3.29 \(\int (a+a \sin (c+d x))^3 \tan ^6(c+d x) \, dx\)

Optimal. Leaf size=180 \[ -\frac{136 a^3 \cos ^3(c+d x)}{15 d}+\frac{136 a^3 \cos (c+d x)}{5 d}+\frac{23 a^6 \sin ^3(c+d x) \cos (c+d x)}{3 d \left (a^3-a^3 \sin (c+d x)\right )}+\frac{a^6 \sin ^5(c+d x) \cos (c+d x)}{5 d (a-a \sin (c+d x))^3}-\frac{13 a^5 \sin ^4(c+d x) \cos (c+d x)}{15 d (a-a \sin (c+d x))^2}+\frac{23 a^3 \sin (c+d x) \cos (c+d x)}{2 d}-\frac{23 a^3 x}{2} \]

[Out]

(-23*a^3*x)/2 + (136*a^3*Cos[c + d*x])/(5*d) - (136*a^3*Cos[c + d*x]^3)/(15*d) + (23*a^3*Cos[c + d*x]*Sin[c +
d*x])/(2*d) + (a^6*Cos[c + d*x]*Sin[c + d*x]^5)/(5*d*(a - a*Sin[c + d*x])^3) - (13*a^5*Cos[c + d*x]*Sin[c + d*
x]^4)/(15*d*(a - a*Sin[c + d*x])^2) + (23*a^6*Cos[c + d*x]*Sin[c + d*x]^3)/(3*d*(a^3 - a^3*Sin[c + d*x]))

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Rubi [A]  time = 0.356664, antiderivative size = 180, normalized size of antiderivative = 1., number of steps used = 9, number of rules used = 7, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.333, Rules used = {2708, 2765, 2977, 2748, 2635, 8, 2633} \[ -\frac{136 a^3 \cos ^3(c+d x)}{15 d}+\frac{136 a^3 \cos (c+d x)}{5 d}+\frac{23 a^6 \sin ^3(c+d x) \cos (c+d x)}{3 d \left (a^3-a^3 \sin (c+d x)\right )}+\frac{a^6 \sin ^5(c+d x) \cos (c+d x)}{5 d (a-a \sin (c+d x))^3}-\frac{13 a^5 \sin ^4(c+d x) \cos (c+d x)}{15 d (a-a \sin (c+d x))^2}+\frac{23 a^3 \sin (c+d x) \cos (c+d x)}{2 d}-\frac{23 a^3 x}{2} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^3*Tan[c + d*x]^6,x]

[Out]

(-23*a^3*x)/2 + (136*a^3*Cos[c + d*x])/(5*d) - (136*a^3*Cos[c + d*x]^3)/(15*d) + (23*a^3*Cos[c + d*x]*Sin[c +
d*x])/(2*d) + (a^6*Cos[c + d*x]*Sin[c + d*x]^5)/(5*d*(a - a*Sin[c + d*x])^3) - (13*a^5*Cos[c + d*x]*Sin[c + d*
x]^4)/(15*d*(a - a*Sin[c + d*x])^2) + (23*a^6*Cos[c + d*x]*Sin[c + d*x]^3)/(3*d*(a^3 - a^3*Sin[c + d*x]))

Rule 2708

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Sin[
e + f*x]^p/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] &&
EqQ[p, 2*m]

Rule 2765

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

Rule 2977

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(e_
.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[((A*b - a*B)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x]
)^n)/(a*f*(2*m + 1)), x] - Dist[1/(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n -
1)*Simp[A*(a*d*n - b*c*(m + 1)) - B*(a*c*m + b*d*n) - d*(a*B*(m - n) + A*b*(m + n + 1))*Sin[e + f*x], x], x],
x] /; FreeQ[{a, b, c, d, e, f, A, B}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ
[m, -2^(-1)] && GtQ[n, 0] && IntegerQ[2*m] && (IntegerQ[2*n] || EqQ[c, 0])

Rule 2748

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Dist[c, Int[(b*S
in[e + f*x])^m, x], x] + Dist[d/b, Int[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2633

Int[sin[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[Expand[(1 - x^2)^((n - 1)/2), x], x], x
, Cos[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[(n - 1)/2, 0]

