3.9 \(\int (a+a \sin (c+d x)) \tan ^4(c+d x) \, dx\)

Optimal. Leaf size=72 \[ -\frac{a \cos (c+d x)}{d}+\frac{a \tan ^3(c+d x)}{3 d}-\frac{a \tan (c+d x)}{d}+\frac{a \sec ^3(c+d x)}{3 d}-\frac{2 a \sec (c+d x)}{d}+a x \]

[Out]

a*x - (a*Cos[c + d*x])/d - (2*a*Sec[c + d*x])/d + (a*Sec[c + d*x]^3)/(3*d) - (a*Tan[c + d*x])/d + (a*Tan[c + d
*x]^3)/(3*d)

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Rubi [A]  time = 0.0743084, antiderivative size = 72, normalized size of antiderivative = 1., number of steps used = 8, 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} \[ -\frac{a \cos (c+d x)}{d}+\frac{a \tan ^3(c+d x)}{3 d}-\frac{a \tan (c+d x)}{d}+\frac{a \sec ^3(c+d x)}{3 d}-\frac{2 a \sec (c+d x)}{d}+a x \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])*Tan[c + d*x]^4,x]

[Out]

a*x - (a*Cos[c + d*x])/d - (2*a*Sec[c + d*x])/d + (a*Sec[c + d*x]^3)/(3*d) - (a*Tan[c + d*x])/d + (a*Tan[c + d
*x]^3)/(3*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 ^4(c+d x) \, dx &=\int \left (a \tan ^4(c+d x)+a \sin (c+d x) \tan ^4(c+d x)\right ) \, dx\\ &=a \int \tan ^4(c+d x) \, dx+a \int \sin (c+d x) \tan ^4(c+d x) \, dx\\ &=\frac{a \tan ^3(c+d x)}{3 d}-a \int \tan ^2(c+d x) \, dx-\frac{a \operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^2}{x^4} \, dx,x,\cos (c+d x)\right )}{d}\\ &=-\frac{a \tan (c+d x)}{d}+\frac{a \tan ^3(c+d x)}{3 d}+a \int 1 \, dx-\frac{a \operatorname{Subst}\left (\int \left (1+\frac{1}{x^4}-\frac{2}{x^2}\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=a x-\frac{a \cos (c+d x)}{d}-\frac{2 a \sec (c+d x)}{d}+\frac{a \sec ^3(c+d x)}{3 d}-\frac{a \tan (c+d x)}{d}+\frac{a \tan ^3(c+d x)}{3 d}\\ \end{align*}

Mathematica [A]  time = 0.0483714, size = 81, normalized size = 1.12 \[ -\frac{a \cos (c+d x)}{d}+\frac{a \tan ^3(c+d x)}{3 d}+\frac{a \tan ^{-1}(\tan (c+d x))}{d}-\frac{a \tan (c+d x)}{d}+\frac{a \sec ^3(c+d x)}{3 d}-\frac{2 a \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])*Tan[c + d*x]^4,x]

[Out]

(a*ArcTan[Tan[c + d*x]])/d - (a*Cos[c + d*x])/d - (2*a*Sec[c + d*x])/d + (a*Sec[c + d*x]^3)/(3*d) - (a*Tan[c +
d*x])/d + (a*Tan[c + d*x]^3)/(3*d)

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Maple [A]  time = 0.057, size = 98, normalized size = 1.4 \begin{align*}{\frac{1}{d} \left ( a \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{\cos \left ( dx+c \right ) }}- \left ({\frac{8}{3}}+ \left ( \sin \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) \cos \left ( dx+c \right ) \right ) +a \left ({\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)^4,x)

[Out]

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

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Maxima [A]  time = 1.82471, size = 88, normalized size = 1.22 \begin{align*} \frac{{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a - a{\left (\frac{6 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} + 3 \, \cos \left (d x + c\right )\right )}}{3 \, 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)^4,x, algorithm="maxima")

[Out]

1/3*((tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a - a*((6*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 + 3*cos(d*x
+ c)))/d

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Fricas [A]  time = 1.4761, size = 224, normalized size = 3.11 \begin{align*} -\frac{3 \, a d x \cos \left (d x + c\right ) - 7 \, a \cos \left (d x + c\right )^{2} -{\left (3 \, a d x \cos \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right )^{2} - 2 \, a\right )} \sin \left (d x + c\right ) - a}{3 \,{\left (d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - d \cos \left (d x + c\right )\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)^4,x, algorithm="fricas")

[Out]

-1/3*(3*a*d*x*cos(d*x + c) - 7*a*cos(d*x + c)^2 - (3*a*d*x*cos(d*x + c) - 3*a*cos(d*x + c)^2 - 2*a)*sin(d*x +
c) - a)/(d*cos(d*x + c)*sin(d*x + c) - d*cos(d*x + c))

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a \left (\int \sin{\left (c + d x \right )} \tan ^{4}{\left (c + d x \right )}\, dx + \int \tan ^{4}{\left (c + d x \right )}\, dx\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)**4,x)

[Out]

a*(Integral(sin(c + d*x)*tan(c + d*x)**4, x) + Integral(tan(c + d*x)**4, x))

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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)^4,x, algorithm="giac")

[Out]

Timed out