3.38 \(\int (a+a \sin (c+d x))^4 \tan ^4(c+d x) \, dx\)

Optimal. Leaf size=143 \[ \frac {4 a^4 \cos ^3(c+d x)}{3 d}-\frac {16 a^4 \cos (c+d x)}{d}-\frac {a^4 \sin ^3(c+d x) \cos (c+d x)}{4 d}-\frac {35 a^4 \sin (c+d x) \cos (c+d x)}{8 d}-\frac {56 a^4 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac {4 a^4 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}+\frac {163 a^4 x}{8} \]

[Out]

163/8*a^4*x-16*a^4*cos(d*x+c)/d+4/3*a^4*cos(d*x+c)^3/d+4/3*a^4*cos(d*x+c)/d/(1-sin(d*x+c))^2-56/3*a^4*cos(d*x+
c)/d/(1-sin(d*x+c))-35/8*a^4*cos(d*x+c)*sin(d*x+c)/d-1/4*a^4*cos(d*x+c)*sin(d*x+c)^3/d

________________________________________________________________________________________

Rubi [A]  time = 0.20, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 13, number of rules used = 7, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {2709, 2650, 2648, 2638, 2635, 8, 2633} \[ \frac {4 a^4 \cos ^3(c+d x)}{3 d}-\frac {16 a^4 \cos (c+d x)}{d}-\frac {a^4 \sin ^3(c+d x) \cos (c+d x)}{4 d}-\frac {35 a^4 \sin (c+d x) \cos (c+d x)}{8 d}-\frac {56 a^4 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac {4 a^4 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}+\frac {163 a^4 x}{8} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(163*a^4*x)/8 - (16*a^4*Cos[c + d*x])/d + (4*a^4*Cos[c + d*x]^3)/(3*d) + (4*a^4*Cos[c + d*x])/(3*d*(1 - Sin[c
+ d*x])^2) - (56*a^4*Cos[c + d*x])/(3*d*(1 - Sin[c + d*x])) - (35*a^4*Cos[c + d*x]*Sin[c + d*x])/(8*d) - (a^4*
Cos[c + d*x]*Sin[c + d*x]^3)/(4*d)

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]

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 2638

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

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2709

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

Rubi steps

\begin {align*} \int (a+a \sin (c+d x))^4 \tan ^4(c+d x) \, dx &=a^4 \int \left (16+\frac {4}{(-1+\sin (c+d x))^2}+\frac {20}{-1+\sin (c+d x)}+12 \sin (c+d x)+8 \sin ^2(c+d x)+4 \sin ^3(c+d x)+\sin ^4(c+d x)\right ) \, dx\\ &=16 a^4 x+a^4 \int \sin ^4(c+d x) \, dx+\left (4 a^4\right ) \int \frac {1}{(-1+\sin (c+d x))^2} \, dx+\left (4 a^4\right ) \int \sin ^3(c+d x) \, dx+\left (8 a^4\right ) \int \sin ^2(c+d x) \, dx+\left (12 a^4\right ) \int \sin (c+d x) \, dx+\left (20 a^4\right ) \int \frac {1}{-1+\sin (c+d x)} \, dx\\ &=16 a^4 x-\frac {12 a^4 \cos (c+d x)}{d}+\frac {4 a^4 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}-\frac {20 a^4 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac {4 a^4 \cos (c+d x) \sin (c+d x)}{d}-\frac {a^4 \cos (c+d x) \sin ^3(c+d x)}{4 d}+\frac {1}{4} \left (3 a^4\right ) \int \sin ^2(c+d x) \, dx-\frac {1}{3} \left (4 a^4\right ) \int \frac {1}{-1+\sin (c+d x)} \, dx+\left (4 a^4\right ) \int 1 \, dx-\frac {\left (4 a^4\right ) \operatorname {Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=20 a^4 x-\frac {16 a^4 \cos (c+d x)}{d}+\frac {4 a^4 \cos ^3(c+d x)}{3 d}+\frac {4 a^4 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}-\frac {56 a^4 \cos (c+d x)}{3 d (1-\sin (c+d x))}-\frac {35 a^4 \cos (c+d x) \sin (c+d x)}{8 d}-\frac {a^4 \cos (c+d x) \sin ^3(c+d x)}{4 d}+\frac {1}{8} \left (3 a^4\right ) \int 1 \, dx\\ &=\frac {163 a^4 x}{8}-\frac {16 a^4 \cos (c+d x)}{d}+\frac {4 a^4 \cos ^3(c+d x)}{3 d}+\frac {4 a^4 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}-\frac {56 a^4 \cos (c+d x)}{3 d (1-\sin (c+d x))}-\frac {35 a^4 \cos (c+d x) \sin (c+d x)}{8 d}-\frac {a^4 \cos (c+d x) \sin ^3(c+d x)}{4 d}\\ \end {align*}

