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

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*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]))

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Rubi [A]  time = 0.0680902, antiderivative size = 115, normalized size of antiderivative = 1., 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 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]

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]))

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*}

Mathematica [A]  time = 0.266773, 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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Maple [A]  time = 0.071, size = 147, normalized size = 1.3 \begin{align*}{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{3\,a \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{3\,a \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{8\,d}}-{\frac{5\,a \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d}}-{\frac{15\,a\sin \left ( dx+c \right ) }{8\,d}}+{\frac{15\,a\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{a \left ( \tan \left ( dx+c \right ) \right ) ^{4}}{4\,d}}-{\frac{a \left ( \tan \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{a\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*sin(d*x+c)^5*(a+a*sin(d*x+c)),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*a*tan(d*x+c)^4/d-1/2*a*tan(d*x+c)^2/d-a*ln(cos(d*x+
c))/d

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Maxima [A]  time = 1.1957, size = 128, normalized size = 1.11 \begin{align*} \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} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^5*(a+a*sin(d*x+c)),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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Fricas [A]  time = 1.57403, size = 410, normalized size = 3.57 \begin{align*} \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 )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^5*(a+a*sin(d*x+c)),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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.42373, size = 136, normalized size = 1.18 \begin{align*} \frac{14 \, a \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 46 \, a \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - 32 \, a \sin \left (d x + c\right ) - \frac{2 \,{\left (7 \, a \sin \left (d x + c\right ) + 5 \, a\right )}}{\sin \left (d x + c\right ) + 1} + \frac{69 \, a \sin \left (d x + c\right )^{2} - 106 \, a \sin \left (d x + c\right ) + 41 \, a}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(14*a*log(abs(sin(d*x + c) + 1)) - 46*a*log(abs(sin(d*x + c) - 1)) - 32*a*sin(d*x + c) - 2*(7*a*sin(d*x +
c) + 5*a)/(sin(d*x + c) + 1) + (69*a*sin(d*x + c)^2 - 106*a*sin(d*x + c) + 41*a)/(sin(d*x + c) - 1)^2)/d