3.2.14 \(\int \frac {(a+a \sec (c+d x))^2}{\sqrt {e \tan (c+d x)}} \, dx\) [114]

Optimal. Leaf size=278 \[ -\frac {a^2 \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d \sqrt {e}}+\frac {a^2 \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d \sqrt {e}}-\frac {a^2 \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d \sqrt {e}}+\frac {a^2 \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d \sqrt {e}}+\frac {2 a^2 F\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{d \sqrt {e \tan (c+d x)}}+\frac {2 a^2 \sqrt {e \tan (c+d x)}}{d e} \]

[Out]

-1/2*a^2*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/d*2^(1/2)/e^(1/2)+1/2*a^2*arctan(1+2^(1/2)*(e*tan(d*x+
c))^(1/2)/e^(1/2))/d*2^(1/2)/e^(1/2)-1/4*a^2*ln(e^(1/2)-2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/d*2^(
1/2)/e^(1/2)+1/4*a^2*ln(e^(1/2)+2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/d*2^(1/2)/e^(1/2)-2*a^2*(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))*sec(d*x+c)*sin(2*d*x+2*c)^(1/2)/
d/(e*tan(d*x+c))^(1/2)+2*a^2*(e*tan(d*x+c))^(1/2)/d/e

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Rubi [A]
time = 0.21, antiderivative size = 278, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 14, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.560, Rules used = {3971, 3557, 335, 217, 1179, 642, 1176, 631, 210, 2694, 2653, 2720, 2687, 32} \begin {gather*} -\frac {a^2 \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d \sqrt {e}}+\frac {a^2 \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d \sqrt {e}}+\frac {2 a^2 \sqrt {e \tan (c+d x)}}{d e}-\frac {a^2 \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d \sqrt {e}}+\frac {a^2 \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d \sqrt {e}}+\frac {2 a^2 \sqrt {\sin (2 c+2 d x)} \sec (c+d x) F\left (\left .c+d x-\frac {\pi }{4}\right |2\right )}{d \sqrt {e \tan (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 32

Int[((a_.) + (b_.)*(x_))^(m_), x_Symbol] :> Simp[(a + b*x)^(m + 1)/(b*(m + 1)), x] /; FreeQ[{a, b, m}, x] && N
eQ[m, -1]

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 2687

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 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 3971

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]

Rubi steps

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

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 32.56, size = 220, normalized size = 0.79 \begin {gather*} \frac {a^2 \cos ^4\left (\frac {1}{2} (c+d x)\right ) \sec ^4\left (\frac {1}{2} \text {ArcTan}(\tan (c+d x))\right ) \left (-2 \sqrt {2} \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )+2 \sqrt {2} \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )-\sqrt {2} \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )+\sqrt {2} \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )+8 \sqrt {\tan (c+d x)}+16 \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\tan ^2(c+d x)\right ) \sqrt {\tan (c+d x)}\right ) \sqrt {\tan (c+d x)}}{4 d \sqrt {e \tan (c+d x)}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a^2*Cos[(c + d*x)/2]^4*Sec[ArcTan[Tan[c + d*x]]/2]^4*(-2*Sqrt[2]*ArcTan[1 - Sqrt[2]*Sqrt[Tan[c + d*x]]] + 2*S
qrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] - Sqrt[2]*Log[1 - Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] + S
qrt[2]*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] + 8*Sqrt[Tan[c + d*x]] + 16*Hypergeometric2F1[1/4, 1
/2, 5/4, -Tan[c + d*x]^2]*Sqrt[Tan[c + d*x]])*Sqrt[Tan[c + d*x]])/(4*d*Sqrt[e*Tan[c + d*x]])

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Maple [C] Result contains complex when optimal does not.
time = 0.30, size = 663, normalized size = 2.38

method	result	size
default	\(-\frac {a^{2} \left (1+\cos \left (d x +c \right )\right )^{2} \left (-1+\cos \left (d x +c \right )\right ) \left (i \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 ) \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 )}}-i \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 ) \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 )}}+\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 ) \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 )}}+\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 ) \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 )}}+2 \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 ) \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 )}}-2 \sqrt {2}\, \cos \left (d x +c \right )+2 \sqrt {2}\right ) \sqrt {2}}{2 d \sin \left (d x +c \right )^{3} \cos \left (d x +c \right ) \sqrt {\frac {e \sin \left (d x +c \right )}{\cos \left (d x +c \right )}}}\)	\(663\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^2/(e*tan(d*x+c))^(1/2),x,method=_RETURNVERBOSE)

[Out]

-1/2*a^2/d*(1+cos(d*x+c))^2*(-1+cos(d*x+c))*(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))*sin(d*x+c)-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))*sin(d*x+c)+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)+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)+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+c
os(d*x+c)-sin(d*x+c))/sin(d*x+c))^(1/2)-2*2^(1/2)*cos(d*x+c)+2*2^(1/2))/sin(d*x+c)^3/cos(d*x+c)/(e*sin(d*x+c)/
cos(d*x+c))^(1/2)*2^(1/2)

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Maxima [A]
time = 0.24, size = 128, normalized size = 0.46 \begin {gather*} \frac {{\left ({\left (2 \, \sqrt {2} \arctan \left (\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} + 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + 2 \, \sqrt {2} \arctan \left (-\frac {1}{2} \, \sqrt {2} {\left (\sqrt {2} - 2 \, \sqrt {\tan \left (d x + c\right )}\right )}\right ) + \sqrt {2} \log \left (\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right ) - \sqrt {2} \log \left (-\sqrt {2} \sqrt {\tan \left (d x + c\right )} + \tan \left (d x + c\right ) + 1\right )\right )} a^{2} + 8 \, a^{2} \sqrt {\tan \left (d x + c\right )}\right )} e^{\left (-\frac {1}{2}\right )}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^2/(e*tan(d*x+c))^(1/2),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)) + tan(d*x + c) + 1) - sqrt(2)*log(-sqrt(2)*s
qrt(tan(d*x + c)) + tan(d*x + c) + 1))*a^2 + 8*a^2*sqrt(tan(d*x + c)))*e^(-1/2)/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))^2/(e*tan(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

Timed out

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{2} \left (\int \frac {1}{\sqrt {e \tan {\left (c + d x \right )}}}\, dx + \int \frac {2 \sec {\left (c + d x \right )}}{\sqrt {e \tan {\left (c + d x \right )}}}\, dx + \int \frac {\sec ^{2}{\left (c + d x \right )}}{\sqrt {e \tan {\left (c + d x \right )}}}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a**2*(Integral(1/sqrt(e*tan(c + d*x)), x) + Integral(2*sec(c + d*x)/sqrt(e*tan(c + d*x)), x) + Integral(sec(c
+ d*x)**2/sqrt(e*tan(c + d*x)), 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))^2/(e*tan(d*x+c))^(1/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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