________________________________________________________________________________________

Mathematica [A]  time = 1.65, size = 252, normalized size = 1.76 \[ \frac {a^4 \left (-11736 c \sin \left (\frac {1}{2} (c+d x)\right )-11736 d x \sin \left (\frac {1}{2} (c+d x)\right )-16488 \sin \left (\frac {1}{2} (c+d x)\right )-3912 c \sin \left (\frac {3}{2} (c+d x)\right )-3912 d x \sin \left (\frac {3}{2} (c+d x)\right )+3704 \sin \left (\frac {3}{2} (c+d x)\right )+885 \sin \left (\frac {5}{2} (c+d x)\right )+129 \sin \left (\frac {7}{2} (c+d x)\right )-23 \sin \left (\frac {9}{2} (c+d x)\right )-3 \sin \left (\frac {11}{2} (c+d x)\right )+24 (489 c+489 d x+209) \cos \left (\frac {1}{2} (c+d x)\right )-24 (163 c+163 d x+453) \cos \left (\frac {3}{2} (c+d x)\right )+885 \cos \left (\frac {5}{2} (c+d x)\right )-129 \cos \left (\frac {7}{2} (c+d x)\right )-23 \cos \left (\frac {9}{2} (c+d x)\right )+3 \cos \left (\frac {11}{2} (c+d x)\right )\right )}{384 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^4*(24*(209 + 489*c + 489*d*x)*Cos[(c + d*x)/2] - 24*(453 + 163*c + 163*d*x)*Cos[(3*(c + d*x))/2] + 885*Cos[
(5*(c + d*x))/2] - 129*Cos[(7*(c + d*x))/2] - 23*Cos[(9*(c + d*x))/2] + 3*Cos[(11*(c + d*x))/2] - 16488*Sin[(c
+ d*x)/2] - 11736*c*Sin[(c + d*x)/2] - 11736*d*x*Sin[(c + d*x)/2] + 3704*Sin[(3*(c + d*x))/2] - 3912*c*Sin[(3
*(c + d*x))/2] - 3912*d*x*Sin[(3*(c + d*x))/2] + 885*Sin[(5*(c + d*x))/2] + 129*Sin[(7*(c + d*x))/2] - 23*Sin[
(9*(c + d*x))/2] - 3*Sin[(11*(c + d*x))/2]))/(384*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3)

________________________________________________________________________________________

fricas [A]  time = 0.43, size = 247, normalized size = 1.73 \[ -\frac {6 \, a^{4} \cos \left (d x + c\right )^{6} - 20 \, a^{4} \cos \left (d x + c\right )^{5} - 85 \, a^{4} \cos \left (d x + c\right )^{4} + 214 \, a^{4} \cos \left (d x + c\right )^{3} + 978 \, a^{4} d x + 32 \, a^{4} - {\left (489 \, a^{4} d x + 721 \, a^{4}\right )} \cos \left (d x + c\right )^{2} + {\left (489 \, a^{4} d x - 962 \, a^{4}\right )} \cos \left (d x + c\right ) - {\left (6 \, a^{4} \cos \left (d x + c\right )^{5} + 26 \, a^{4} \cos \left (d x + c\right )^{4} - 59 \, a^{4} \cos \left (d x + c\right )^{3} + 978 \, a^{4} d x - 273 \, a^{4} \cos \left (d x + c\right )^{2} - 32 \, a^{4} + {\left (489 \, a^{4} d x - 994 \, a^{4}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{24 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/24*(6*a^4*cos(d*x + c)^6 - 20*a^4*cos(d*x + c)^5 - 85*a^4*cos(d*x + c)^4 + 214*a^4*cos(d*x + c)^3 + 978*a^4
*d*x + 32*a^4 - (489*a^4*d*x + 721*a^4)*cos(d*x + c)^2 + (489*a^4*d*x - 962*a^4)*cos(d*x + c) - (6*a^4*cos(d*x
+ c)^5 + 26*a^4*cos(d*x + c)^4 - 59*a^4*cos(d*x + c)^3 + 978*a^4*d*x - 273*a^4*cos(d*x + c)^2 - 32*a^4 + (489
*a^4*d*x - 994*a^4)*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^2 - d*cos(d*x + c) + (d*cos(d*x + c) + 2*d)*si
n(d*x + c) - 2*d)

