3.823 \(\int \frac{\tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=69 \[ \frac{\tan ^5(c+d x)}{5 a d}-\frac{\sec ^5(c+d x)}{5 a d}+\frac{2 \sec ^3(c+d x)}{3 a d}-\frac{\sec (c+d x)}{a d} \]

[Out]

-(Sec[c + d*x]/(a*d)) + (2*Sec[c + d*x]^3)/(3*a*d) - Sec[c + d*x]^5/(5*a*d) + Tan[c + d*x]^5/(5*a*d)

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Rubi [A]  time = 0.0956588, antiderivative size = 69, 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, 2606, 194} \[ \frac{\tan ^5(c+d x)}{5 a d}-\frac{\sec ^5(c+d x)}{5 a d}+\frac{2 \sec ^3(c+d x)}{3 a d}-\frac{\sec (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[c + d*x]/(a*d)) + (2*Sec[c + d*x]^3)/(3*a*d) - Sec[c + d*x]^5/(5*a*d) + Tan[c + d*x]^5/(5*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 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 194

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Int[ExpandIntegrand[(a + b*x^n)^p, x], x] /; FreeQ[{a, b}, x]
&& IGtQ[n, 0] && IGtQ[p, 0]

Rubi steps

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

Mathematica [A]  time = 0.269237, size = 106, normalized size = 1.54 \[ -\frac{\sec ^3(c+d x) (-64 \sin (c+d x)-178 \sin (2 (c+d x))+192 \sin (3 (c+d x))-89 \sin (4 (c+d x))-534 \cos (c+d x)+288 \cos (2 (c+d x))-178 \cos (3 (c+d x))+24 \cos (4 (c+d x))+200)}{960 a d (\sin (c+d x)+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(Sec[c + d*x]^3*(200 - 534*Cos[c + d*x] + 288*Cos[2*(c + d*x)] - 178*Cos[3*(c + d*x)] + 24*Cos[4*(c + d*x)] -
64*Sin[c + d*x] - 178*Sin[2*(c + d*x)] + 192*Sin[3*(c + d*x)] - 89*Sin[4*(c + d*x)]))/(960*a*d*(1 + Sin[c + d
*x]))

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Maple [B]  time = 0.086, size = 130, normalized size = 1.9 \begin{align*} 32\,{\frac{1}{da} \left ( -{\frac{1}{192\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{3}}}-{\frac{1}{128\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) -1 \right ) ^{2}}}+{\frac{3}{256\,\tan \left ( 1/2\,dx+c/2 \right ) -256}}-{\frac{1}{80\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{5}}}+1/32\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{-4}-{\frac{1}{96\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{3}}}-{\frac{1}{64\, \left ( \tan \left ( 1/2\,dx+c/2 \right ) +1 \right ) ^{2}}}-{\frac{3}{256\,\tan \left ( 1/2\,dx+c/2 \right ) +256}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

32/d/a*(-1/192/(tan(1/2*d*x+1/2*c)-1)^3-1/128/(tan(1/2*d*x+1/2*c)-1)^2+3/256/(tan(1/2*d*x+1/2*c)-1)-1/80/(tan(
1/2*d*x+1/2*c)+1)^5+1/32/(tan(1/2*d*x+1/2*c)+1)^4-1/96/(tan(1/2*d*x+1/2*c)+1)^3-1/64/(tan(1/2*d*x+1/2*c)+1)^2-
3/256/(tan(1/2*d*x+1/2*c)+1))

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Maxima [B]  time = 1.03692, size = 289, normalized size = 4.19 \begin{align*} -\frac{16 \,{\left (\frac{2 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{2 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{6 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + 1\right )}}{15 \,{\left (a + \frac{2 \, a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{2 \, a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{6 \, a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{6 \, a \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{2 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{2 \, a \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac{a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-16/15*(2*sin(d*x + c)/(cos(d*x + c) + 1) - 2*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 6*sin(d*x + c)^3/(cos(d*x
+ c) + 1)^3 + 1)/((a + 2*a*sin(d*x + c)/(cos(d*x + c) + 1) - 2*a*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 6*a*sin
(d*x + c)^3/(cos(d*x + c) + 1)^3 + 6*a*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 2*a*sin(d*x + c)^6/(cos(d*x + c)
+ 1)^6 - 2*a*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - a*sin(d*x + c)^8/(cos(d*x + c) + 1)^8)*d)

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Fricas [A]  time = 1.37555, size = 194, normalized size = 2.81 \begin{align*} -\frac{3 \, \cos \left (d x + c\right )^{4} + 6 \, \cos \left (d x + c\right )^{2} + 4 \,{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} \sin \left (d x + c\right ) - 1}{15 \,{\left (a d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/15*(3*cos(d*x + c)^4 + 6*cos(d*x + c)^2 + 4*(3*cos(d*x + c)^2 - 1)*sin(d*x + c) - 1)/(a*d*cos(d*x + c)^3*si
n(d*x + c) + a*d*cos(d*x + c)^3)

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

[Out]

Timed out

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Giac [A]  time = 1.28909, size = 162, normalized size = 2.35 \begin{align*} \frac{\frac{5 \,{\left (9 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 24 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 11\right )}}{a{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}} - \frac{45 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 240 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 490 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 320 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 73}{a{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{5}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/120*(5*(9*tan(1/2*d*x + 1/2*c)^2 - 24*tan(1/2*d*x + 1/2*c) + 11)/(a*(tan(1/2*d*x + 1/2*c) - 1)^3) - (45*tan(
1/2*d*x + 1/2*c)^4 + 240*tan(1/2*d*x + 1/2*c)^3 + 490*tan(1/2*d*x + 1/2*c)^2 + 320*tan(1/2*d*x + 1/2*c) + 73)/
(a*(tan(1/2*d*x + 1/2*c) + 1)^5))/d