3.2.18 \(\int \frac {(e \tan (c+d x))^{11/2}}{a+a \sec (c+d x)} \, dx\) [118]

Optimal. Leaf size=330 \[ \frac {e^{11/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}-\frac {e^{11/2} \text {ArcTan}\left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {e^{11/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}-\frac {e^{11/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}+\frac {5 e^6 F\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{21 a d \sqrt {e \tan (c+d x)}}+\frac {2 e^5 (21-5 \sec (c+d x)) \sqrt {e \tan (c+d x)}}{21 a d}-\frac {2 e^3 (7-5 \sec (c+d x)) (e \tan (c+d x))^{5/2}}{35 a d} \]

[Out]

1/2*e^(11/2)*arctan(1-2^(1/2)*(e*tan(d*x+c))^(1/2)/e^(1/2))/a/d*2^(1/2)-1/2*e^(11/2)*arctan(1+2^(1/2)*(e*tan(d
*x+c))^(1/2)/e^(1/2))/a/d*2^(1/2)+1/4*e^(11/2)*ln(e^(1/2)-2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/a/d
*2^(1/2)-1/4*e^(11/2)*ln(e^(1/2)+2^(1/2)*(e*tan(d*x+c))^(1/2)+e^(1/2)*tan(d*x+c))/a/d*2^(1/2)-5/21*e^6*(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)/a/
d/(e*tan(d*x+c))^(1/2)+2/21*e^5*(21-5*sec(d*x+c))*(e*tan(d*x+c))^(1/2)/a/d-2/35*e^3*(7-5*sec(d*x+c))*(e*tan(d*
x+c))^(5/2)/a/d

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Rubi [A]
time = 0.30, antiderivative size = 330, 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 = {3973, 3966, 3969, 3557, 335, 217, 1179, 642, 1176, 631, 210, 2694, 2653, 2720} \begin {gather*} \frac {e^{11/2} \text {ArcTan}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}-\frac {e^{11/2} \text {ArcTan}\left (\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}+1\right )}{\sqrt {2} a d}+\frac {e^{11/2} \log \left (\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d}-\frac {e^{11/2} \log \left (\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}+\sqrt {e}\right )}{2 \sqrt {2} a d}+\frac {5 e^6 \sqrt {\sin (2 c+2 d x)} \sec (c+d x) F\left (\left .c+d x-\frac {\pi }{4}\right |2\right )}{21 a d \sqrt {e \tan (c+d x)}}+\frac {2 e^5 (21-5 \sec (c+d x)) \sqrt {e \tan (c+d x)}}{21 a d}-\frac {2 e^3 (7-5 \sec (c+d x)) (e \tan (c+d x))^{5/2}}{35 a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(e*Tan[c + d*x])^(11/2)/(a + a*Sec[c + d*x]),x]

[Out]

(e^(11/2)*ArcTan[1 - (Sqrt[2]*Sqrt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*a*d) - (e^(11/2)*ArcTan[1 + (Sqrt[2]*Sq
rt[e*Tan[c + d*x]])/Sqrt[e]])/(Sqrt[2]*a*d) + (e^(11/2)*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] - Sqrt[2]*Sqrt[e*Ta
n[c + d*x]]])/(2*Sqrt[2]*a*d) - (e^(11/2)*Log[Sqrt[e] + Sqrt[e]*Tan[c + d*x] + Sqrt[2]*Sqrt[e*Tan[c + d*x]]])/
(2*Sqrt[2]*a*d) + (5*e^6*EllipticF[c - Pi/4 + d*x, 2]*Sec[c + d*x]*Sqrt[Sin[2*c + 2*d*x]])/(21*a*d*Sqrt[e*Tan[
c + d*x]]) + (2*e^5*(21 - 5*Sec[c + d*x])*Sqrt[e*Tan[c + d*x]])/(21*a*d) - (2*e^3*(7 - 5*Sec[c + d*x])*(e*Tan[
c + d*x])^(5/2))/(35*a*d)

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 3973

Int[(cot[(c_.) + (d_.)*(x_)]*(e_.))^(m_)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_), x_Symbol] :> Dist[a^(2*n
)/e^(2*n), Int[(e*Cot[c + d*x])^(m + 2*n)/(-a + b*Csc[c + d*x])^n, x], x] /; FreeQ[{a, b, c, d, e, m}, x] && E
qQ[a^2 - b^2, 0] && ILtQ[n, 0]

