3.1.29 \(\int (a+a \sin (c+d x))^3 \tan ^6(c+d x) \, dx\) [29]

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

[Out]

-23/2*a^3*x+136/5*a^3*cos(d*x+c)/d-136/15*a^3*cos(d*x+c)^3/d+23/2*a^3*cos(d*x+c)*sin(d*x+c)/d+1/5*a^6*cos(d*x+
c)*sin(d*x+c)^5/d/(a-a*sin(d*x+c))^3-13/15*a^5*cos(d*x+c)*sin(d*x+c)^4/d/(a-a*sin(d*x+c))^2+23/3*a^6*cos(d*x+c
)*sin(d*x+c)^3/d/(a^3-a^3*sin(d*x+c))

________________________________________________________________________________________

Rubi [A]
time = 0.25, antiderivative size = 180, normalized size of antiderivative = 1.00, 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 = {2787, 2844, 3056, 2827, 2715, 8, 2713} \begin {gather*} \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 {136 a^3 \cos ^3(c+d x)}{15 d}+\frac {136 a^3 \cos (c+d x)}{5 d}+\frac {23 a^3 \sin (c+d x) \cos (c+d x)}{2 d}-\frac {23 a^3 x}{2}+\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 )} \end {gather*}

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 8

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

Rule 2713

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]

Rule 2715

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] && Integ
erQ[2*n]

Rule 2787

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 2827

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 2844

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 3056

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])

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 ) \text {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 = 3.24, size = 243, normalized size = 1.35 \begin {gather*} \frac {(a+a \sin (c+d x))^3 \left (-690 (c+d x)+405 \cos (c+d x)-5 \cos (3 (c+d x))+\frac {12}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}-\frac {112}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+\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 {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 {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 )}+45 \sin (2 (c+d x))\right )}{60 d \left (\cos \left (\frac {1}{2} (c+d x)\right )+\sin \left (\frac {1}{2} (c+d x)\right )\right )^6} \end {gather*}

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)

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(358\) vs. \(2(166)=332\).
time = 0.27, size = 359, normalized size = 1.99

method	result	size
risch	\(-\frac {23 a^{3} x}{2}-\frac {a^{3} {\mathrm e}^{3 i \left (d x +c \right )}}{24 d}-\frac {3 i a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{8 d}+\frac {27 a^{3} {\mathrm e}^{i \left (d x +c \right )}}{8 d}+\frac {27 a^{3} {\mathrm e}^{-i \left (d x +c \right )}}{8 d}+\frac {3 i a^{3} {\mathrm e}^{-2 i \left (d x +c \right )}}{8 d}-\frac {a^{3} {\mathrm e}^{-3 i \left (d x +c \right )}}{24 d}+\frac {-\frac {464 a^{3} {\mathrm e}^{2 i \left (d x +c \right )}}{3}-108 i a^{3} {\mathrm e}^{3 i \left (d x +c \right )}+\frac {304 i a^{3} {\mathrm e}^{i \left (d x +c \right )}}{3}+30 a^{3} {\mathrm e}^{4 i \left (d x +c \right )}+\frac {394 a^{3}}{15}}{\left ({\mathrm e}^{i \left (d x +c \right )}-i\right )^{5} d}\)	\(195\)
derivativedivides	\(\frac {a^{3} \left (\frac {\sin ^{10}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}-\frac {\sin ^{10}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}+\frac {7 \left (\sin ^{10}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}+\frac {7 \left (\frac {128}{35}+\sin ^{8}\left (d x +c \right )+\frac {8 \left (\sin ^{6}\left (d x +c \right )\right )}{7}+\frac {48 \left (\sin ^{4}\left (d x +c \right )\right )}{35}+\frac {64 \left (\sin ^{2}\left (d x +c \right )\right )}{35}\right ) \cos \left (d x +c \right )}{3}\right )+3 a^{3} \left (\frac {\sin ^{9}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}-\frac {4 \left (\sin ^{9}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}+\frac {8 \left (\sin ^{9}\left (d x +c \right )\right )}{5 \cos \left (d x +c \right )}+\frac {8 \left (\sin ^{7}\left (d x +c \right )+\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{6}+\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}+\frac {35 \sin \left (d x +c \right )}{16}\right ) \cos \left (d x +c \right )}{5}-\frac {7 d x}{2}-\frac {7 c}{2}\right )+3 a^{3} \left (\frac {\sin ^{8}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}-\frac {\sin ^{8}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{3}}+\frac {\sin ^{8}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {16}{5}+\sin ^{6}\left (d x +c \right )+\frac {6 \left (\sin ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (d x +c \right )\right )}{5}\right ) \cos \left (d x +c \right )\right )+a^{3} \left (\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}+\tan \left (d x +c \right )-d x -c \right )}{d}\)	\(359\)
default	\(\frac {a^{3} \left (\frac {\sin ^{10}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}-\frac {\sin ^{10}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}+\frac {7 \left (\sin ^{10}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}+\frac {7 \left (\frac {128}{35}+\sin ^{8}\left (d x +c \right )+\frac {8 \left (\sin ^{6}\left (d x +c \right )\right )}{7}+\frac {48 \left (\sin ^{4}\left (d x +c \right )\right )}{35}+\frac {64 \left (\sin ^{2}\left (d x +c \right )\right )}{35}\right ) \cos \left (d x +c \right )}{3}\right )+3 a^{3} \left (\frac {\sin ^{9}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}-\frac {4 \left (\sin ^{9}\left (d x +c \right )\right )}{15 \cos \left (d x +c \right )^{3}}+\frac {8 \left (\sin ^{9}\left (d x +c \right )\right )}{5 \cos \left (d x +c \right )}+\frac {8 \left (\sin ^{7}\left (d x +c \right )+\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{6}+\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}+\frac {35 \sin \left (d x +c \right )}{16}\right ) \cos \left (d x +c \right )}{5}-\frac {7 d x}{2}-\frac {7 c}{2}\right )+3 a^{3} \left (\frac {\sin ^{8}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{5}}-\frac {\sin ^{8}\left (d x +c \right )}{5 \cos \left (d x +c \right )^{3}}+\frac {\sin ^{8}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {16}{5}+\sin ^{6}\left (d x +c \right )+\frac {6 \left (\sin ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (d x +c \right )\right )}{5}\right ) \cos \left (d x +c \right )\right )+a^{3} \left (\frac {\left (\tan ^{5}\left (d x +c \right )\right )}{5}-\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}+\tan \left (d x +c \right )-d x -c \right )}{d}\)	\(359\)