Rubi steps

\begin{align*} \int (a+a \sin (c+d x))^3 \tan ^6(c+d x) \, dx &=a^6 \int \frac{\sin ^6(c+d x)}{(a-a \sin (c+d x))^3} \, dx\\ &=\frac{a^6 \cos (c+d x) \sin ^5(c+d x)}{5 d (a-a \sin (c+d x))^3}+\frac{1}{5} a^4 \int \frac{\sin ^4(c+d x) (-5 a-8 a \sin (c+d x))}{(a-a \sin (c+d x))^2} \, dx\\ &=\frac{a^6 \cos (c+d x) \sin ^5(c+d x)}{5 d (a-a \sin (c+d x))^3}-\frac{13 a^5 \cos (c+d x) \sin ^4(c+d x)}{15 d (a-a \sin (c+d x))^2}-\frac{1}{15} a^2 \int \frac{\sin ^3(c+d x) \left (-52 a^2-63 a^2 \sin (c+d x)\right )}{a-a \sin (c+d x)} \, dx\\ &=\frac{a^6 \cos (c+d x) \sin ^5(c+d x)}{5 d (a-a \sin (c+d x))^3}-\frac{13 a^5 \cos (c+d x) \sin ^4(c+d x)}{15 d (a-a \sin (c+d x))^2}+\frac{23 a^4 \cos (c+d x) \sin ^3(c+d x)}{3 d (a-a \sin (c+d x))}+\frac{1}{15} \int \sin ^2(c+d x) \left (-345 a^3-408 a^3 \sin (c+d x)\right ) \, dx\\ &=\frac{a^6 \cos (c+d x) \sin ^5(c+d x)}{5 d (a-a \sin (c+d x))^3}-\frac{13 a^5 \cos (c+d x) \sin ^4(c+d x)}{15 d (a-a \sin (c+d x))^2}+\frac{23 a^4 \cos (c+d x) \sin ^3(c+d x)}{3 d (a-a \sin (c+d x))}-\left (23 a^3\right ) \int \sin ^2(c+d x) \, dx-\frac{1}{5} \left (136 a^3\right ) \int \sin ^3(c+d x) \, dx\\ &=\frac{23 a^3 \cos (c+d x) \sin (c+d x)}{2 d}+\frac{a^6 \cos (c+d x) \sin ^5(c+d x)}{5 d (a-a \sin (c+d x))^3}-\frac{13 a^5 \cos (c+d x) \sin ^4(c+d x)}{15 d (a-a \sin (c+d x))^2}+\frac{23 a^4 \cos (c+d x) \sin ^3(c+d x)}{3 d (a-a \sin (c+d x))}-\frac{1}{2} \left (23 a^3\right ) \int 1 \, dx+\frac{\left (136 a^3\right ) \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{5 d}\\ &=-\frac{23 a^3 x}{2}+\frac{136 a^3 \cos (c+d x)}{5 d}-\frac{136 a^3 \cos ^3(c+d x)}{15 d}+\frac{23 a^3 \cos (c+d x) \sin (c+d x)}{2 d}+\frac{a^6 \cos (c+d x) \sin ^5(c+d x)}{5 d (a-a \sin (c+d x))^3}-\frac{13 a^5 \cos (c+d x) \sin ^4(c+d x)}{15 d (a-a \sin (c+d x))^2}+\frac{23 a^4 \cos (c+d x) \sin ^3(c+d x)}{3 d (a-a \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 5.61378, size = 243, normalized size = 1.35 \[ \frac{(a \sin (c+d x)+a)^3 \left (-690 (c+d x)+45 \sin (2 (c+d x))+405 \cos (c+d x)-5 \cos (3 (c+d x))+\frac{1576 \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}-\frac{224 \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{24 \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^5}-\frac{112}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+\frac{12}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^4}\right )}{60 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^3*Tan[c + d*x]^6,x]

[Out]

((a + a*Sin[c + d*x])^3*(-690*(c + d*x) + 405*Cos[c + d*x] - 5*Cos[3*(c + d*x)] + 12/(Cos[(c + d*x)/2] - Sin[(
c + d*x)/2])^4 - 112/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (24*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(
c + d*x)/2])^5 - (224*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3 + (1576*Sin[(c + d*x)/2])/(Cos
[(c + d*x)/2] - Sin[(c + d*x)/2]) + 45*Sin[2*(c + d*x)]))/(60*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)

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Maple [B]  time = 0.131, size = 359, normalized size = 2. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))^3*tan(d*x+c)^6,x)

[Out]

1/d*(a^3*(1/5*sin(d*x+c)^10/cos(d*x+c)^5-1/3*sin(d*x+c)^10/cos(d*x+c)^3+7/3*sin(d*x+c)^10/cos(d*x+c)+7/3*(128/
35+sin(d*x+c)^8+8/7*sin(d*x+c)^6+48/35*sin(d*x+c)^4+64/35*sin(d*x+c)^2)*cos(d*x+c))+3*a^3*(1/5*sin(d*x+c)^9/co
s(d*x+c)^5-4/15*sin(d*x+c)^9/cos(d*x+c)^3+8/5*sin(d*x+c)^9/cos(d*x+c)+8/5*(sin(d*x+c)^7+7/6*sin(d*x+c)^5+35/24
*sin(d*x+c)^3+35/16*sin(d*x+c))*cos(d*x+c)-7/2*d*x-7/2*c)+3*a^3*(1/5*sin(d*x+c)^8/cos(d*x+c)^5-1/5*sin(d*x+c)^
8/cos(d*x+c)^3+sin(d*x+c)^8/cos(d*x+c)+(16/5+sin(d*x+c)^6+6/5*sin(d*x+c)^4+8/5*sin(d*x+c)^2)*cos(d*x+c))+a^3*(
1/5*tan(d*x+c)^5-1/3*tan(d*x+c)^3+tan(d*x+c)-d*x-c))

