3.2.6 \(\int (a+a \sec (c+d x)) \sqrt {e \tan (c+d x)} \, dx\) [106]

Optimal. Leaf size=272 \[ -\frac {a \sqrt {e} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}+\frac {a \sqrt {e} \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}+\frac {a \sqrt {e} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d}-\frac {a \sqrt {e} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d}-\frac {2 a \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}+\frac {2 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e} \]

[Out]

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

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Rubi [A]
time = 0.17, antiderivative size = 272, normalized size of antiderivative = 1.00, 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 = {3969, 3557, 335, 303, 1176, 631, 210, 1179, 642, 2693, 2695, 2652, 2719} \begin {gather*} -\frac {a \sqrt {e} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}+\frac {a \sqrt {e} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} d}+\frac {a \sqrt {e} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d}-\frac {a \sqrt {e} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} d}+\frac {2 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {2 a \cos (c+d x) E\left (\left .c+d x-\frac {\pi }{4}\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 303

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 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 2652

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 2693

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 2695

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 2719

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(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 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]

Rubi steps

\begin {align*} \int (a+a \sec (c+d x)) \sqrt {e \tan (c+d x)} \, dx &=a \int \sqrt {e \tan (c+d x)} \, dx+a \int \sec (c+d x) \sqrt {e \tan (c+d x)} \, dx\\ &=\frac {2 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-(2 a) \int \cos (c+d x) \sqrt {e \tan (c+d x)} \, dx+\frac {(a e) \text {Subst}\left (\int \frac {\sqrt {x}}{e^2+x^2} \, dx,x,e \tan (c+d x)\right )}{d}\\ &=\frac {2 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}+\frac {(2 a e) \text {Subst}\left (\int \frac {x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{d}-\frac {\left (2 a \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)} \, dx}{\sqrt {\sin (c+d x)}}\\ &=\frac {2 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}-\frac {(a e) \text {Subst}\left (\int \frac {e-x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{d}+\frac {(a e) \text {Subst}\left (\int \frac {e+x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{d}-\frac {\left (2 a \cos (c+d x) \sqrt {e \tan (c+d x)}\right ) \int \sqrt {\sin (2 c+2 d x)} \, dx}{\sqrt {\sin (2 c+2 d x)}}\\ &=-\frac {2 a \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}+\frac {2 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}+\frac {\left (a \sqrt {e}\right ) \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}+\frac {\left (a \sqrt {e}\right ) \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}+\frac {(a e) \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 e) \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 \sqrt {e} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d}-\frac {a \sqrt {e} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d}-\frac {2 a \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}+\frac {2 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{d e}+\frac {\left (a \sqrt {e}\right ) \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}-\frac {\left (a \sqrt {e}\right ) \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}\\ &=-\frac {a \sqrt {e} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}+\frac {a \sqrt {e} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} d}+\frac {a \sqrt {e} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d}-\frac {a \sqrt {e} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} d}-\frac {2 a \cos (c+d x) E\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sqrt {e \tan (c+d x)}}{d \sqrt {\sin (2 c+2 d x)}}+\frac {2 a \cos (c+d x) (e \tan (c+d x))^{3/2}}{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 = 1.48, size = 182, normalized size = 0.67 \begin {gather*} -\frac {a (1+\cos (c+d x)) \csc (c+d x) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \sqrt {e \tan (c+d x)} \left (3 \sqrt {\sec ^2(c+d x)} \left (-4 \sin ^2(c+d x)+\text {ArcSin}(\cos (c+d x)-\sin (c+d x)) \sqrt {\sin (2 (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 )+8 \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\tan ^2(c+d x)\right ) \tan ^2(c+d x)\right )}{12 d \sqrt {\sec ^2(c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(a*(1 + Cos[c + d*x])*Csc[c + d*x]*Sec[(c + d*x)/2]^2*Sqrt[e*Tan[c + d*x]]*(3*Sqrt[Sec[c + d*x]^2]*(-4*S
in[c + d*x]^2 + ArcSin[Cos[c + d*x] - Sin[c + d*x]]*Sqrt[Sin[2*(c + d*x)]] + Log[Cos[c + d*x] + Sin[c + d*x] +
Sqrt[Sin[2*(c + d*x)]]]*Sqrt[Sin[2*(c + d*x)]]) + 8*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[c + d*x]^2]*Tan[c +
d*x]^2))/(d*Sqrt[Sec[c + d*x]^2])

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Maple [C] Result contains complex when optimal does not.
time = 0.25, size = 1429, normalized size = 5.25

method	result	size
default	\(\text {Expression too large to display}\)	\(1429\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*a/d*(e*sin(d*x+c)/cos(d*x+c))^(1/2)*(1+cos(d*x+c))^2*(-1+cos(d*x+c))^2*(I*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)*El
lipticPi((-(-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*cos(d*x+c)*((-1+cos(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)*(-(-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)*((-1+cos(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)*(-(-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)*((-1+cos(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)*(-(-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))+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
)*EllipticF((-(-1+cos(d*x+c)-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))-4*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)*Ellipti
cE((-(-1+cos(d*x+c)-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))+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))-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))+(-(-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))+(-(-1+cos(d*x+c)-sin(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))^(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))+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)*EllipticF((-(-1+cos(d*x+c)-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2
))-4*((-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)*EllipticE((-(-1+cos(d*x+c)-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))+2*2^(1/2)*cos(d*x
+c)-2*2^(1/2))/sin(d*x+c)^5*2^(1/2)

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Maxima [A]
time = 0.23, size = 111, normalized size = 0.41 \begin {gather*} \frac {{\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 e^{\frac {1}{2}}}{4 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))*(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)*sq
rt(tan(d*x + c)) + tan(d*x + c) + 1))*a*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))*(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 \left (\int \sqrt {e \tan {\left (c + d x \right )}}\, dx + \int \sqrt {e \tan {\left (c + d x \right )}} \sec {\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))*(e*tan(d*x+c))**(1/2),x)

[Out]

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

[Out]

integrate((a*sec(d*x + c) + a)*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 \sqrt {e\,\mathrm {tan}\left (c+d\,x\right )}\,\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right ) \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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