3.1.53 \(\int \frac {\tan ^6(c+d x)}{a+a \sin (c+d x)} \, dx\) [53]

Optimal. Leaf size=84 \[ \frac {\sec (c+d x)}{a d}-\frac {\sec ^3(c+d x)}{a d}+\frac {3 \sec ^5(c+d x)}{5 a d}-\frac {\sec ^7(c+d x)}{7 a d}+\frac {\tan ^7(c+d x)}{7 a d} \]

[Out]

sec(d*x+c)/a/d-sec(d*x+c)^3/a/d+3/5*sec(d*x+c)^5/a/d-1/7*sec(d*x+c)^7/a/d+1/7*tan(d*x+c)^7/a/d

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Rubi [A]
time = 0.07, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2785, 2687, 30, 2686, 200} \begin {gather*} \frac {\tan ^7(c+d x)}{7 a d}-\frac {\sec ^7(c+d x)}{7 a d}+\frac {3 \sec ^5(c+d x)}{5 a d}-\frac {\sec ^3(c+d x)}{a d}+\frac {\sec (c+d x)}{a d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Tan[c + d*x]^6/(a + a*Sin[c + d*x]),x]

[Out]

Sec[c + d*x]/(a*d) - Sec[c + d*x]^3/(a*d) + (3*Sec[c + d*x]^5)/(5*a*d) - Sec[c + d*x]^7/(7*a*d) + Tan[c + d*x]
^7/(7*a*d)

Rule 30

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

Rule 200

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rule 2686

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

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 2785

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

Rubi steps

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

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Mathematica [A]
time = 0.21, size = 146, normalized size = 1.74 \begin {gather*} \frac {\sec ^5(c+d x) (2912-7620 \cos (c+d x)+3760 \cos (2 (c+d x))-3810 \cos (3 (c+d x))+1440 \cos (4 (c+d x))-762 \cos (5 (c+d x))+80 \cos (6 (c+d x))+2432 \sin (c+d x)-1905 \sin (2 (c+d x))+320 \sin (3 (c+d x))-1524 \sin (4 (c+d x))+960 \sin (5 (c+d x))-381 \sin (6 (c+d x)))}{17920 a d (1+\sin (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Tan[c + d*x]^6/(a + a*Sin[c + d*x]),x]

[Out]

(Sec[c + d*x]^5*(2912 - 7620*Cos[c + d*x] + 3760*Cos[2*(c + d*x)] - 3810*Cos[3*(c + d*x)] + 1440*Cos[4*(c + d*
x)] - 762*Cos[5*(c + d*x)] + 80*Cos[6*(c + d*x)] + 2432*Sin[c + d*x] - 1905*Sin[2*(c + d*x)] + 320*Sin[3*(c +
d*x)] - 1524*Sin[4*(c + d*x)] + 960*Sin[5*(c + d*x)] - 381*Sin[6*(c + d*x)]))/(17920*a*d*(1 + Sin[c + d*x]))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(174\) vs. \(2(78)=156\).
time = 0.20, size = 175, normalized size = 2.08

method	result	size
risch	\(\frac {-\frac {10 \,{\mathrm e}^{i \left (d x +c \right )}}{7}+\frac {52 i {\mathrm e}^{6 i \left (d x +c \right )}}{5}+\frac {52 i {\mathrm e}^{4 i \left (d x +c \right )}}{7}+6 i {\mathrm e}^{8 i \left (d x +c \right )}+\frac {22 i {\mathrm e}^{2 i \left (d x +c \right )}}{7}-\frac {52 \,{\mathrm e}^{5 i \left (d x +c \right )}}{35}+\frac {6 \,{\mathrm e}^{3 i \left (d x +c \right )}}{7}+\frac {36 \,{\mathrm e}^{7 i \left (d x +c \right )}}{5}+2 \,{\mathrm e}^{9 i \left (d x +c \right )}+2 i {\mathrm e}^{10 i \left (d x +c \right )}+2 \,{\mathrm e}^{11 i \left (d x +c \right )}+\frac {2 i}{7}}{\left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{5} \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )^{7} d a}\)	\(166\)
derivativedivides	\(\frac {-\frac {1}{10 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {5}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {9}{10 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {3}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {5}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d a}\)	\(175\)
default	\(\frac {-\frac {1}{10 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{5}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{4}}+\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )^{2}}-\frac {5}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}-\frac {2}{7 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{7}}+\frac {1}{\left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{6}}-\frac {9}{10 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{5}}-\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{4}}+\frac {1}{4 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{3}}+\frac {3}{8 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )^{2}}+\frac {5}{16 \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}}{d a}\)	\(175\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^6/(a+a*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

