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

Optimal. Leaf size=115 \[ \frac {a^3}{8 d (a-a \sin (c+d x))^2}-\frac {a^2}{d (a-a \sin (c+d x))}+\frac {a^2}{8 d (a \sin (c+d x)+a)}-\frac {a \sin (c+d x)}{d}-\frac {23 a \log (1-\sin (c+d x))}{16 d}+\frac {7 a \log (\sin (c+d x)+1)}{16 d} \]

[Out]

-23/16*a*ln(1-sin(d*x+c))/d+7/16*a*ln(1+sin(d*x+c))/d-a*sin(d*x+c)/d+1/8*a^3/d/(a-a*sin(d*x+c))^2-a^2/d/(a-a*s
in(d*x+c))+1/8*a^2/d/(a+a*sin(d*x+c))

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Rubi [A]  time = 0.07, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2707, 88} \[ \frac {a^3}{8 d (a-a \sin (c+d x))^2}-\frac {a^2}{d (a-a \sin (c+d x))}+\frac {a^2}{8 d (a \sin (c+d x)+a)}-\frac {a \sin (c+d x)}{d}-\frac {23 a \log (1-\sin (c+d x))}{16 d}+\frac {7 a \log (\sin (c+d x)+1)}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-23*a*Log[1 - Sin[c + d*x]])/(16*d) + (7*a*Log[1 + Sin[c + d*x]])/(16*d) - (a*Sin[c + d*x])/d + a^3/(8*d*(a -
a*Sin[c + d*x])^2) - a^2/(d*(a - a*Sin[c + d*x])) + a^2/(8*d*(a + a*Sin[c + d*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 2707

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

Rubi steps

\begin {align*} \int (a+a \sin (c+d x)) \tan ^5(c+d x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {x^5}{(a-x)^3 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {\operatorname {Subst}\left (\int \left (-1+\frac {a^3}{4 (a-x)^3}-\frac {a^2}{(a-x)^2}+\frac {23 a}{16 (a-x)}-\frac {a^2}{8 (a+x)^2}+\frac {7 a}{16 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=-\frac {23 a \log (1-\sin (c+d x))}{16 d}+\frac {7 a \log (1+\sin (c+d x))}{16 d}-\frac {a \sin (c+d x)}{d}+\frac {a^3}{8 d (a-a \sin (c+d x))^2}-\frac {a^2}{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 = 123, normalized size = 1.07 \[ -\frac {a \sin (c+d x) \tan ^4(c+d x)}{d}-\frac {a \left (-\tan ^4(c+d x)+2 \tan ^2(c+d x)+4 \log (\cos (c+d x))\right )}{4 d}-\frac {5 a \left (6 \tan (c+d x) \sec ^3(c+d x)-8 \tan ^3(c+d x) \sec (c+d x)-3 \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a*Sin[c + d*x]*Tan[c + d*x]^4)/d) - (a*(4*Log[Cos[c + d*x]] + 2*Tan[c + d*x]^2 - Tan[c + d*x]^4))/(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]*T
an[c + d*x])))/(8*d)

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fricas [A]  time = 0.44, size = 159, normalized size = 1.38 \[ \frac {16 \, a \cos \left (d x + c\right )^{4} + 2 \, a \cos \left (d x + c\right )^{2} + 7 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 23 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (8 \, a \cos \left (d x + c\right )^{2} + a\right )} \sin \left (d x + c\right ) - 6 \, a}{16 \, {\left (d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(16*a*cos(d*x + c)^4 + 2*a*cos(d*x + c)^2 + 7*(a*cos(d*x + c)^2*sin(d*x + c) - a*cos(d*x + c)^2)*log(sin(
d*x + c) + 1) - 23*(a*cos(d*x + c)^2*sin(d*x + c) - a*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) + 2*(8*a*cos(d*x
+ c)^2 + a)*sin(d*x + c) - 6*a)/(d*cos(d*x + c)^2*sin(d*x + c) - d*cos(d*x + c)^2)

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giac [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((a+a*sin(d*x+c))*tan(d*x+c)^5,x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.15, size = 147, normalized size = 1.28 \[ \frac {a \left (\sin ^{7}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {3 a \left (\sin ^{7}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}-\frac {3 a \left (\sin ^{5}\left (d x +c \right )\right )}{8 d}-\frac {5 a \left (\sin ^{3}\left (d x +c \right )\right )}{8 d}-\frac {15 a \sin \left (d x +c \right )}{8 d}+\frac {15 a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {a \left (\tan ^{4}\left (d x +c \right )\right )}{4 d}-\frac {a \left (\tan ^{2}\left (d x +c \right )\right )}{2 d}-\frac {a \ln \left (\cos \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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))+1/4/d*a*tan(d*x+c)^4-1/2/d*a*tan(d*x+c)^2-1/d*a*ln(cos(
d*x+c))

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maxima [A]  time = 0.33, size = 95, normalized size = 0.83 \[ \frac {7 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) - 23 \, a \log \left (\sin \left (d x + c\right ) - 1\right ) - 16 \, a \sin \left (d x + c\right ) + \frac {2 \, {\left (9 \, a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - 6 \, a\right )}}{\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )^{2} - \sin \left (d x + c\right ) + 1}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(7*a*log(sin(d*x + c) + 1) - 23*a*log(sin(d*x + c) - 1) - 16*a*sin(d*x + c) + 2*(9*a*sin(d*x + c)^2 - a*s
in(d*x + c) - 6*a)/(sin(d*x + c)^3 - sin(d*x + c)^2 - sin(d*x + c) + 1))/d

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mupad [B]  time = 6.63, size = 235, normalized size = 2.04 \[ \frac {-\frac {15\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7}{4}+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6}{2}+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5}{4}-5\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3}{4}+\frac {11\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2}{2}-\frac {15\,a\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}{4}}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-2\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}-\frac {23\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}{8\,d}+\frac {7\,a\,\ln \left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+1\right )}{8\,d}+\frac {a\,\ln \left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((11*a*tan(c/2 + (d*x)/2)^2)/2 - (15*a*tan(c/2 + (d*x)/2))/4 + (11*a*tan(c/2 + (d*x)/2)^3)/4 - 5*a*tan(c/2 + (
d*x)/2)^4 + (11*a*tan(c/2 + (d*x)/2)^5)/4 + (11*a*tan(c/2 + (d*x)/2)^6)/2 - (15*a*tan(c/2 + (d*x)/2)^7)/4)/(d*
(2*tan(c/2 + (d*x)/2)^3 - 2*tan(c/2 + (d*x)/2) - 2*tan(c/2 + (d*x)/2)^4 + 2*tan(c/2 + (d*x)/2)^5 - 2*tan(c/2 +
(d*x)/2)^7 + tan(c/2 + (d*x)/2)^8 + 1)) - (23*a*log(tan(c/2 + (d*x)/2) - 1))/(8*d) + (7*a*log(tan(c/2 + (d*x)
/2) + 1))/(8*d) + (a*log(tan(c/2 + (d*x)/2)^2 + 1))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int \sin {\left (c + d x \right )} \tan ^{5}{\left (c + d x \right )}\, dx + \int \tan ^{5}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))*tan(d*x+c)**5,x)

[Out]

a*(Integral(sin(c + d*x)*tan(c + d*x)**5, x) + Integral(tan(c + d*x)**5, x))

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