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*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)

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Rubi [A]  time = 0.198685, antiderivative size = 143, normalized size of antiderivative = 1., 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 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])

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 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 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, 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))^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 = 1.65143, 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)

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Maple [B]  time = 0.092, size = 360, normalized size = 2.5 \begin{align*}{\frac{1}{d} \left ({a}^{4} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{9}}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-2\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{9}}{\cos \left ( dx+c \right ) }}-2\, \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{7}+7/6\, \left ( \sin \left ( dx+c \right ) \right ) ^{5}+{\frac{35\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{24}}+{\frac{35\,\sin \left ( dx+c \right ) }{16}} \right ) \cos \left ( dx+c \right ) +{\frac{35\,dx}{8}}+{\frac{35\,c}{8}} \right ) +4\,{a}^{4} \left ( 1/3\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-5/3\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{\cos \left ( dx+c \right ) }}-5/3\, \left ({\frac{16}{5}}+ \left ( \sin \left ( dx+c \right ) \right ) ^{6}+6/5\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}+8/5\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) \right ) +6\,{a}^{4} \left ( 1/3\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-4/3\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{\cos \left ( dx+c \right ) }}-4/3\, \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{5}+5/4\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}+{\frac{15\,\sin \left ( dx+c \right ) }{8}} \right ) \cos \left ( dx+c \right ) +5/2\,dx+5/2\,c \right ) +4\,{a}^{4} \left ( 1/3\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{\cos \left ( dx+c \right ) }}- \left ( 8/3+ \left ( \sin \left ( dx+c \right ) \right ) ^{4}+4/3\, \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) \right ) +{a}^{4} \left ({\frac{ \left ( \tan \left ( dx+c \right ) \right ) ^{3}}{3}}-\tan \left ( dx+c \right ) +dx+c \right ) \right ) } \end{align*}

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))

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Maxima [A]  time = 1.67263, size = 321, normalized size = 2.24 \begin{align*} \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} \end{align*}

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

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Fricas [A]  time = 1.56258, size = 620, normalized size = 4.34 \begin{align*} -\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 )}} \end{align*}

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)

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

[Out]

Timed out