3.851 \(\int \sin (c+d x) (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx\)

Optimal. Leaf size=133 \[ \frac {a^3}{8 d (a-a \sin (c+d x))^2}-\frac {5 a^2}{4 d (a-a \sin (c+d x))}-\frac {a^2}{8 d (a \sin (c+d x)+a)}-\frac {a \sin ^2(c+d x)}{2 d}-\frac {a \sin (c+d x)}{d}-\frac {39 a \log (1-\sin (c+d x))}{16 d}-\frac {9 a \log (\sin (c+d x)+1)}{16 d} \]

[Out]

-39/16*a*ln(1-sin(d*x+c))/d-9/16*a*ln(1+sin(d*x+c))/d-a*sin(d*x+c)/d-1/2*a*sin(d*x+c)^2/d+1/8*a^3/d/(a-a*sin(d
*x+c))^2-5/4*a^2/d/(a-a*sin(d*x+c))-1/8*a^2/d/(a+a*sin(d*x+c))

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Rubi [A]  time = 0.11, antiderivative size = 133, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, Rules used = {2836, 12, 88} \[ \frac {a^3}{8 d (a-a \sin (c+d x))^2}-\frac {5 a^2}{4 d (a-a \sin (c+d x))}-\frac {a^2}{8 d (a \sin (c+d x)+a)}-\frac {a \sin ^2(c+d x)}{2 d}-\frac {a \sin (c+d x)}{d}-\frac {39 a \log (1-\sin (c+d x))}{16 d}-\frac {9 a \log (\sin (c+d x)+1)}{16 d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]*(a + a*Sin[c + d*x])*Tan[c + d*x]^5,x]

[Out]

(-39*a*Log[1 - Sin[c + d*x]])/(16*d) - (9*a*Log[1 + Sin[c + d*x]])/(16*d) - (a*Sin[c + d*x])/d - (a*Sin[c + d*
x]^2)/(2*d) + a^3/(8*d*(a - a*Sin[c + d*x])^2) - (5*a^2)/(4*d*(a - a*Sin[c + d*x])) - a^2/(8*d*(a + a*Sin[c +
d*x]))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 2836

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)
*(x_)])^(n_.), x_Symbol] :> Dist[1/(b^p*f), Subst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps

\begin {align*} \int \sin (c+d x) (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx &=\frac {a^5 \operatorname {Subst}\left (\int \frac {x^6}{a^6 (a-x)^3 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x^6}{(a-x)^3 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-a+\frac {a^4}{4 (a-x)^3}-\frac {5 a^3}{4 (a-x)^2}+\frac {39 a^2}{16 (a-x)}-x+\frac {a^3}{8 (a+x)^2}-\frac {9 a^2}{16 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{a d}\\ &=-\frac {39 a \log (1-\sin (c+d x))}{16 d}-\frac {9 a \log (1+\sin (c+d x))}{16 d}-\frac {a \sin (c+d x)}{d}-\frac {a \sin ^2(c+d x)}{2 d}+\frac {a^3}{8 d (a-a \sin (c+d x))^2}-\frac {5 a^2}{4 d (a-a \sin (c+d x))}-\frac {a^2}{8 d (a+a \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.45, size = 133, normalized size = 1.00 \[ -\frac {a \sin (c+d x) \tan ^4(c+d x)}{d}-\frac {a \left (2 \sin ^2(c+d x)-\sec ^4(c+d x)+6 \sec ^2(c+d x)+12 \log (\cos (c+d x))\right )}{4 d}-\frac {5 a \left (6 \tan (c+d x) \sec ^3(c+d x)-8 \tan ^3(c+d x) \sec (c+d x)-3 \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]*(a + a*Sin[c + d*x])*Tan[c + d*x]^5,x]

[Out]

-1/4*(a*(12*Log[Cos[c + d*x]] + 6*Sec[c + d*x]^2 - Sec[c + d*x]^4 + 2*Sin[c + d*x]^2))/d - (a*Sin[c + d*x]*Tan
[c + d*x]^4)/d - (5*a*(6*Sec[c + d*x]^3*Tan[c + d*x] - 8*Sec[c + d*x]*Tan[c + d*x]^3 - 3*(ArcTanh[Sin[c + d*x]
] + Sec[c + d*x]*Tan[c + d*x])))/(8*d)

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fricas [A]  time = 0.49, size = 172, normalized size = 1.29 \[ \frac {8 \, a \cos \left (d x + c\right )^{4} + 6 \, a \cos \left (d x + c\right )^{2} - 9 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 39 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (4 \, a \cos \left (d x + c\right )^{4} + 6 \, a \cos \left (d x + c\right )^{2} - 3 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{16 \, {\left (d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^6*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/16*(8*a*cos(d*x + c)^4 + 6*a*cos(d*x + c)^2 - 9*(a*cos(d*x + c)^2*sin(d*x + c) - a*cos(d*x + c)^2)*log(sin(d
*x + c) + 1) - 39*(a*cos(d*x + c)^2*sin(d*x + c) - a*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) + 2*(4*a*cos(d*x +
c)^4 + 6*a*cos(d*x + c)^2 - 3*a)*sin(d*x + c) + 2*a)/(d*cos(d*x + c)^2*sin(d*x + c) - d*cos(d*x + c)^2)

