3.798 \(\int \sec (c+d x) (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx\)

Optimal. Leaf size=60 \[ \frac{a \tan ^3(c+d x)}{3 d}-\frac{a \tan (c+d x)}{d}+\frac{a \sec ^3(c+d x)}{3 d}-\frac{a \sec (c+d x)}{d}+a x \]

[Out]

a*x - (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 = 0.0978578, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used = {2838, 2606, 3473, 8} \[ \frac{a \tan ^3(c+d x)}{3 d}-\frac{a \tan (c+d x)}{d}+\frac{a \sec ^3(c+d x)}{3 d}-\frac{a \sec (c+d x)}{d}+a x \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]*(a + a*Sin[c + d*x])*Tan[c + d*x]^3,x]

[Out]

a*x - (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 2838

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.)*((a_) + (b_.)*sin[(e_.) + (f_.)
*(x_)]), x_Symbol] :> Dist[a, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^n, x], x] + Dist[b/d, Int[(g*Cos[e + f*x
])^p*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x]

Rule 2606

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Dist[a/f, Subst[
Int[(a*x)^(m - 1)*(-1 + x^2)^((n - 1)/2), x], x, Sec[e + f*x]], x] /; FreeQ[{a, e, f, m}, x] && IntegerQ[(n -
1)/2] &&  !(IntegerQ[m/2] && LtQ[0, m, n + 1])

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]

Rubi steps

\begin{align*} \int \sec (c+d x) (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx &=a \int \sec (c+d x) \tan ^3(c+d x) \, dx+a \int \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 \left (-1+x^2\right ) \, dx,x,\sec (c+d x)\right )}{d}\\ &=-\frac{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}+a \int 1 \, dx\\ &=a x-\frac{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 = 0.041551, size = 69, normalized size = 1.15 \[ \frac{a \tan ^3(c+d x)}{3 d}+\frac{a \tan ^{-1}(\tan (c+d x))}{d}-\frac{a \tan (c+d x)}{d}+\frac{a \sec ^3(c+d x)}{3 d}-\frac{a \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]*(a + a*Sin[c + d*x])*Tan[c + d*x]^3,x]

[Out]

(a*ArcTan[Tan[c + d*x]])/d - (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 = 0.064, size = 88, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ( a \left ({\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3}}-\tan \left ( dx+c \right ) +dx+c \right ) +a \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{3\,\cos \left ( dx+c \right ) }}-{\frac{ \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) }{3}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*sin(d*x+c)^3*(a+a*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.53314, size = 74, normalized size = 1.23 \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 - \frac{{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a}{\cos \left (d x + c\right )^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.42534, size = 196, normalized size = 3.27 \begin{align*} -\frac{3 \, a d x \cos \left (d x + c\right ) - 4 \, a \cos \left (d x + c\right )^{2} -{\left (3 \, a d x \cos \left (d x + c\right ) + a\right )} \sin \left (d x + c\right ) + 2 \, 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(sec(d*x+c)^4*sin(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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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)**4*sin(d*x+c)**3*(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.23333, size = 100, normalized size = 1.67 \begin{align*} \frac{6 \,{\left (d x + c\right )} a + \frac{3 \, a}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1} + \frac{9 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 24 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 11 \, a}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^3*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/6*(6*(d*x + c)*a + 3*a/(tan(1/2*d*x + 1/2*c) + 1) + (9*a*tan(1/2*d*x + 1/2*c)^2 - 24*a*tan(1/2*d*x + 1/2*c)
+ 11*a)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d