3.241 \(\int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \cot (c+d x)}} \, dx\)

Optimal. Leaf size=339 \[ \frac {4 a^2 \sin (c+d x)}{d \sqrt {e \cot (c+d x)}}-\frac {a^2 \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d \sqrt {\tan (c+d x)} \sqrt {e \cot (c+d x)}}+\frac {a^2 \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d \sqrt {\tan (c+d x)} \sqrt {e \cot (c+d x)}}+\frac {2 a^2 \tan (c+d x)}{3 d \sqrt {e \cot (c+d x)}}+\frac {a^2 \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \sqrt {\tan (c+d x)} \sqrt {e \cot (c+d x)}}-\frac {a^2 \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \sqrt {\tan (c+d x)} \sqrt {e \cot (c+d x)}}-\frac {4 a^2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right )}{d \sqrt {\sin (2 c+2 d x)} \sqrt {e \cot (c+d x)}} \]

[Out]

4*a^2*sin(d*x+c)/d/(e*cot(d*x+c))^(1/2)+4*a^2*cos(d*x+c)*(sin(c+1/4*Pi+d*x)^2)^(1/2)/sin(c+1/4*Pi+d*x)*Ellipti
cE(cos(c+1/4*Pi+d*x),2^(1/2))/d/(e*cot(d*x+c))^(1/2)/sin(2*d*x+2*c)^(1/2)+1/2*a^2*arctan(-1+2^(1/2)*tan(d*x+c)
^(1/2))/d*2^(1/2)/(e*cot(d*x+c))^(1/2)/tan(d*x+c)^(1/2)+1/2*a^2*arctan(1+2^(1/2)*tan(d*x+c)^(1/2))/d*2^(1/2)/(
e*cot(d*x+c))^(1/2)/tan(d*x+c)^(1/2)+1/4*a^2*ln(1-2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)/(e*cot(d*x+c)
)^(1/2)/tan(d*x+c)^(1/2)-1/4*a^2*ln(1+2^(1/2)*tan(d*x+c)^(1/2)+tan(d*x+c))/d*2^(1/2)/(e*cot(d*x+c))^(1/2)/tan(
d*x+c)^(1/2)+2/3*a^2*tan(d*x+c)/d/(e*cot(d*x+c))^(1/2)

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Rubi [A]  time = 0.32, antiderivative size = 339, normalized size of antiderivative = 1.00, number of steps used = 20, number of rules used = 16, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.640, Rules used = {3900, 3886, 3476, 329, 297, 1162, 617, 204, 1165, 628, 2613, 2615, 2572, 2639, 2607, 30} \[ \frac {4 a^2 \sin (c+d x)}{d \sqrt {e \cot (c+d x)}}-\frac {a^2 \tan ^{-1}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )}{\sqrt {2} d \sqrt {\tan (c+d x)} \sqrt {e \cot (c+d x)}}+\frac {a^2 \tan ^{-1}\left (\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{\sqrt {2} d \sqrt {\tan (c+d x)} \sqrt {e \cot (c+d x)}}+\frac {2 a^2 \tan (c+d x)}{3 d \sqrt {e \cot (c+d x)}}+\frac {a^2 \log \left (\tan (c+d x)-\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \sqrt {\tan (c+d x)} \sqrt {e \cot (c+d x)}}-\frac {a^2 \log \left (\tan (c+d x)+\sqrt {2} \sqrt {\tan (c+d x)}+1\right )}{2 \sqrt {2} d \sqrt {\tan (c+d x)} \sqrt {e \cot (c+d x)}}-\frac {4 a^2 \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right )}{d \sqrt {\sin (2 c+2 d x)} \sqrt {e \cot (c+d x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(4*a^2*Sin[c + d*x])/(d*Sqrt[e*Cot[c + d*x]]) - (4*a^2*Cos[c + d*x]*EllipticE[c - Pi/4 + d*x, 2])/(d*Sqrt[e*Co
t[c + d*x]]*Sqrt[Sin[2*c + 2*d*x]]) - (a^2*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d*Sqrt[e*Cot[c + d
*x]]*Sqrt[Tan[c + d*x]]) + (a^2*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]])/(Sqrt[2]*d*Sqrt[e*Cot[c + d*x]]*Sqrt[T
an[c + d*x]]) + (a^2*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d*Sqrt[e*Cot[c + d*x]]*Sqr
t[Tan[c + d*x]]) - (a^2*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]])/(2*Sqrt[2]*d*Sqrt[e*Cot[c + d*x]]*
Sqrt[Tan[c + d*x]]) + (2*a^2*Tan[c + d*x])/(3*d*Sqrt[e*Cot[c + d*x]])

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

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 297

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

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 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 2572

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

Rule 2607

Int[sec[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[1/f, Subst[Int[(b*x)
^n*(1 + x^2)^(m/2 - 1), x], x, Tan[e + f*x]], x] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2] &&  !(IntegerQ[(n
- 1)/2] && LtQ[0, n, m - 1])

Rule 2613

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(a^2*(a*Sec[
e + f*x])^(m - 2)*(b*Tan[e + f*x])^(n + 1))/(b*f*(m + n - 1)), x] + Dist[(a^2*(m - 2))/(m + n - 1), Int[(a*Sec
[e + f*x])^(m - 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[
n, 1/2])) && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 2615

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

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, 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 3886

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Int[ExpandI
ntegrand[(e*Cot[c + d*x])^m, (a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && IGtQ[n, 0]

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]