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giac [A]  time = 0.26, size = 113, normalized size = 0.85 \[ -\frac {16 \, a \sin \left (d x + c\right )^{2} + 18 \, a \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + 78 \, a \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + 32 \, a \sin \left (d x + c\right ) - \frac {2 \, {\left (9 \, a \sin \left (d x + c\right ) + 7 \, a\right )}}{\sin \left (d x + c\right ) + 1} - \frac {117 \, a \sin \left (d x + c\right )^{2} - 194 \, a \sin \left (d x + c\right ) + 81 \, a}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^6*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/32*(16*a*sin(d*x + c)^2 + 18*a*log(abs(sin(d*x + c) + 1)) + 78*a*log(abs(sin(d*x + c) - 1)) + 32*a*sin(d*x
+ c) - 2*(9*a*sin(d*x + c) + 7*a)/(sin(d*x + c) + 1) - (117*a*sin(d*x + c)^2 - 194*a*sin(d*x + c) + 81*a)/(sin
(d*x + c) - 1)^2)/d

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maple [A]  time = 0.28, size = 205, normalized size = 1.54 \[ \frac {a \left (\sin ^{8}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {a \left (\sin ^{8}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}-\frac {a \left (\sin ^{6}\left (d x +c \right )\right )}{2 d}-\frac {3 a \left (\sin ^{4}\left (d x +c \right )\right )}{4 d}-\frac {3 a \left (\sin ^{2}\left (d x +c \right )\right )}{2 d}-\frac {3 a \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {a \left (\sin ^{7}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {3 a \left (\sin ^{7}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}-\frac {3 a \left (\sin ^{5}\left (d x +c \right )\right )}{8 d}-\frac {5 a \left (\sin ^{3}\left (d x +c \right )\right )}{8 d}-\frac {15 a \sin \left (d x +c \right )}{8 d}+\frac {15 a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*sin(d*x+c)^6*(a+a*sin(d*x+c)),x)

[Out]

1/4/d*a*sin(d*x+c)^8/cos(d*x+c)^4-1/2/d*a*sin(d*x+c)^8/cos(d*x+c)^2-1/2*a*sin(d*x+c)^6/d-3/4*a*sin(d*x+c)^4/d-
3/2*a*sin(d*x+c)^2/d-3*a*ln(cos(d*x+c))/d+1/4/d*a*sin(d*x+c)^7/cos(d*x+c)^4-3/8/d*a*sin(d*x+c)^7/cos(d*x+c)^2-
3/8*a*sin(d*x+c)^5/d-5/8*a*sin(d*x+c)^3/d-15/8*a*sin(d*x+c)/d+15/8/d*a*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.32, size = 106, normalized size = 0.80 \[ -\frac {8 \, a \sin \left (d x + c\right )^{2} + 9 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) + 39 \, a \log \left (\sin \left (d x + c\right ) - 1\right ) + 16 \, a \sin \left (d x + c\right ) - \frac {2 \, {\left (9 \, a \sin \left (d x + c\right )^{2} + 3 \, a \sin \left (d x + c\right ) - 10 \, a\right )}}{\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )^{2} - \sin \left (d x + c\right ) + 1}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^6*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/16*(8*a*sin(d*x + c)^2 + 9*a*log(sin(d*x + c) + 1) + 39*a*log(sin(d*x + c) - 1) + 16*a*sin(d*x + c) - 2*(9*
a*sin(d*x + c)^2 + 3*a*sin(d*x + c) - 10*a)/(sin(d*x + c)^3 - sin(d*x + c)^2 - sin(d*x + c) + 1))/d

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mupad [B]  time = 9.82, size = 286, normalized size = 2.15 \[ \frac {-\frac {15\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9}{4}+\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8}{2}+7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7-\frac {7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{2}-\frac {7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4}{2}+7\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\frac {3\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}-\frac {15\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+4\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}-\frac {9\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{8\,d}-\frac {39\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{8\,d}+\frac {3\,a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)^6*(a + a*sin(c + d*x)))/cos(c + d*x)^5,x)

[Out]

((3*a*tan(c/2 + (d*x)/2)^2)/2 - (15*a*tan(c/2 + (d*x)/2))/4 + 7*a*tan(c/2 + (d*x)/2)^3 - (7*a*tan(c/2 + (d*x)/
2)^4)/2 + (11*a*tan(c/2 + (d*x)/2)^5)/2 - (7*a*tan(c/2 + (d*x)/2)^6)/2 + 7*a*tan(c/2 + (d*x)/2)^7 + (3*a*tan(c
/2 + (d*x)/2)^8)/2 - (15*a*tan(c/2 + (d*x)/2)^9)/4)/(d*(tan(c/2 + (d*x)/2)^2 - 2*tan(c/2 + (d*x)/2) - 2*tan(c/
2 + (d*x)/2)^4 + 4*tan(c/2 + (d*x)/2)^5 - 2*tan(c/2 + (d*x)/2)^6 + tan(c/2 + (d*x)/2)^8 - 2*tan(c/2 + (d*x)/2)
^9 + tan(c/2 + (d*x)/2)^10 + 1)) - (9*a*log(tan(c/2 + (d*x)/2) + 1))/(8*d) - (39*a*log(tan(c/2 + (d*x)/2) - 1)
)/(8*d) + (3*a*log(tan(c/2 + (d*x)/2)^2 + 1))/d

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sympy [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(sec(d*x+c)**5*sin(d*x+c)**6*(a+a*sin(d*x+c)),x)

[Out]

Timed out

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