3.8 \(\int (a+a \sin (c+d x)) \tan ^6(c+d x) \, dx\)

Optimal. Leaf size=101 \[ \frac{a \cos (c+d x)}{d}+\frac{a \tan ^5(c+d x)}{5 d}-\frac{a \tan ^3(c+d x)}{3 d}+\frac{a \tan (c+d x)}{d}+\frac{a \sec ^5(c+d x)}{5 d}-\frac{a \sec ^3(c+d x)}{d}+\frac{3 a \sec (c+d x)}{d}-a x \]

[Out]

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

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Rubi [A]  time = 0.0922464, antiderivative size = 101, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 5, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.263, Rules used = {2710, 3473, 8, 2590, 270} \[ \frac{a \cos (c+d x)}{d}+\frac{a \tan ^5(c+d x)}{5 d}-\frac{a \tan ^3(c+d x)}{3 d}+\frac{a \tan (c+d x)}{d}+\frac{a \sec ^5(c+d x)}{5 d}-\frac{a \sec ^3(c+d x)}{d}+\frac{3 a \sec (c+d x)}{d}-a x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2710

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Int[Expan
dIntegrand[(g*Tan[e + f*x])^p, (a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2
, 0] && IGtQ[m, 0]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 270

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

Rubi steps

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

Mathematica [A]  time = 0.0654162, size = 110, normalized size = 1.09 \[ \frac{a \cos (c+d x)}{d}+\frac{a \tan ^5(c+d x)}{5 d}-\frac{a \tan ^3(c+d x)}{3 d}-\frac{a \tan ^{-1}(\tan (c+d x))}{d}+\frac{a \tan (c+d x)}{d}+\frac{a \sec ^5(c+d x)}{5 d}-\frac{a \sec ^3(c+d x)}{d}+\frac{3 a \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.071, size = 135, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ( a \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{5}}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{5\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{\cos \left ( dx+c \right ) }}+ \left ({\frac{16}{5}}+ \left ( \sin \left ( dx+c \right ) \right ) ^{6}+{\frac{6\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{5}}+{\frac{8\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{5}} \right ) \cos \left ( dx+c \right ) \right ) +a \left ({\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{5}}{5}}-{\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3}}+\tan \left ( dx+c \right ) -dx-c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a*(1/5*sin(d*x+c)^8/cos(d*x+c)^5-1/5*sin(d*x+c)^8/cos(d*x+c)^3+sin(d*x+c)^8/cos(d*x+c)+(16/5+sin(d*x+c)^6
+6/5*sin(d*x+c)^4+8/5*sin(d*x+c)^2)*cos(d*x+c))+a*(1/5*tan(d*x+c)^5-1/3*tan(d*x+c)^3+tan(d*x+c)-d*x-c))

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Maxima [A]  time = 1.65939, size = 117, normalized size = 1.16 \begin{align*} \frac{{\left (3 \, \tan \left (d x + c\right )^{5} - 5 \, \tan \left (d x + c\right )^{3} - 15 \, d x - 15 \, c + 15 \, \tan \left (d x + c\right )\right )} a + 3 \, a{\left (\frac{15 \, \cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{2} + 1}{\cos \left (d x + c\right )^{5}} + 5 \, \cos \left (d x + c\right )\right )}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*((3*tan(d*x + c)^5 - 5*tan(d*x + c)^3 - 15*d*x - 15*c + 15*tan(d*x + c))*a + 3*a*((15*cos(d*x + c)^4 - 5*
cos(d*x + c)^2 + 1)/cos(d*x + c)^5 + 5*cos(d*x + c)))/d

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Fricas [A]  time = 1.4669, size = 300, normalized size = 2.97 \begin{align*} \frac{15 \, a d x \cos \left (d x + c\right )^{3} - 38 \, a \cos \left (d x + c\right )^{4} - 11 \, a \cos \left (d x + c\right )^{2} -{\left (15 \, a d x \cos \left (d x + c\right )^{3} - 15 \, a \cos \left (d x + c\right )^{4} - 22 \, a \cos \left (d x + c\right )^{2} + 4 \, a\right )} \sin \left (d x + c\right ) + a}{15 \,{\left (d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/15*(15*a*d*x*cos(d*x + c)^3 - 38*a*cos(d*x + c)^4 - 11*a*cos(d*x + c)^2 - (15*a*d*x*cos(d*x + c)^3 - 15*a*co
s(d*x + c)^4 - 22*a*cos(d*x + c)^2 + 4*a)*sin(d*x + c) + a)/(d*cos(d*x + c)^3*sin(d*x + c) - d*cos(d*x + c)^3)

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

[Out]

Timed out

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

[Out]

Timed out