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Maxima [A]  time = 1.648, size = 282, normalized size = 1.57 \begin{align*} \frac{3 \,{\left (6 \, \tan \left (d x + c\right )^{5} - 20 \, \tan \left (d x + c\right )^{3} - 105 \, d x - 105 \, c + \frac{15 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} + 90 \, \tan \left (d x + c\right )\right )} a^{3} + 2 \,{\left (3 \, \tan \left (d x + c\right )^{5} - 5 \, \tan \left (d x + c\right )^{3} - 15 \, d x - 15 \, c + 15 \, \tan \left (d x + c\right )\right )} a^{3} - 2 \,{\left (5 \, \cos \left (d x + c\right )^{3} - \frac{90 \, \cos \left (d x + c\right )^{4} - 20 \, \cos \left (d x + c\right )^{2} + 3}{\cos \left (d x + c\right )^{5}} - 60 \, \cos \left (d x + c\right )\right )} a^{3} + 18 \, a^{3}{\left (\frac{15 \, \cos \left (d x + c\right )^{4} - 5 \, \cos \left (d x + c\right )^{2} + 1}{\cos \left (d x + c\right )^{5}} + 5 \, \cos \left (d x + c\right )\right )}}{30 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^3*tan(d*x+c)^6,x, algorithm="maxima")

[Out]

1/30*(3*(6*tan(d*x + c)^5 - 20*tan(d*x + c)^3 - 105*d*x - 105*c + 15*tan(d*x + c)/(tan(d*x + c)^2 + 1) + 90*ta
n(d*x + c))*a^3 + 2*(3*tan(d*x + c)^5 - 5*tan(d*x + c)^3 - 15*d*x - 15*c + 15*tan(d*x + c))*a^3 - 2*(5*cos(d*x
+ c)^3 - (90*cos(d*x + c)^4 - 20*cos(d*x + c)^2 + 3)/cos(d*x + c)^5 - 60*cos(d*x + c))*a^3 + 18*a^3*((15*cos(
d*x + c)^4 - 5*cos(d*x + c)^2 + 1)/cos(d*x + c)^5 + 5*cos(d*x + c)))/d

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Fricas [A]  time = 1.59798, size = 734, normalized size = 4.08 \begin{align*} -\frac{10 \, a^{3} \cos \left (d x + c\right )^{6} - 15 \, a^{3} \cos \left (d x + c\right )^{5} - 140 \, a^{3} \cos \left (d x + c\right )^{4} - 1380 \, a^{3} d x +{\left (345 \, a^{3} d x - 839 \, a^{3}\right )} \cos \left (d x + c\right )^{3} + 6 \, a^{3} +{\left (1035 \, a^{3} d x + 668 \, a^{3}\right )} \cos \left (d x + c\right )^{2} - 6 \,{\left (115 \, a^{3} d x - 233 \, a^{3}\right )} \cos \left (d x + c\right ) -{\left (10 \, a^{3} \cos \left (d x + c\right )^{5} + 25 \, a^{3} \cos \left (d x + c\right )^{4} - 115 \, a^{3} \cos \left (d x + c\right )^{3} - 1380 \, a^{3} d x - 6 \, a^{3} +{\left (345 \, a^{3} d x + 724 \, a^{3}\right )} \cos \left (d x + c\right )^{2} - 6 \,{\left (115 \, a^{3} d x - 232 \, a^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{30 \,{\left (d \cos \left (d x + c\right )^{3} + 3 \, d \cos \left (d x + c\right )^{2} - 2 \, d \cos \left (d x + c\right ) -{\left (d \cos \left (d x + c\right )^{2} - 2 \, d \cos \left (d x + c\right ) - 4 \, d\right )} \sin \left (d x + c\right ) - 4 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^3*tan(d*x+c)^6,x, algorithm="fricas")

[Out]

-1/30*(10*a^3*cos(d*x + c)^6 - 15*a^3*cos(d*x + c)^5 - 140*a^3*cos(d*x + c)^4 - 1380*a^3*d*x + (345*a^3*d*x -
839*a^3)*cos(d*x + c)^3 + 6*a^3 + (1035*a^3*d*x + 668*a^3)*cos(d*x + c)^2 - 6*(115*a^3*d*x - 233*a^3)*cos(d*x
+ c) - (10*a^3*cos(d*x + c)^5 + 25*a^3*cos(d*x + c)^4 - 115*a^3*cos(d*x + c)^3 - 1380*a^3*d*x - 6*a^3 + (345*a
^3*d*x + 724*a^3)*cos(d*x + c)^2 - 6*(115*a^3*d*x - 232*a^3)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^3 + 3
*d*cos(d*x + c)^2 - 2*d*cos(d*x + c) - (d*cos(d*x + c)^2 - 2*d*cos(d*x + c) - 4*d)*sin(d*x + c) - 4*d)

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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((a+a*sin(d*x+c))**3*tan(d*x+c)**6,x)

[Out]

Timed out

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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))^3*tan(d*x+c)^6,x, algorithm="giac")

[Out]

Timed out