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

Optimal. Leaf size=71 \[ \frac{a^2}{2 d (a-a \sin (c+d x))}+\frac{a \sin (c+d x)}{d}+\frac{5 a \log (1-\sin (c+d x))}{4 d}-\frac{a \log (\sin (c+d x)+1)}{4 d} \]

[Out]

(5*a*Log[1 - Sin[c + d*x]])/(4*d) - (a*Log[1 + Sin[c + d*x]])/(4*d) + (a*Sin[c + d*x])/d + a^2/(2*d*(a - a*Sin
[c + d*x]))

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Rubi [A]  time = 0.048039, antiderivative size = 71, 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} \[ \frac{a^2}{2 d (a-a \sin (c+d x))}+\frac{a \sin (c+d x)}{d}+\frac{5 a \log (1-\sin (c+d x))}{4 d}-\frac{a \log (\sin (c+d x)+1)}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(5*a*Log[1 - Sin[c + d*x]])/(4*d) - (a*Log[1 + Sin[c + d*x]])/(4*d) + (a*Sin[c + d*x])/d + a^2/(2*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 ^3(c+d x) \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^3}{(a-x)^2 (a+x)} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{\operatorname{Subst}\left (\int \left (1+\frac{a^2}{2 (a-x)^2}-\frac{5 a}{4 (a-x)}-\frac{a}{4 (a+x)}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{5 a \log (1-\sin (c+d x))}{4 d}-\frac{a \log (1+\sin (c+d x))}{4 d}+\frac{a \sin (c+d x)}{d}+\frac{a^2}{2 d (a-a \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.108376, size = 77, normalized size = 1.08 \[ -\frac{a \sin (c+d x) \tan ^2(c+d x)}{d}+\frac{a \left (\tan ^2(c+d x)+2 \log (\cos (c+d x))\right )}{2 d}-\frac{3 a \left (\tanh ^{-1}(\sin (c+d x))-\tan (c+d x) \sec (c+d x)\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.056, size = 96, normalized size = 1.4 \begin{align*}{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{a \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d}}+{\frac{3\,a\sin \left ( dx+c \right ) }{2\,d}}-{\frac{3\,a\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,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((a+a*sin(d*x+c))*tan(d*x+c)^3,x)

[Out]

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

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Maxima [A]  time = 0.99153, size = 69, normalized size = 0.97 \begin{align*} -\frac{a \log \left (\sin \left (d x + c\right ) + 1\right ) - 5 \, a \log \left (\sin \left (d x + c\right ) - 1\right ) - 4 \, a \sin \left (d x + c\right ) + \frac{2 \, a}{\sin \left (d x + c\right ) - 1}}{4 \, 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)^3,x, algorithm="maxima")

[Out]

-1/4*(a*log(sin(d*x + c) + 1) - 5*a*log(sin(d*x + c) - 1) - 4*a*sin(d*x + c) + 2*a/(sin(d*x + c) - 1))/d

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Fricas [A]  time = 1.53161, size = 224, normalized size = 3.15 \begin{align*} -\frac{4 \, a \cos \left (d x + c\right )^{2} +{\left (a \sin \left (d x + c\right ) - a\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 5 \,{\left (a \sin \left (d x + c\right ) - a\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, a \sin \left (d x + c\right ) - 2 \, a}{4 \,{\left (d \sin \left (d x + c\right ) - d\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)^3,x, algorithm="fricas")

[Out]

-1/4*(4*a*cos(d*x + c)^2 + (a*sin(d*x + c) - a)*log(sin(d*x + c) + 1) - 5*(a*sin(d*x + c) - a)*log(-sin(d*x +
c) + 1) + 4*a*sin(d*x + c) - 2*a)/(d*sin(d*x + c) - d)

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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 ^{3}{\left (c + d x \right )}\, dx + \int \tan ^{3}{\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)**3,x)

[Out]

a*(Integral(sin(c + d*x)*tan(c + d*x)**3, x) + Integral(tan(c + d*x)**3, 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)^3,x, algorithm="giac")

[Out]

Timed out