3.177 \(\int \frac {\tan ^6(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx\)

Optimal. Leaf size=189 \[ \frac {2 a^4 \tan ^9(c+d x)}{9 d (a \sec (c+d x)+a)^{9/2}}+\frac {6 a^3 \tan ^7(c+d x)}{7 d (a \sec (c+d x)+a)^{7/2}}+\frac {2 a^2 \tan ^5(c+d x)}{5 d (a \sec (c+d x)+a)^{5/2}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d}-\frac {2 a \tan ^3(c+d x)}{3 d (a \sec (c+d x)+a)^{3/2}}+\frac {2 \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}} \]

[Out]

-2*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/d/a^(1/2)+2*tan(d*x+c)/d/(a+a*sec(d*x+c))^(1/2)-2/3*a*tan
(d*x+c)^3/d/(a+a*sec(d*x+c))^(3/2)+2/5*a^2*tan(d*x+c)^5/d/(a+a*sec(d*x+c))^(5/2)+6/7*a^3*tan(d*x+c)^7/d/(a+a*s
ec(d*x+c))^(7/2)+2/9*a^4*tan(d*x+c)^9/d/(a+a*sec(d*x+c))^(9/2)

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Rubi [A]  time = 0.10, antiderivative size = 189, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3887, 461, 203} \[ \frac {2 a^4 \tan ^9(c+d x)}{9 d (a \sec (c+d x)+a)^{9/2}}+\frac {6 a^3 \tan ^7(c+d x)}{7 d (a \sec (c+d x)+a)^{7/2}}+\frac {2 a^2 \tan ^5(c+d x)}{5 d (a \sec (c+d x)+a)^{5/2}}-\frac {2 a \tan ^3(c+d x)}{3 d (a \sec (c+d x)+a)^{3/2}}-\frac {2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{\sqrt {a} d}+\frac {2 \tan (c+d x)}{d \sqrt {a \sec (c+d x)+a}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^6/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(-2*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(Sqrt[a]*d) + (2*Tan[c + d*x])/(d*Sqrt[a + a*Sec[
c + d*x]]) - (2*a*Tan[c + d*x]^3)/(3*d*(a + a*Sec[c + d*x])^(3/2)) + (2*a^2*Tan[c + d*x]^5)/(5*d*(a + a*Sec[c
+ d*x])^(5/2)) + (6*a^3*Tan[c + d*x]^7)/(7*d*(a + a*Sec[c + d*x])^(7/2)) + (2*a^4*Tan[c + d*x]^9)/(9*d*(a + a*
Sec[c + d*x])^(9/2))

Rule 203

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

Rule 461

Int[(((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_))/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Int[ExpandIntegr
and[((e*x)^m*(a + b*x^n)^p)/(c + d*x^n), x], x] /; FreeQ[{a, b, c, d, e, m}, x] && NeQ[b*c - a*d, 0] && IGtQ[n
, 0] && IGtQ[p, 0] && (IntegerQ[m] || IGtQ[2*(m + 1), 0] ||  !RationalQ[m])

Rule 3887

Int[cot[(c_.) + (d_.)*(x_)]^(m_.)*(csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_))^(n_.), x_Symbol] :> Dist[(-2*a^(m/2 +
n + 1/2))/d, Subst[Int[(x^m*(2 + a*x^2)^(m/2 + n - 1/2))/(1 + a*x^2), x], x, Cot[c + d*x]/Sqrt[a + b*Csc[c +
d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0] && IntegerQ[m/2] && IntegerQ[n - 1/2]