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,method=_RETURNVERBOSE)

[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))

________________________________________________________________________________________

Maxima [A]
time = 0.50, size = 209, normalized size = 1.16 \begin {gather*} \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 {gather*}

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

________________________________________________________________________________________

Fricas [A]
time = 0.36, size = 289, normalized size = 1.61 \begin {gather*} -\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 {gather*}

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)

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{3} \left (\int 3 \sin {\left (c + d x \right )} \tan ^{6}{\left (c + d x \right )}\, dx + \int 3 \sin ^{2}{\left (c + d x \right )} \tan ^{6}{\left (c + d x \right )}\, dx + \int \sin ^{3}{\left (c + d x \right )} \tan ^{6}{\left (c + d x \right )}\, dx + \int \tan ^{6}{\left (c + d x \right )}\, dx\right ) \end {gather*}

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]

a**3*(Integral(3*sin(c + d*x)*tan(c + d*x)**6, x) + Integral(3*sin(c + d*x)**2*tan(c + d*x)**6, x) + Integral(
sin(c + d*x)**3*tan(c + d*x)**6, x) + Integral(tan(c + d*x)**6, x))

________________________________________________________________________________________

Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

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

________________________________________________________________________________________

Mupad [B]
time = 11.05, size = 438, normalized size = 2.43 \begin {gather*} -\frac {23\,a^3\,x}{2}-\frac {\frac {23\,a^3\,\left (c+d\,x\right )}{2}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {115\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (1725\,c+1725\,d\,x-4750\right )}{30}\right )-\frac {a^3\,\left (345\,c+345\,d\,x-1088\right )}{30}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (\frac {115\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (1725\,c+1725\,d\,x-690\right )}{30}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {299\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (4485\,c+4485\,d\,x-3450\right )}{30}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {299\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (4485\,c+4485\,d\,x-10694\right )}{30}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {575\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (8625\,c+8625\,d\,x-8740\right )}{30}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {575\,a^3\,\left (c+d\,x\right )}{2}-\frac {a^3\,\left (8625\,c+8625\,d\,x-18460\right )}{30}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (437\,a^3\,\left (c+d\,x\right )-\frac {a^3\,\left (13110\,c+13110\,d\,x-16100\right )}{30}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (437\,a^3\,\left (c+d\,x\right )-\frac {a^3\,\left (13110\,c+13110\,d\,x-25244\right )}{30}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (529\,a^3\,\left (c+d\,x\right )-\frac {a^3\,\left (15870\,c+15870\,d\,x-23368\right )}{30}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (529\,a^3\,\left (c+d\,x\right )-\frac {a^3\,\left (15870\,c+15870\,d\,x-26680\right )}{30}\right )}{d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^5\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^3} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (23*a^3*x)/2 - ((23*a^3*(c + d*x))/2 - tan(c/2 + (d*x)/2)*((115*a^3*(c + d*x))/2 - (a^3*(1725*c + 1725*d*x -
4750))/30) - (a^3*(345*c + 345*d*x - 1088))/30 + tan(c/2 + (d*x)/2)^10*((115*a^3*(c + d*x))/2 - (a^3*(1725*c
+ 1725*d*x - 690))/30) - tan(c/2 + (d*x)/2)^9*((299*a^3*(c + d*x))/2 - (a^3*(4485*c + 4485*d*x - 3450))/30) +
tan(c/2 + (d*x)/2)^2*((299*a^3*(c + d*x))/2 - (a^3*(4485*c + 4485*d*x - 10694))/30) + tan(c/2 + (d*x)/2)^8*((5
75*a^3*(c + d*x))/2 - (a^3*(8625*c + 8625*d*x - 8740))/30) - tan(c/2 + (d*x)/2)^3*((575*a^3*(c + d*x))/2 - (a^
3*(8625*c + 8625*d*x - 18460))/30) - tan(c/2 + (d*x)/2)^7*(437*a^3*(c + d*x) - (a^3*(13110*c + 13110*d*x - 161
00))/30) + tan(c/2 + (d*x)/2)^4*(437*a^3*(c + d*x) - (a^3*(13110*c + 13110*d*x - 25244))/30) + tan(c/2 + (d*x)
/2)^6*(529*a^3*(c + d*x) - (a^3*(15870*c + 15870*d*x - 23368))/30) - tan(c/2 + (d*x)/2)^5*(529*a^3*(c + d*x) -
(a^3*(15870*c + 15870*d*x - 26680))/30))/(d*(tan(c/2 + (d*x)/2) - 1)^5*(tan(c/2 + (d*x)/2)^2 + 1)^3)

________________________________________________________________________________________