3.854 \(\int \sec ^2(c+d x) (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx\)

Optimal. Leaf size=84 \[ \frac{a^3}{8 d (a-a \sin (c+d x))^2}-\frac{a^2}{2 d (a-a \sin (c+d x))}+\frac{a^2}{8 d (a \sin (c+d x)+a)}+\frac{3 a \tanh ^{-1}(\sin (c+d x))}{8 d} \]

[Out]

(3*a*ArcTanh[Sin[c + d*x]])/(8*d) + a^3/(8*d*(a - a*Sin[c + d*x])^2) - a^2/(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 = 0.101589, antiderivative size = 84, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {2836, 12, 88, 206} \[ \frac{a^3}{8 d (a-a \sin (c+d x))^2}-\frac{a^2}{2 d (a-a \sin (c+d x))}+\frac{a^2}{8 d (a \sin (c+d x)+a)}+\frac{3 a \tanh ^{-1}(\sin (c+d x))}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a + a*Sin[c + d*x])*Tan[c + d*x]^3,x]

[Out]

(3*a*ArcTanh[Sin[c + d*x]])/(8*d) + a^3/(8*d*(a - a*Sin[c + d*x])^2) - a^2/(2*d*(a - a*Sin[c + d*x])) + a^2/(8
*d*(a + a*Sin[c + d*x]))

Rule 2836

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

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, 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 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \sec ^2(c+d x) (a+a \sin (c+d x)) \tan ^3(c+d x) \, dx &=\frac{a^5 \operatorname{Subst}\left (\int \frac{x^3}{a^3 (a-x)^3 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^2 \operatorname{Subst}\left (\int \frac{x^3}{(a-x)^3 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^2 \operatorname{Subst}\left (\int \left (\frac{a}{4 (a-x)^3}-\frac{1}{2 (a-x)^2}-\frac{1}{8 (a+x)^2}+\frac{3}{8 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^3}{8 d (a-a \sin (c+d x))^2}-\frac{a^2}{2 d (a-a \sin (c+d x))}+\frac{a^2}{8 d (a+a \sin (c+d x))}+\frac{\left (3 a^2\right ) \operatorname{Subst}\left (\int \frac{1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{8 d}\\ &=\frac{3 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^3}{8 d (a-a \sin (c+d x))^2}-\frac{a^2}{2 d (a-a \sin (c+d x))}+\frac{a^2}{8 d (a+a \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.201194, size = 84, normalized size = 1. \[ \frac{a \tan ^4(c+d x)}{4 d}+\frac{a \tan ^3(c+d x) \sec (c+d x)}{d}-\frac{a \left (6 \tan (c+d x) \sec ^3(c+d x)-3 \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(a + a*Sin[c + d*x])*Tan[c + d*x]^3,x]

[Out]

(a*Sec[c + d*x]*Tan[c + d*x]^3)/d + (a*Tan[c + d*x]^4)/(4*d) - (a*(6*Sec[c + d*x]^3*Tan[c + d*x] - 3*(ArcTanh[
Sin[c + d*x]] + Sec[c + d*x]*Tan[c + d*x])))/(8*d)

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Maple [A]  time = 0.067, size = 114, normalized size = 1.4 \begin{align*}{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d}}-{\frac{3\,a\sin \left ( dx+c \right ) }{8\,d}}+{\frac{3\,a\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*sin(d*x+c)^3*(a+a*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.02721, size = 116, normalized size = 1.38 \begin{align*} \frac{3 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a \log \left (\sin \left (d x + c\right ) - 1\right ) + \frac{2 \,{\left (5 \, a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - 2 \, 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)^3*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/16*(3*a*log(sin(d*x + c) + 1) - 3*a*log(sin(d*x + c) - 1) + 2*(5*a*sin(d*x + c)^2 - a*sin(d*x + c) - 2*a)/(s
in(d*x + c)^3 - sin(d*x + c)^2 - sin(d*x + c) + 1))/d

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Fricas [A]  time = 1.35258, size = 350, normalized size = 4.17 \begin{align*} \frac{10 \, a \cos \left (d x + c\right )^{2} + 3 \,{\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 ) - 3 \,{\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 \, a \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)^3*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/16*(10*a*cos(d*x + c)^2 + 3*(a*cos(d*x + c)^2*sin(d*x + c) - a*cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 3*(a*
cos(d*x + c)^2*sin(d*x + c) - a*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) + 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)**3*(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.33009, size = 122, normalized size = 1.45 \begin{align*} \frac{6 \, a \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 6 \, a \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (3 \, a \sin \left (d x + c\right ) + a\right )}}{\sin \left (d x + c\right ) + 1} + \frac{9 \, a \sin \left (d x + c\right )^{2} - 2 \, a \sin \left (d x + c\right ) - 3 \, 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)^3*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/32*(6*a*log(abs(sin(d*x + c) + 1)) - 6*a*log(abs(sin(d*x + c) - 1)) - 2*(3*a*sin(d*x + c) + a)/(sin(d*x + c)
+ 1) + (9*a*sin(d*x + c)^2 - 2*a*sin(d*x + c) - 3*a)/(sin(d*x + c) - 1)^2)/d