________________________________________________________________________________________

giac [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

________________________________________________________________________________________

maple [B]  time = 0.41, size = 360, normalized size = 2.52 \[ \frac {a^{4} \left (\frac {\sin ^{9}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {2 \left (\sin ^{9}\left (d x +c \right )\right )}{\cos \left (d x +c \right )}-2 \left (\sin ^{7}\left (d x +c \right )+\frac {7 \left (\sin ^{5}\left (d x +c \right )\right )}{6}+\frac {35 \left (\sin ^{3}\left (d x +c \right )\right )}{24}+\frac {35 \sin \left (d x +c \right )}{16}\right ) \cos \left (d x +c \right )+\frac {35 d x}{8}+\frac {35 c}{8}\right )+4 a^{4} \left (\frac {\sin ^{8}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {5 \left (\sin ^{8}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {5 \left (\frac {16}{5}+\sin ^{6}\left (d x +c \right )+\frac {6 \left (\sin ^{4}\left (d x +c \right )\right )}{5}+\frac {8 \left (\sin ^{2}\left (d x +c \right )\right )}{5}\right ) \cos \left (d x +c \right )}{3}\right )+6 a^{4} \left (\frac {\sin ^{7}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {4 \left (\sin ^{7}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )}-\frac {4 \left (\sin ^{5}\left (d x +c \right )+\frac {5 \left (\sin ^{3}\left (d x +c \right )\right )}{4}+\frac {15 \sin \left (d x +c \right )}{8}\right ) \cos \left (d x +c \right )}{3}+\frac {5 d x}{2}+\frac {5 c}{2}\right )+4 a^{4} \left (\frac {\sin ^{6}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+a^{4} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.42, size = 238, normalized size = 1.66 \[ \frac {32 \, {\left (\cos \left (d x + c\right )^{3} - \frac {9 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} - 9 \, \cos \left (d x + c\right )\right )} a^{4} + {\left (8 \, \tan \left (d x + c\right )^{3} + 105 \, d x + 105 \, c - \frac {3 \, {\left (13 \, \tan \left (d x + c\right )^{3} + 11 \, \tan \left (d x + c\right )\right )}}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1} - 72 \, \tan \left (d x + c\right )\right )} a^{4} + 24 \, {\left (2 \, \tan \left (d x + c\right )^{3} + 15 \, d x + 15 \, c - \frac {3 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 12 \, \tan \left (d x + c\right )\right )} a^{4} + 8 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{4} - 32 \, a^{4} {\left (\frac {6 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} + 3 \, \cos \left (d x + c\right )\right )}}{24 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/24*(32*(cos(d*x + c)^3 - (9*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 - 9*cos(d*x + c))*a^4 + (8*tan(d*x + c)^3 + 1
05*d*x + 105*c - 3*(13*tan(d*x + c)^3 + 11*tan(d*x + c))/(tan(d*x + c)^4 + 2*tan(d*x + c)^2 + 1) - 72*tan(d*x
+ c))*a^4 + 24*(2*tan(d*x + c)^3 + 15*d*x + 15*c - 3*tan(d*x + c)/(tan(d*x + c)^2 + 1) - 12*tan(d*x + c))*a^4
+ 8*(tan(d*x + c)^3 + 3*d*x + 3*c - 3*tan(d*x + c))*a^4 - 32*a^4*((6*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 + 3*co
s(d*x + c)))/d