Rubi steps

\begin {align*} \int \frac {\tan ^6(c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx &=-\frac {\left (2 a^3\right ) \operatorname {Subst}\left (\int \frac {x^6 \left (2+a x^2\right )^2}{1+a x^2} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}\\ &=-\frac {\left (2 a^3\right ) \operatorname {Subst}\left (\int \left (\frac {1}{a^3}-\frac {x^2}{a^2}+\frac {x^4}{a}+3 x^6+a x^8-\frac {1}{a^3 \left (1+a x^2\right )}\right ) \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}\\ &=\frac {2 \tan (c+d x)}{d \sqrt {a+a \sec (c+d x)}}-\frac {2 a \tan ^3(c+d x)}{3 d (a+a \sec (c+d x))^{3/2}}+\frac {2 a^2 \tan ^5(c+d x)}{5 d (a+a \sec (c+d x))^{5/2}}+\frac {6 a^3 \tan ^7(c+d x)}{7 d (a+a \sec (c+d x))^{7/2}}+\frac {2 a^4 \tan ^9(c+d x)}{9 d (a+a \sec (c+d x))^{9/2}}+\frac {2 \operatorname {Subst}\left (\int \frac {1}{1+a x^2} \, dx,x,-\frac {\tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{d}\\ &=-\frac {2 \tan ^{-1}\left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{\sqrt {a} d}+\frac {2 \tan (c+d x)}{d \sqrt {a+a \sec (c+d x)}}-\frac {2 a \tan ^3(c+d x)}{3 d (a+a \sec (c+d x))^{3/2}}+\frac {2 a^2 \tan ^5(c+d x)}{5 d (a+a \sec (c+d x))^{5/2}}+\frac {6 a^3 \tan ^7(c+d x)}{7 d (a+a \sec (c+d x))^{7/2}}+\frac {2 a^4 \tan ^9(c+d x)}{9 d (a+a \sec (c+d x))^{9/2}}\\ \end {align*}

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Mathematica [C]  time = 19.22, size = 467, normalized size = 2.47 \[ \frac {\cos \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \left (\frac {1532}{315} \sin \left (\frac {1}{2} (c+d x)\right )+\frac {4}{9} \sin \left (\frac {1}{2} (c+d x)\right ) \sec ^4(c+d x)-\frac {4}{63} \sin \left (\frac {1}{2} (c+d x)\right ) \sec ^3(c+d x)-\frac {176}{105} \sin \left (\frac {1}{2} (c+d x)\right ) \sec ^2(c+d x)+\frac {136}{315} \sin \left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)\right )}{d \sqrt {a (\sec (c+d x)+1)}}+\frac {16 \left (-3-2 \sqrt {2}\right ) \cos \left (\frac {1}{2} (c+d x)\right ) \sqrt {\frac {\left (10-7 \sqrt {2}\right ) \cos \left (\frac {1}{2} (c+d x)\right )-5 \sqrt {2}+7}{\cos \left (\frac {1}{2} (c+d x)\right )+1}} \sqrt {\frac {-\left (\left (\sqrt {2}-2\right ) \cos \left (\frac {1}{2} (c+d x)\right )\right )+\sqrt {2}-1}{\cos \left (\frac {1}{2} (c+d x)\right )+1}} \left (\left (\sqrt {2}-2\right ) \cos \left (\frac {1}{2} (c+d x)\right )-\sqrt {2}+1\right ) \cos ^4\left (\frac {1}{4} (c+d x)\right ) \sqrt {-\tan ^2\left (\frac {1}{4} (c+d x)\right )-2 \sqrt {2}+3} \sec ^2(c+d x) \sqrt {\left (\left (2+\sqrt {2}\right ) \cos \left (\frac {1}{2} (c+d x)\right )-\sqrt {2}-1\right ) \sec ^2\left (\frac {1}{4} (c+d x)\right )} \left (F\left (\sin ^{-1}\left (\frac {\tan \left (\frac {1}{4} (c+d x)\right )}{\sqrt {3-2 \sqrt {2}}}\right )|17-12 \sqrt {2}\right )-2 \Pi \left (-3+2 \sqrt {2};\sin ^{-1}\left (\frac {\tan \left (\frac {1}{4} (c+d x)\right )}{\sqrt {3-2 \sqrt {2}}}\right )|17-12 \sqrt {2}\right )\right )}{d \sqrt {a (\sec (c+d x)+1)}} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[Tan[c + d*x]^6/Sqrt[a + a*Sec[c + d*x]],x]

[Out]

