3.30 \(\int (a+a \sin (c+d x))^3 \tan ^4(c+d x) \, dx\)

Optimal. Leaf size=119 \[ \frac{a^3 \cos ^3(c+d x)}{3 d}-\frac{6 a^3 \cos (c+d x)}{d}-\frac{3 a^3 \sin (c+d x) \cos (c+d x)}{2 d}-\frac{25 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac{2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}+\frac{17 a^3 x}{2} \]

[Out]

(17*a^3*x)/2 - (6*a^3*Cos[c + d*x])/d + (a^3*Cos[c + d*x]^3)/(3*d) + (2*a^3*Cos[c + d*x])/(3*d*(1 - Sin[c + d*
x])^2) - (25*a^3*Cos[c + d*x])/(3*d*(1 - Sin[c + d*x])) - (3*a^3*Cos[c + d*x]*Sin[c + d*x])/(2*d)

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Rubi [A]  time = 0.194183, antiderivative size = 119, normalized size of antiderivative = 1., number of steps used = 10, 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} \[ \frac{a^3 \cos ^3(c+d x)}{3 d}-\frac{6 a^3 \cos (c+d x)}{d}-\frac{3 a^3 \sin (c+d x) \cos (c+d x)}{2 d}-\frac{25 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))}+\frac{2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}+\frac{17 a^3 x}{2} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^3*Tan[c + d*x]^4,x]

[Out]

(17*a^3*x)/2 - (6*a^3*Cos[c + d*x])/d + (a^3*Cos[c + d*x]^3)/(3*d) + (2*a^3*Cos[c + d*x])/(3*d*(1 - Sin[c + d*
x])^2) - (25*a^3*Cos[c + d*x])/(3*d*(1 - Sin[c + d*x])) - (3*a^3*Cos[c + d*x]*Sin[c + d*x])/(2*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))^3 \tan ^4(c+d x) \, dx &=a^4 \int \left (\frac{7}{a}+\frac{2}{a (-1+\sin (c+d x))^2}+\frac{9}{a (-1+\sin (c+d x))}+\frac{5 \sin (c+d x)}{a}+\frac{3 \sin ^2(c+d x)}{a}+\frac{\sin ^3(c+d x)}{a}\right ) \, dx\\ &=7 a^3 x+a^3 \int \sin ^3(c+d x) \, dx+\left (2 a^3\right ) \int \frac{1}{(-1+\sin (c+d x))^2} \, dx+\left (3 a^3\right ) \int \sin ^2(c+d x) \, dx+\left (5 a^3\right ) \int \sin (c+d x) \, dx+\left (9 a^3\right ) \int \frac{1}{-1+\sin (c+d x)} \, dx\\ &=7 a^3 x-\frac{5 a^3 \cos (c+d x)}{d}+\frac{2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}-\frac{9 a^3 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac{3 a^3 \cos (c+d x) \sin (c+d x)}{2 d}-\frac{1}{3} \left (2 a^3\right ) \int \frac{1}{-1+\sin (c+d x)} \, dx+\frac{1}{2} \left (3 a^3\right ) \int 1 \, dx-\frac{a^3 \operatorname{Subst}\left (\int \left (1-x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac{17 a^3 x}{2}-\frac{6 a^3 \cos (c+d x)}{d}+\frac{a^3 \cos ^3(c+d x)}{3 d}+\frac{2 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))^2}-\frac{25 a^3 \cos (c+d x)}{3 d (1-\sin (c+d x))}-\frac{3 a^3 \cos (c+d x) \sin (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 2.25007, size = 177, normalized size = 1.49 \[ \frac{(a \sin (c+d x)+a)^3 \left (102 (c+d x)-9 \sin (2 (c+d x))-69 \cos (c+d x)+\cos (3 (c+d x))-\frac{200 \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{16 \sin \left (\frac{1}{2} (c+d x)\right )}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{8}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}\right )}{12 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^3*Tan[c + d*x]^4,x]

[Out]