Rubi steps

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

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Mathematica [C]  time = 11.46, size = 221, normalized size = 0.65 \[ \frac {a^2 \sin \left (\frac {1}{2} (c+d x)\right ) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \sec ^4\left (\frac {1}{2} \cot ^{-1}(\cot (c+d x))\right ) \left (8 \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {7}{4};-\tan ^2(c+d x)\right )+4 \, _2F_1\left (-\frac {3}{4},1;\frac {1}{4};-\cot ^2(c+d x)\right )+3 \sqrt {2} \cot ^{\frac {3}{2}}(c+d x) \left (\log \left (\cot (c+d x)-\sqrt {2} \sqrt {\cot (c+d x)}+1\right )-\log \left (\cot (c+d x)+\sqrt {2} \sqrt {\cot (c+d x)}+1\right )+2 \tan ^{-1}\left (1-\sqrt {2} \sqrt {\cot (c+d x)}\right )-2 \tan ^{-1}\left (\sqrt {2} \sqrt {\cot (c+d x)}+1\right )\right )\right )}{3 d \sqrt {e \cot (c+d x)}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*Cos[(c + d*x)/2]^5*(4*Hypergeometric2F1[-3/4, 1, 1/4, -Cot[c + d*x]^2] + 8*Hypergeometric2F1[1/2, 3/4, 7/
4, -Tan[c + d*x]^2] + 3*Sqrt[2]*Cot[c + d*x]^(3/2)*(2*ArcTan[1 - Sqrt[2]*Sqrt[Cot[c + d*x]]] - 2*ArcTan[1 + Sq
rt[2]*Sqrt[Cot[c + d*x]]] + Log[1 - Sqrt[2]*Sqrt[Cot[c + d*x]] + Cot[c + d*x]] - Log[1 + Sqrt[2]*Sqrt[Cot[c +
d*x]] + Cot[c + d*x]]))*Sec[c + d*x]*Sec[ArcCot[Cot[c + d*x]]/2]^4*Sin[(c + d*x)/2])/(3*d*Sqrt[e*Cot[c + d*x]]
)

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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+a*sec(d*x+c))^2/(e*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 \[ \int \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{2}}{\sqrt {e \cot \left (d x + c\right )}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [C]  time = 2.35, size = 1480, normalized size = 4.37 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*a^2/d*(1+cos(d*x+c))^2*(-1+cos(d*x+c))^2*(3*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))*((-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)*cos(d*x+c)^2-3*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))*((-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)*cos(d*x+c)^2+3*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))*cos(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)-3*I*cos(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)-3*((-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)*cos(d*x+c)^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))-3*((-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)*cos(d*x+c)^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))+24*((-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)*cos(d*x+c)^2*EllipticE(((1-cos(d*x+c)+sin
(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))-12*((-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)*cos(d*x+c)^2*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/s
in(d*x+c))^(1/2),1/2*2^(1/2))-3*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)*((-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)+si
n(d*x+c))/sin(d*x+c))^(1/2)-3*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))*c
os(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)+24*EllipticE(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*cos(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)-12*EllipticF(((1-cos(d*x+c)+sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))*cos(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
)-14*cos(d*x+c)^2*2^(1/2)+12*cos(d*x+c)*2^(1/2)+2*2^(1/2))/cos(d*x+c)/sin(d*x+c)^5/(e*cos(d*x+c)/sin(d*x+c))^(
1/2)*2^(1/2)

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maxima [A]  time = 0.51, size = 194, normalized size = 0.57 \[ -\frac {3 \, a^{2} e {\left (\frac {2 \, \sqrt {2} \arctan \left (\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} + 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{e^{\frac {3}{2}}} + \frac {2 \, \sqrt {2} \arctan \left (-\frac {\sqrt {2} {\left (\sqrt {2} \sqrt {e} - 2 \, \sqrt {\frac {e}{\tan \left (d x + c\right )}}\right )}}{2 \, \sqrt {e}}\right )}{e^{\frac {3}{2}}} + \frac {\sqrt {2} \log \left (\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{e^{\frac {3}{2}}} - \frac {\sqrt {2} \log \left (-\sqrt {2} \sqrt {e} \sqrt {\frac {e}{\tan \left (d x + c\right )}} + e + \frac {e}{\tan \left (d x + c\right )}\right )}{e^{\frac {3}{2}}}\right )} - \frac {8 \, a^{2} \tan \left (d x + c\right )}{\sqrt {\frac {e}{\tan \left (d x + c\right )}}}}{12 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12*(3*a^2*e*(2*sqrt(2)*arctan(1/2*sqrt(2)*(sqrt(2)*sqrt(e) + 2*sqrt(e/tan(d*x + c)))/sqrt(e))/e^(3/2) + 2*s
qrt(2)*arctan(-1/2*sqrt(2)*(sqrt(2)*sqrt(e) - 2*sqrt(e/tan(d*x + c)))/sqrt(e))/e^(3/2) + sqrt(2)*log(sqrt(2)*s
qrt(e)*sqrt(e/tan(d*x + c)) + e + e/tan(d*x + c))/e^(3/2) - sqrt(2)*log(-sqrt(2)*sqrt(e)*sqrt(e/tan(d*x + c))
+ e + e/tan(d*x + c))/e^(3/2)) - 8*a^2*tan(d*x + c)/sqrt(e/tan(d*x + c)))/d

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^2}{\sqrt {e\,\mathrm {cot}\left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int \frac {1}{\sqrt {e \cot {\left (c + d x \right )}}}\, dx + \int \frac {2 \sec {\left (c + d x \right )}}{\sqrt {e \cot {\left (c + d x \right )}}}\, dx + \int \frac {\sec ^{2}{\left (c + d x \right )}}{\sqrt {e \cot {\left (c + d x \right )}}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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