Rubi steps

\begin {align*} \int \frac {(e \tan (c+d x))^{11/2}}{a+a \sec (c+d x)} \, dx &=\frac {e^2 \int (-a+a \sec (c+d x)) (e \tan (c+d x))^{7/2} \, dx}{a^2}\\ &=-\frac {2 e^3 (7-5 \sec (c+d x)) (e \tan (c+d x))^{5/2}}{35 a d}-\frac {\left (2 e^4\right ) \int \left (-\frac {7 a}{2}+\frac {5}{2} a \sec (c+d x)\right ) (e \tan (c+d x))^{3/2} \, dx}{7 a^2}\\ &=\frac {2 e^5 (21-5 \sec (c+d x)) \sqrt {e \tan (c+d x)}}{21 a d}-\frac {2 e^3 (7-5 \sec (c+d x)) (e \tan (c+d x))^{5/2}}{35 a d}+\frac {\left (4 e^6\right ) \int \frac {-\frac {21 a}{4}+\frac {5}{4} a \sec (c+d x)}{\sqrt {e \tan (c+d x)}} \, dx}{21 a^2}\\ &=\frac {2 e^5 (21-5 \sec (c+d x)) \sqrt {e \tan (c+d x)}}{21 a d}-\frac {2 e^3 (7-5 \sec (c+d x)) (e \tan (c+d x))^{5/2}}{35 a d}+\frac {\left (5 e^6\right ) \int \frac {\sec (c+d x)}{\sqrt {e \tan (c+d x)}} \, dx}{21 a}-\frac {e^6 \int \frac {1}{\sqrt {e \tan (c+d x)}} \, dx}{a}\\ &=\frac {2 e^5 (21-5 \sec (c+d x)) \sqrt {e \tan (c+d x)}}{21 a d}-\frac {2 e^3 (7-5 \sec (c+d x)) (e \tan (c+d x))^{5/2}}{35 a d}-\frac {e^7 \text {Subst}\left (\int \frac {1}{\sqrt {x} \left (e^2+x^2\right )} \, dx,x,e \tan (c+d x)\right )}{a d}+\frac {\left (5 e^6 \sqrt {\sin (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)} \sqrt {\sin (c+d x)}} \, dx}{21 a \sqrt {\cos (c+d x)} \sqrt {e \tan (c+d x)}}\\ &=\frac {2 e^5 (21-5 \sec (c+d x)) \sqrt {e \tan (c+d x)}}{21 a d}-\frac {2 e^3 (7-5 \sec (c+d x)) (e \tan (c+d x))^{5/2}}{35 a d}-\frac {\left (2 e^7\right ) \text {Subst}\left (\int \frac {1}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a d}+\frac {\left (5 e^6 \sec (c+d x) \sqrt {\sin (2 c+2 d x)}\right ) \int \frac {1}{\sqrt {\sin (2 c+2 d x)}} \, dx}{21 a \sqrt {e \tan (c+d x)}}\\ &=\frac {5 e^6 F\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{21 a d \sqrt {e \tan (c+d x)}}+\frac {2 e^5 (21-5 \sec (c+d x)) \sqrt {e \tan (c+d x)}}{21 a d}-\frac {2 e^3 (7-5 \sec (c+d x)) (e \tan (c+d x))^{5/2}}{35 a d}-\frac {e^6 \text {Subst}\left (\int \frac {e-x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a d}-\frac {e^6 \text {Subst}\left (\int \frac {e+x^2}{e^2+x^4} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{a d}\\ &=\frac {5 e^6 F\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{21 a d \sqrt {e \tan (c+d x)}}+\frac {2 e^5 (21-5 \sec (c+d x)) \sqrt {e \tan (c+d x)}}{21 a d}-\frac {2 e^3 (7-5 \sec (c+d x)) (e \tan (c+d x))^{5/2}}{35 a d}+\frac {e^{11/2} \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} a d}+\frac {e^{11/2} \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} a d}-\frac {e^6 \text {Subst}\left (\int \frac {1}{e-\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 a d}-\frac {e^6 \text {Subst}\left (\int \frac {1}{e+\sqrt {2} \sqrt {e} x+x^2} \, dx,x,\sqrt {e \tan (c+d x)}\right )}{2 a d}\\ &=\frac {e^{11/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}-\frac {e^{11/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}+\frac {5 e^6 F\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{21 a d \sqrt {e \tan (c+d x)}}+\frac {2 e^5 (21-5 \sec (c+d x)) \sqrt {e \tan (c+d x)}}{21 a d}-\frac {2 e^3 (7-5 \sec (c+d x)) (e \tan (c+d x))^{5/2}}{35 a d}-\frac {e^{11/2} \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} a d}+\frac {e^{11/2} \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} a d}\\ &=\frac {e^{11/2} \tan ^{-1}\left (1-\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}-\frac {e^{11/2} \tan ^{-1}\left (1+\frac {\sqrt {2} \sqrt {e \tan (c+d x)}}{\sqrt {e}}\right )}{\sqrt {2} a d}+\frac {e^{11/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)-\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}-\frac {e^{11/2} \log \left (\sqrt {e}+\sqrt {e} \tan (c+d x)+\sqrt {2} \sqrt {e \tan (c+d x)}\right )}{2 \sqrt {2} a d}+\frac {5 e^6 F\left (\left .c-\frac {\pi }{4}+d x\right |2\right ) \sec (c+d x) \sqrt {\sin (2 c+2 d x)}}{21 a d \sqrt {e \tan (c+d x)}}+\frac {2 e^5 (21-5 \sec (c+d x)) \sqrt {e \tan (c+d x)}}{21 a d}-\frac {2 e^3 (7-5 \sec (c+d x)) (e \tan (c+d x))^{5/2}}{35 a d}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 48.74, size = 332, normalized size = 1.01 \begin {gather*} \frac {e^5 \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (1+\sqrt {\sec ^2(c+d x)}\right ) \sqrt {e \tan (c+d x)} \left (70 \sqrt {2} \text {ArcTan}\left (1-\sqrt {2} \sqrt {\tan (c+d x)}\right )-70 \sqrt {2} \text {ArcTan}\left (1+\sqrt {2} \sqrt {\tan (c+d x)}\right )+35 \sqrt {2} \log \left (1-\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )-35 \sqrt {2} \log \left (1+\sqrt {2} \sqrt {\tan (c+d x)}+\tan (c+d x)\right )+280 \sqrt {\tan (c+d x)}-320 \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {5}{4};-\tan ^2(c+d x)\right ) \sqrt {\tan (c+d x)}+280 \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\tan ^2(c+d x)\right ) \sqrt {\tan (c+d x)}+40 \sqrt {\sec ^2(c+d x)} \sqrt {\tan (c+d x)}-56 \tan ^{\frac {5}{2}}(c+d x)+40 \sqrt {\sec ^2(c+d x)} \tan ^{\frac {5}{2}}(c+d x)\right )}{70 a d (1+\sec (c+d x))^2 \sqrt {\tan (c+d x)}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[(e*Tan[c + d*x])^(11/2)/(a + a*Sec[c + d*x]),x]

