3.46 \(\int \frac{\tan ^3(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=82 \[ \frac{\tan ^4(c+d x)}{4 a d}-\frac{3 \tanh ^{-1}(\sin (c+d x))}{8 a d}-\frac{\tan ^3(c+d x) \sec (c+d x)}{4 a d}+\frac{3 \tan (c+d x) \sec (c+d x)}{8 a d} \]

[Out]

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

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Rubi [A]  time = 0.115661, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2706, 2607, 30, 2611, 3770} \[ \frac{\tan ^4(c+d x)}{4 a d}-\frac{3 \tanh ^{-1}(\sin (c+d x))}{8 a d}-\frac{\tan ^3(c+d x) \sec (c+d x)}{4 a d}+\frac{3 \tan (c+d x) \sec (c+d x)}{8 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2706

Int[((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.)/((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[1/a, Int[S
ec[e + f*x]^2*(g*Tan[e + f*x])^p, x], x] - Dist[1/(b*g), Int[Sec[e + f*x]*(g*Tan[e + f*x])^(p + 1), x], x] /;
FreeQ[{a, b, e, f, g, p}, x] && EqQ[a^2 - b^2, 0] && NeQ[p, -1]

Rule 2607

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

Rule 30

Int[(x_)^(m_.), x_Symbol] :> Simp[x^(m + 1)/(m + 1), x] /; FreeQ[m, x] && NeQ[m, -1]

Rule 2611

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(a*Sec[e
+ f*x])^m*(b*Tan[e + f*x])^(n - 1))/(f*(m + n - 1)), x] - Dist[(b^2*(n - 1))/(m + n - 1), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 3770

Int[csc[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[ArcTanh[Cos[c + d*x]]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.164595, size = 54, normalized size = 0.66 \[ -\frac{\frac{1}{\sin (c+d x)-1}+\frac{4}{\sin (c+d x)+1}-\frac{1}{(\sin (c+d x)+1)^2}+3 \tanh ^{-1}(\sin (c+d x))}{8 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(3*ArcTanh[Sin[c + d*x]] + (-1 + Sin[c + d*x])^(-1) - (1 + Sin[c + d*x])^(-2) + 4/(1 + Sin[c + d*x]))/(8*a*d)

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Maple [A]  time = 0.053, size = 90, normalized size = 1.1 \begin{align*} -{\frac{1}{8\,da \left ( \sin \left ( dx+c \right ) -1 \right ) }}+{\frac{3\,\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{16\,da}}+{\frac{1}{8\,da \left ( 1+\sin \left ( dx+c \right ) \right ) ^{2}}}-{\frac{1}{2\,da \left ( 1+\sin \left ( dx+c \right ) \right ) }}-{\frac{3\,\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{16\,da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(d*x+c)^3/(a+a*sin(d*x+c)),x)

[Out]

-1/8/a/d/(sin(d*x+c)-1)+3/16/a/d*ln(sin(d*x+c)-1)+1/8/a/d/(1+sin(d*x+c))^2-1/2/a/d/(1+sin(d*x+c))-3/16*ln(1+si
n(d*x+c))/a/d

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Maxima [A]  time = 0.999972, size = 120, normalized size = 1.46 \begin{align*} -\frac{\frac{2 \,{\left (5 \, \sin \left (d x + c\right )^{2} + \sin \left (d x + c\right ) - 2\right )}}{a \sin \left (d x + c\right )^{3} + a \sin \left (d x + c\right )^{2} - a \sin \left (d x + c\right ) - a} + \frac{3 \, \log \left (\sin \left (d x + c\right ) + 1\right )}{a} - \frac{3 \, \log \left (\sin \left (d x + c\right ) - 1\right )}{a}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.34329, size = 338, normalized size = 4.12 \begin{align*} -\frac{10 \, \cos \left (d x + c\right )^{2} + 3 \,{\left (\cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (\cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, \sin \left (d x + c\right ) - 6}{16 \,{\left (a d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/16*(10*cos(d*x + c)^2 + 3*(cos(d*x + c)^2*sin(d*x + c) + cos(d*x + c)^2)*log(sin(d*x + c) + 1) - 3*(cos(d*x
+ c)^2*sin(d*x + c) + cos(d*x + c)^2)*log(-sin(d*x + c) + 1) - 2*sin(d*x + c) - 6)/(a*d*cos(d*x + c)^2*sin(d*
x + c) + a*d*cos(d*x + c)^2)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \frac{\int \frac{\tan ^{3}{\left (c + d x \right )}}{\sin{\left (c + d x \right )} + 1}\, dx}{a} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(tan(d*x+c)**3/(a+a*sin(d*x+c)),x)

[Out]

Integral(tan(c + d*x)**3/(sin(c + d*x) + 1), x)/a

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Giac [A]  time = 2.11468, size = 130, normalized size = 1.59 \begin{align*} -\frac{\frac{6 \, \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a} - \frac{6 \, \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a} + \frac{2 \,{\left (3 \, \sin \left (d x + c\right ) - 1\right )}}{a{\left (\sin \left (d x + c\right ) - 1\right )}} - \frac{9 \, \sin \left (d x + c\right )^{2} + 2 \, \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(tan(d*x+c)^3/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

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