((a + a*Sin[c + d*x])^3*(102*(c + d*x) - 69*Cos[c + d*x] + Cos[3*(c + d*x)] + 8/(Cos[(c + d*x)/2] - Sin[(c + d
*x)/2])^2 + (16*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3 - (200*Sin[(c + d*x)/2])/(Cos[(c + d
*x)/2] - Sin[(c + d*x)/2]) - 9*Sin[2*(c + d*x)]))/(12*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)

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Maple [B]  time = 0.085, size = 266, normalized size = 2.2 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{5\, \left ( \sin \left ( dx+c \right ) \right ) ^{8}}{3\,\cos \left ( dx+c \right ) }}-{\frac{5\,\cos \left ( dx+c \right ) }{3} \left ({\frac{16}{5}}+ \left ( \sin \left ( dx+c \right ) \right ) ^{6}+{\frac{6\, \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{5}}+{\frac{8\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{5}} \right ) } \right ) +3\,{a}^{3} \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 ) +3\,{a}^{3} \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}^{3} \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))^3*tan(d*x+c)^4,x)

[Out]

1/d*(a^3*(1/3*sin(d*x+c)^8/cos(d*x+c)^3-5/3*sin(d*x+c)^8/cos(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))+3*a^3*(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)+3*a^3*(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^3*(1/3*tan(d*x+c)^3-tan(d*x+c)+d*x+c))

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Maxima [A]  time = 1.61188, size = 223, normalized size = 1.87 \begin{align*} \frac{2 \,{\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^{3} + 3 \,{\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^{3} + 2 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{3} - 6 \, a^{3}{\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 )}}{6 \, 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)^4,x, algorithm="maxima")

[Out]

1/6*(2*(cos(d*x + c)^3 - (9*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 - 9*cos(d*x + c))*a^3 + 3*(2*tan(d*x + c)^3 + 1
5*d*x + 15*c - 3*tan(d*x + c)/(tan(d*x + c)^2 + 1) - 12*tan(d*x + c))*a^3 + 2*(tan(d*x + c)^3 + 3*d*x + 3*c -
3*tan(d*x + c))*a^3 - 6*a^3*((6*cos(d*x + c)^2 - 1)/cos(d*x + c)^3 + 3*cos(d*x + c)))/d

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Fricas [B]  time = 1.5826, size = 539, normalized size = 4.53 \begin{align*} \frac{2 \, a^{3} \cos \left (d x + c\right )^{5} + 7 \, a^{3} \cos \left (d x + c\right )^{4} - 22 \, a^{3} \cos \left (d x + c\right )^{3} - 102 \, a^{3} d x - 4 \, a^{3} +{\left (51 \, a^{3} d x + 77 \, a^{3}\right )} \cos \left (d x + c\right )^{2} -{\left (51 \, a^{3} d x - 100 \, a^{3}\right )} \cos \left (d x + c\right ) +{\left (2 \, a^{3} \cos \left (d x + c\right )^{4} - 5 \, a^{3} \cos \left (d x + c\right )^{3} + 102 \, a^{3} d x - 27 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3} +{\left (51 \, a^{3} d x - 104 \, a^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{6 \,{\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))^3*tan(d*x+c)^4,x, algorithm="fricas")

[Out]

1/6*(2*a^3*cos(d*x + c)^5 + 7*a^3*cos(d*x + c)^4 - 22*a^3*cos(d*x + c)^3 - 102*a^3*d*x - 4*a^3 + (51*a^3*d*x +
77*a^3)*cos(d*x + c)^2 - (51*a^3*d*x - 100*a^3)*cos(d*x + c) + (2*a^3*cos(d*x + c)^4 - 5*a^3*cos(d*x + c)^3 +
102*a^3*d*x - 27*a^3*cos(d*x + c)^2 - 4*a^3 + (51*a^3*d*x - 104*a^3)*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)*sin(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))**3*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))^3*tan(d*x+c)^4,x, algorithm="giac")

[Out]

Timed out