[Out]

(e^5*Cos[(c + d*x)/2]^2*Sec[c + d*x]*(1 + Sqrt[Sec[c + d*x]^2])*Sqrt[e*Tan[c + d*x]]*(70*Sqrt[2]*ArcTan[1 - Sq
rt[2]*Sqrt[Tan[c + d*x]]] - 70*Sqrt[2]*ArcTan[1 + Sqrt[2]*Sqrt[Tan[c + d*x]]] + 35*Sqrt[2]*Log[1 - Sqrt[2]*Sqr
t[Tan[c + d*x]] + Tan[c + d*x]] - 35*Sqrt[2]*Log[1 + Sqrt[2]*Sqrt[Tan[c + d*x]] + Tan[c + d*x]] + 280*Sqrt[Tan
[c + d*x]] - 320*Hypergeometric2F1[-1/2, 1/4, 5/4, -Tan[c + d*x]^2]*Sqrt[Tan[c + d*x]] + 280*Hypergeometric2F1
[1/4, 1/2, 5/4, -Tan[c + d*x]^2]*Sqrt[Tan[c + d*x]] + 40*Sqrt[Sec[c + d*x]^2]*Sqrt[Tan[c + d*x]] - 56*Tan[c +
d*x]^(5/2) + 40*Sqrt[Sec[c + d*x]^2]*Tan[c + d*x]^(5/2)))/(70*a*d*(1 + Sec[c + d*x])^2*Sqrt[Tan[c + d*x]])