(Cos[(c + d*x)/2]*Sec[c + d*x]*((1532*Sin[(c + d*x)/2])/315 + (136*Sec[c + d*x]*Sin[(c + d*x)/2])/315 - (176*S
ec[c + d*x]^2*Sin[(c + d*x)/2])/105 - (4*Sec[c + d*x]^3*Sin[(c + d*x)/2])/63 + (4*Sec[c + d*x]^4*Sin[(c + d*x)
/2])/9))/(d*Sqrt[a*(1 + Sec[c + d*x])]) + (16*(-3 - 2*Sqrt[2])*Cos[(c + d*x)/4]^4*Cos[(c + d*x)/2]*Sqrt[(7 - 5
*Sqrt[2] + (10 - 7*Sqrt[2])*Cos[(c + d*x)/2])/(1 + Cos[(c + d*x)/2])]*Sqrt[(-1 + Sqrt[2] - (-2 + Sqrt[2])*Cos[
(c + d*x)/2])/(1 + Cos[(c + d*x)/2])]*(1 - Sqrt[2] + (-2 + Sqrt[2])*Cos[(c + d*x)/2])*(EllipticF[ArcSin[Tan[(c
+ d*x)/4]/Sqrt[3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]] - 2*EllipticPi[-3 + 2*Sqrt[2], ArcSin[Tan[(c + d*x)/4]/Sqrt[
3 - 2*Sqrt[2]]], 17 - 12*Sqrt[2]])*Sqrt[(-1 - Sqrt[2] + (2 + Sqrt[2])*Cos[(c + d*x)/2])*Sec[(c + d*x)/4]^2]*Se
c[c + d*x]^2*Sqrt[3 - 2*Sqrt[2] - Tan[(c + d*x)/4]^2])/(d*Sqrt[a*(1 + Sec[c + d*x])])

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fricas [A]  time = 0.75, size = 355, normalized size = 1.88 \[ \left [-\frac {315 \, {\left (\cos \left (d x + c\right )^{5} + \cos \left (d x + c\right )^{4}\right )} \sqrt {-a} \log \left (\frac {2 \, a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) + a \cos \left (d x + c\right ) - a}{\cos \left (d x + c\right ) + 1}\right ) - 2 \, {\left (383 \, \cos \left (d x + c\right )^{4} + 34 \, \cos \left (d x + c\right )^{3} - 132 \, \cos \left (d x + c\right )^{2} - 5 \, \cos \left (d x + c\right ) + 35\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{315 \, {\left (a d \cos \left (d x + c\right )^{5} + a d \cos \left (d x + c\right )^{4}\right )}}, \frac {2 \, {\left (315 \, {\left (\cos \left (d x + c\right )^{5} + \cos \left (d x + c\right )^{4}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + {\left (383 \, \cos \left (d x + c\right )^{4} + 34 \, \cos \left (d x + c\right )^{3} - 132 \, \cos \left (d x + c\right )^{2} - 5 \, \cos \left (d x + c\right ) + 35\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )\right )}}{315 \, {\left (a d \cos \left (d x + c\right )^{5} + a d \cos \left (d x + c\right )^{4}\right )}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^6/(a+a*sec(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

[-1/315*(315*(cos(d*x + c)^5 + cos(d*x + c)^4)*sqrt(-a)*log((2*a*cos(d*x + c)^2 - 2*sqrt(-a)*sqrt((a*cos(d*x +
c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) - 2*(383*cos(d*x +
c)^4 + 34*cos(d*x + c)^3 - 132*cos(d*x + c)^2 - 5*cos(d*x + c) + 35)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*s
in(d*x + c))/(a*d*cos(d*x + c)^5 + a*d*cos(d*x + c)^4), 2/315*(315*(cos(d*x + c)^5 + cos(d*x + c)^4)*sqrt(a)*a
rctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + (383*cos(d*x + c)^4 + 34*
cos(d*x + c)^3 - 132*cos(d*x + c)^2 - 5*cos(d*x + c) + 35)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c
))/(a*d*cos(d*x + c)^5 + a*d*cos(d*x + c)^4)]

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giac [B]  time = 9.45, size = 353, normalized size = 1.87 \[ -\frac {\sqrt {2} {\left (\frac {315 \, \sqrt {2} \sqrt {-a} \log \left (\frac {{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} - 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}{{\left | 2 \, {\left (\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}\right )}^{2} + 4 \, \sqrt {2} {\left | a \right |} - 6 \, a \right |}}\right )}{{\left | a \right |} \mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} + \frac {4 \, {\left (\frac {315 \, a^{4}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - {\left (\frac {1470 \, a^{4}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - {\left (\frac {2772 \, a^{4}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} + {\left (\frac {257 \, a^{4} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )} - \frac {1314 \, a^{4}}{\mathrm {sgn}\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{4} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}\right )}}{630 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^6/(a+a*sec(d*x+c))^(1/2),x, algorithm="giac")