128/d/a*(-1/1280/(tan(1/2*d*x+1/2*c)-1)^5-1/512/(tan(1/2*d*x+1/2*c)-1)^4+1/512/(tan(1/2*d*x+1/2*c)-1)^2-5/2048
/(tan(1/2*d*x+1/2*c)-1)-1/448/(tan(1/2*d*x+1/2*c)+1)^7+1/128/(tan(1/2*d*x+1/2*c)+1)^6-9/1280/(tan(1/2*d*x+1/2*
c)+1)^5-1/512/(tan(1/2*d*x+1/2*c)+1)^4+1/512/(tan(1/2*d*x+1/2*c)+1)^3+3/1024/(tan(1/2*d*x+1/2*c)+1)^2+5/2048/(
tan(1/2*d*x+1/2*c)+1))

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 338 vs. \(2 (78) = 156\).
time = 0.31, size = 338, normalized size = 4.02 \begin {gather*} \frac {32 \, {\left (\frac {2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {4 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {10 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {20 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + 1\right )}}{35 \, {\left (a + \frac {2 \, a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {4 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac {10 \, a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac {5 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {20 \, a \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {20 \, a \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac {5 \, a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} + \frac {10 \, a \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + \frac {4 \, a \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}} - \frac {2 \, a \sin \left (d x + c\right )^{11}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{11}} - \frac {a \sin \left (d x + c\right )^{12}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{12}}\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

32/35*(2*sin(d*x + c)/(cos(d*x + c) + 1) - 4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 10*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 + 5*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 20*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 1)/((a + 2*a*si
n(d*x + c)/(cos(d*x + c) + 1) - 4*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 10*a*sin(d*x + c)^3/(cos(d*x + c) +
1)^3 + 5*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 20*a*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 - 20*a*sin(d*x + c)^
7/(cos(d*x + c) + 1)^7 - 5*a*sin(d*x + c)^8/(cos(d*x + c) + 1)^8 + 10*a*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 +
4*a*sin(d*x + c)^10/(cos(d*x + c) + 1)^10 - 2*a*sin(d*x + c)^11/(cos(d*x + c) + 1)^11 - a*sin(d*x + c)^12/(cos
(d*x + c) + 1)^12)*d)

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Fricas [A]
time = 0.35, size = 95, normalized size = 1.13 \begin {gather*} \frac {5 \, \cos \left (d x + c\right )^{6} + 15 \, \cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{2} + 2 \, {\left (15 \, \cos \left (d x + c\right )^{4} - 10 \, \cos \left (d x + c\right )^{2} + 3\right )} \sin \left (d x + c\right ) + 1}{35 \, {\left (a d \cos \left (d x + c\right )^{5} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{5}\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/35*(5*cos(d*x + c)^6 + 15*cos(d*x + c)^4 - 5*cos(d*x + c)^2 + 2*(15*cos(d*x + c)^4 - 10*cos(d*x + c)^2 + 3)*
sin(d*x + c) + 1)/(a*d*cos(d*x + c)^5*sin(d*x + c) + a*d*cos(d*x + c)^5)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\tan ^{6}{\left (c + d x \right )}}{\sin {\left (c + d x \right )} + 1}\, dx}{a} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**6/(a+a*sin(d*x+c)),x)

[Out]

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

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Giac [B] Leaf count of result is larger than twice the leaf count of optimal. 172 vs. \(2 (78) = 156\).
time = 8.84, size = 172, normalized size = 2.05 \begin {gather*} -\frac {\frac {7 \, {\left (25 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 120 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 210 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 140 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 33\right )}}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{5}} - \frac {175 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{6} + 1260 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 3815 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 6020 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4641 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1792 \, \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 281}{a {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1\right )}^{7}}}{560 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/560*(7*(25*tan(1/2*d*x + 1/2*c)^4 - 120*tan(1/2*d*x + 1/2*c)^3 + 210*tan(1/2*d*x + 1/2*c)^2 - 140*tan(1/2*d
*x + 1/2*c) + 33)/(a*(tan(1/2*d*x + 1/2*c) - 1)^5) - (175*tan(1/2*d*x + 1/2*c)^6 + 1260*tan(1/2*d*x + 1/2*c)^5
+ 3815*tan(1/2*d*x + 1/2*c)^4 + 6020*tan(1/2*d*x + 1/2*c)^3 + 4641*tan(1/2*d*x + 1/2*c)^2 + 1792*tan(1/2*d*x
+ 1/2*c) + 281)/(a*(tan(1/2*d*x + 1/2*c) + 1)^7))/d

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Mupad [B]
time = 8.49, size = 99, normalized size = 1.18 \begin {gather*} -\frac {32\,\left (20\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+5\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4-10\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{35\,a\,d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^5\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}^7} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^6/(a + a*sin(c + d*x)),x)

[Out]

-(32*(2*tan(c/2 + (d*x)/2) - 4*tan(c/2 + (d*x)/2)^2 - 10*tan(c/2 + (d*x)/2)^3 + 5*tan(c/2 + (d*x)/2)^4 + 20*ta
n(c/2 + (d*x)/2)^5 + 1))/(35*a*d*(tan(c/2 + (d*x)/2) - 1)^5*(tan(c/2 + (d*x)/2) + 1)^7)

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