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Maple [C] Result contains complex when optimal does not.
time = 0.25, size = 744, normalized size = 2.25

method	result	size
default	\(\frac {\left (-1+\cos \left (d x +c \right )\right ) \left (105 i \sqrt {-\frac {-1+\cos \left (d x +c \right )-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (\cos ^{3}\left (d x +c \right )\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 )}}\, \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 )-105 i \sqrt {-\frac {-1+\cos \left (d x +c \right )-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (\cos ^{3}\left (d x +c \right )\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 )}}\, \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 )-260 \sqrt {-\frac {-1+\cos \left (d x +c \right )-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (\cos ^{3}\left (d x +c \right )\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 )}}\, \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 )+105 \sqrt {-\frac {-1+\cos \left (d x +c \right )-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (\cos ^{3}\left (d x +c \right )\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 )}}\, \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 )+105 \sqrt {-\frac {-1+\cos \left (d x +c \right )-\sin \left (d x +c \right )}{\sin \left (d x +c \right )}}\, \left (\cos ^{3}\left (d x +c \right )\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 )}}\, \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 )+252 \left (\cos ^{4}\left (d x +c \right )\right ) \sqrt {2}-332 \left (\cos ^{3}\left (d x +c \right )\right ) \sqrt {2}+38 \left (\cos ^{2}\left (d x +c \right )\right ) \sqrt {2}+72 \sqrt {2}\, \cos \left (d x +c \right )-30 \sqrt {2}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \left (1+\cos \left (d x +c \right )\right )^{2} \left (\frac {e \sin \left (d x +c \right )}{\cos \left (d x +c \right )}\right )^{\frac {11}{2}} \sqrt {2}}{210 a d \sin \left (d x +c \right )^{9}}\)	\(744\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*tan(d*x+c))^(11/2)/(a+a*sec(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

1/210/a/d*(-1+cos(d*x+c))*(105*I*(-(-1+cos(d*x+c)-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3*sin(d*x+c)*((-1+c
os(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))-105*I*(-(-1+cos(d*x+c)-sin(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^
3*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)*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))-260*(-(-1+cos(d*x+c)-sin(d*x+c))/sin(d*x+c))^
(1/2)*cos(d*x+c)^3*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)
*EllipticF((-(-1+cos(d*x+c)-sin(d*x+c))/sin(d*x+c))^(1/2),1/2*2^(1/2))+105*(-(-1+cos(d*x+c)-sin(d*x+c))/sin(d*
x+c))^(1/2)*cos(d*x+c)^3*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)*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))+105*(-(-1+cos(d*x+c)-s
in(d*x+c))/sin(d*x+c))^(1/2)*cos(d*x+c)^3*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)*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))+252*c
os(d*x+c)^4*2^(1/2)-332*cos(d*x+c)^3*2^(1/2)+38*cos(d*x+c)^2*2^(1/2)+72*2^(1/2)*cos(d*x+c)-30*2^(1/2))*cos(d*x
+c)^2*(1+cos(d*x+c))^2*(e*sin(d*x+c)/cos(d*x+c))^(11/2)/sin(d*x+c)^9*2^(1/2)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((e*tan(d*x+c))^(11/2)/(a+a*sec(d*x+c)),x, algorithm="maxima")

[Out]

e^(11/2)*integrate(tan(d*x + c)^(11/2)/(a*sec(d*x + c) + a), x)

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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((e*tan(d*x+c))^(11/2)/(a+a*sec(d*x+c)),x, algorithm="fricas")

[Out]

Timed out

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Sympy [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((e*tan(d*x+c))**(11/2)/(a+a*sec(d*x+c)),x)

[Out]

Timed out

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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((e*tan(d*x+c))^(11/2)/(a+a*sec(d*x+c)),x, algorithm="giac")

[Out]

integrate((e*tan(d*x + c))^(11/2)/(a*sec(d*x + c) + a), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\cos \left (c+d\,x\right )\,{\left (e\,\mathrm {tan}\left (c+d\,x\right )\right )}^{11/2}}{a\,\left (\cos \left (c+d\,x\right )+1\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((e*tan(c + d*x))^(11/2)/(a + a/cos(c + d*x)),x)

[Out]

int((cos(c + d*x)*(e*tan(c + d*x))^(11/2))/(a*(cos(c + d*x) + 1)), x)

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