[Out]

-1/630*sqrt(2)*(315*sqrt(2)*sqrt(-a)*log(abs(2*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2
+ a))^2 - 4*sqrt(2)*abs(a) - 6*a)/abs(2*(sqrt(-a)*tan(1/2*d*x + 1/2*c) - sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a))
^2 + 4*sqrt(2)*abs(a) - 6*a))/(abs(a)*sgn(tan(1/2*d*x + 1/2*c)^2 - 1)) + 4*(315*a^4/sgn(tan(1/2*d*x + 1/2*c)^2
- 1) - (1470*a^4/sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - (2772*a^4/sgn(tan(1/2*d*x + 1/2*c)^2 - 1) + (257*a^4*tan(1
/2*d*x + 1/2*c)^2/sgn(tan(1/2*d*x + 1/2*c)^2 - 1) - 1314*a^4/sgn(tan(1/2*d*x + 1/2*c)^2 - 1))*tan(1/2*d*x + 1/
2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)^2)*tan(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^4
*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d

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maple [B]  time = 1.40, size = 480, normalized size = 2.54 \[ \frac {\sqrt {\frac {a \left (1+\cos \left (d x +c \right )\right )}{\cos \left (d x +c \right )}}\, \left (315 \sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {9}{2}} \left (\cos ^{4}\left (d x +c \right )\right ) \sin \left (d x +c \right )+1260 \sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {9}{2}} \left (\cos ^{3}\left (d x +c \right )\right ) \sin \left (d x +c \right )+1890 \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {9}{2}} \sqrt {2}\, \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+1260 \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {9}{2}} \sqrt {2}\, \cos \left (d x +c \right ) \sin \left (d x +c \right )+315 \sqrt {2}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \left (-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}\right )^{\frac {9}{2}} \sin \left (d x +c \right )-12256 \left (\cos ^{5}\left (d x +c \right )\right )+11168 \left (\cos ^{4}\left (d x +c \right )\right )+5312 \left (\cos ^{3}\left (d x +c \right )\right )-4064 \left (\cos ^{2}\left (d x +c \right )\right )-1280 \cos \left (d x +c \right )+1120\right )}{5040 d \sin \left (d x +c \right ) \cos \left (d x +c \right )^{4} a} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^6/(a+a*sec(d*x+c))^(1/2),x)

[Out]

1/5040/d*(a*(1+cos(d*x+c))/cos(d*x+c))^(1/2)*(315*2^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin
(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*cos(d*x+c)^4*sin(d*x+c)+1260*2^(1/2)*arctanh(
1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*c
os(d*x+c)^3*sin(d*x+c)+1890*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-
2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*2^(1/2)*cos(d*x+c)^2*sin(d*x+c)+1260*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+
c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(9/2)*2^(1/2)*cos(d*x+c)*sin(d*x+c)+3
15*2^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*(-2*cos(d*x+c)/(1+c
os(d*x+c)))^(9/2)*sin(d*x+c)-12256*cos(d*x+c)^5+11168*cos(d*x+c)^4+5312*cos(d*x+c)^3-4064*cos(d*x+c)^2-1280*co
s(d*x+c)+1120)/sin(d*x+c)/cos(d*x+c)^4/a

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maxima [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(tan(d*x+c)^6/(a+a*sec(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {tan}\left (c+d\,x\right )}^6}{\sqrt {a+\frac {a}{\cos \left (c+d\,x\right )}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^6/(a + a/cos(c + d*x))^(1/2),x)

[Out]

int(tan(c + d*x)^6/(a + a/cos(c + d*x))^(1/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{6}{\left (c + d x \right )}}{\sqrt {a \left (\sec {\left (c + d x \right )} + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**6/(a+a*sec(d*x+c))**(1/2),x)

[Out]

Integral(tan(c + d*x)**6/sqrt(a*(sec(c + d*x) + 1)), x)

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