________________________________________________________________________________________

mupad [B]  time = 11.05, size = 437, normalized size = 3.06 \[ \frac {163\,a^4\,x}{8}+\frac {\frac {163\,a^4\,\left (c+d\,x\right )}{8}-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {489\,a^4\,\left (c+d\,x\right )}{8}-\frac {a^4\,\left (1467\,c+1467\,d\,x-3630\right )}{24}\right )-\frac {a^4\,\left (489\,c+489\,d\,x-1536\right )}{24}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^{10}\,\left (\frac {489\,a^4\,\left (c+d\,x\right )}{8}-\frac {a^4\,\left (1467\,c+1467\,d\,x-978\right )}{24}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^9\,\left (\frac {1141\,a^4\,\left (c+d\,x\right )}{8}-\frac {a^4\,\left (3423\,c+3423\,d\,x-2934\right )}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {1141\,a^4\,\left (c+d\,x\right )}{8}-\frac {a^4\,\left (3423\,c+3423\,d\,x-7818\right )}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^8\,\left (\frac {2119\,a^4\,\left (c+d\,x\right )}{8}-\frac {a^4\,\left (6357\,c+6357\,d\,x-6520\right )}{24}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {2119\,a^4\,\left (c+d\,x\right )}{8}-\frac {a^4\,\left (6357\,c+6357\,d\,x-13448\right )}{24}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^7\,\left (\frac {1467\,a^4\,\left (c+d\,x\right )}{4}-\frac {a^4\,\left (8802\,c+8802\,d\,x-11736\right )}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {1467\,a^4\,\left (c+d\,x\right )}{4}-\frac {a^4\,\left (8802\,c+8802\,d\,x-15912\right )}{24}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (\frac {1793\,a^4\,\left (c+d\,x\right )}{4}-\frac {a^4\,\left (10758\,c+10758\,d\,x-15364\right )}{24}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (\frac {1793\,a^4\,\left (c+d\,x\right )}{4}-\frac {a^4\,\left (10758\,c+10758\,d\,x-18428\right )}{24}\right )}{d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^4} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(163*a^4*x)/8 + ((163*a^4*(c + d*x))/8 - tan(c/2 + (d*x)/2)*((489*a^4*(c + d*x))/8 - (a^4*(1467*c + 1467*d*x -
3630))/24) - (a^4*(489*c + 489*d*x - 1536))/24 + tan(c/2 + (d*x)/2)^10*((489*a^4*(c + d*x))/8 - (a^4*(1467*c
+ 1467*d*x - 978))/24) - tan(c/2 + (d*x)/2)^9*((1141*a^4*(c + d*x))/8 - (a^4*(3423*c + 3423*d*x - 2934))/24) +
tan(c/2 + (d*x)/2)^2*((1141*a^4*(c + d*x))/8 - (a^4*(3423*c + 3423*d*x - 7818))/24) + tan(c/2 + (d*x)/2)^8*((
2119*a^4*(c + d*x))/8 - (a^4*(6357*c + 6357*d*x - 6520))/24) - tan(c/2 + (d*x)/2)^3*((2119*a^4*(c + d*x))/8 -
(a^4*(6357*c + 6357*d*x - 13448))/24) - tan(c/2 + (d*x)/2)^7*((1467*a^4*(c + d*x))/4 - (a^4*(8802*c + 8802*d*x
- 11736))/24) + tan(c/2 + (d*x)/2)^4*((1467*a^4*(c + d*x))/4 - (a^4*(8802*c + 8802*d*x - 15912))/24) + tan(c/
2 + (d*x)/2)^6*((1793*a^4*(c + d*x))/4 - (a^4*(10758*c + 10758*d*x - 15364))/24) - tan(c/2 + (d*x)/2)^5*((1793
*a^4*(c + d*x))/4 - (a^4*(10758*c + 10758*d*x - 18428))/24))/(d*(tan(c/2 + (d*x)/2) - 1)^3*(tan(c/2 + (d*x)/2)
^2 + 1)^4)

________________________________________________________________________________________

sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))**4*tan(d*x+c)**4,x)

[Out]

Timed out

________________________________________________________________________________________