d x +c \right )-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticPi \left (\sqrt {-\frac {-1+\cos \left (d x +c \right )-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )+4 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {-1+\cos \left (d x +c \right )-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticF \left (\sqrt {-\frac {-1+\cos \left (d x +c \right )-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {\sqrt {2}}{2}\right )-3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {-1+\cos \left (d x +c \right )-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticPi \left (\sqrt {-\frac {-1+\cos \left (d x +c \right )-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}-\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-3 \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {-1+\cos \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {\frac {-1+\cos \left (d x +c \right )+\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \sqrt {-\frac {-1+\cos \left (d x +c \right )-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \EllipticPi \left (\sqrt {-\frac {-1+\cos \left (d x +c \right )-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}, \frac {1}{2}+\frac {i}{2}, \frac {\sqrt {2}}{2}\right )-6 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+4 \sqrt {2}\, \cos \left (d x +c \right )+2 \sqrt {2}\right ) \left (1+\cos \left (d x +c \right )\right )^{2} \sqrt {2}}{6 d \left (\frac {e \cos \left (d x +c \right )}{\sin \left (d x +c \right )}\right )^{\frac {3}{2}} \sin \left (d x +c \right )^{5}}\)	\(698\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))/(e*cot(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/6*a/d*(-1+cos(d*x+c))*(-3*I*((-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)*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))*cos(d*x+c)*sin(d*x+c)+3*I*((-1+cos(d*x+c))/sin(d*x+c))^(1/2)*((-1+cos(d*x+c)+sin(d*x+c))/s
in(d*x+c))^(1/2)*(-(-1+cos(d*x+c)-sin(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))*cos(d*x+c)*sin(d*x+c)+4*cos(d*x+c)*sin(d*x+c)*((-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)*EllipticF((
-(-1+cos(d*x+c)-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))-3*cos(d*x+c)*sin(d*x+c)*((-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)*EllipticP
i((-(-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))-3*cos(d*x+c)*sin(d*x+c)*((-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)*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))-6*cos(d*x+c)^2*2^(1/2)+4*
2^(1/2)*cos(d*x+c)+2*2^(1/2))*(1+cos(d*x+c))^2/(e*cos(d*x+c)/sin(d*x+c))^(3/2)/sin(d*x+c)^5*2^(1/2)

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Maxima [A]
time = 0.24, size = 125, normalized size = 0.39 \begin {gather*} \frac {{\left (2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) + 2 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - \frac {2}{\sqrt {\tan \left (d x + c\right )}}\right )}\right ) - \sqrt {2} \log \left (\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + \sqrt {2} \log \left (-\frac {\sqrt {2}}{\sqrt {\tan \left (d x + c\right )}} + \frac {1}{\tan \left (d x + c\right )} + 1\right ) + 8 \, \sqrt {\tan \left (d x + c\right )}\right )} a e^{\left (-\frac {3}{2}\right )}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))/(e*cot(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

1/4*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2) + 2/sqrt(tan(d*x + c)))) + 2*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2) -
2/sqrt(tan(d*x + c)))) - sqrt(2)*log(sqrt(2)/sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) + sqrt(2)*log(-sqrt(2)/
sqrt(tan(d*x + c)) + 1/tan(d*x + c) + 1) + 8*sqrt(tan(d*x + c)))*a*e^(-3/2)/d

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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((a+a*sec(d*x+c))/(e*cot(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*} a \left (\int \frac {1}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx + \int \frac {\sec {\left (c + d x \right )}}{\left (e \cot {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))/(e*cot(d*x+c))**(3/2),x)

[Out]

a*(Integral((e*cot(c + d*x))**(-3/2), x) + Integral(sec(c + d*x)/(e*cot(c + d*x))**(3/2), x))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))/(e*cot(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)/(e*cot(d*x + c))^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {a+\frac {a}{\cos \left (c+d\,x\right )}}{{\left (e\,\mathrm {cot}\left (c+d\,x\right )\right )}^{3/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))/(e*cot(c + d*x))^(3/2),x)

[Out]

int((a + a/cos(c + d*x))/(e*cot(c + d*x))^(3/2), x)

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