3.39 \(\int (a+a \sin (c+d x))^4 \tan ^2(c+d x) \, dx\)

Optimal. Leaf size=113 \[ -\frac{4 a^4 \cos ^3(c+d x)}{3 d}+\frac{12 a^4 \cos (c+d x)}{d}+\frac{a^4 \sin ^3(c+d x) \cos (c+d x)}{4 d}+\frac{31 a^4 \sin (c+d x) \cos (c+d x)}{8 d}+\frac{8 a^4 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac{95 a^4 x}{8} \]

[Out]

(-95*a^4*x)/8 + (12*a^4*Cos[c + d*x])/d - (4*a^4*Cos[c + d*x]^3)/(3*d) + (8*a^4*Cos[c + d*x])/(d*(1 - Sin[c +
d*x])) + (31*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 = 0.160916, antiderivative size = 113, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 6, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.286, Rules used = {2709, 2648, 2638, 2635, 8, 2633} \[ -\frac{4 a^4 \cos ^3(c+d x)}{3 d}+\frac{12 a^4 \cos (c+d x)}{d}+\frac{a^4 \sin ^3(c+d x) \cos (c+d x)}{4 d}+\frac{31 a^4 \sin (c+d x) \cos (c+d x)}{8 d}+\frac{8 a^4 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac{95 a^4 x}{8} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^4*Tan[c + d*x]^2,x]

[Out]

(-95*a^4*x)/8 + (12*a^4*Cos[c + d*x])/d - (4*a^4*Cos[c + d*x]^3)/(3*d) + (8*a^4*Cos[c + d*x])/(d*(1 - Sin[c +
d*x])) + (31*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 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 ^2(c+d x) \, dx &=a^2 \int \left (-8 a^2-\frac{8 a^2}{-1+\sin (c+d x)}-8 a^2 \sin (c+d x)-7 a^2 \sin ^2(c+d x)-4 a^2 \sin ^3(c+d x)-a^2 \sin ^4(c+d x)\right ) \, dx\\ &=-8 a^4 x-a^4 \int \sin ^4(c+d x) \, dx-\left (4 a^4\right ) \int \sin ^3(c+d x) \, dx-\left (7 a^4\right ) \int \sin ^2(c+d x) \, dx-\left (8 a^4\right ) \int \frac{1}{-1+\sin (c+d x)} \, dx-\left (8 a^4\right ) \int \sin (c+d x) \, dx\\ &=-8 a^4 x+\frac{8 a^4 \cos (c+d x)}{d}+\frac{8 a^4 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac{7 a^4 \cos (c+d x) \sin (c+d x)}{2 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}{2} \left (7 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}\\ &=-\frac{23 a^4 x}{2}+\frac{12 a^4 \cos (c+d x)}{d}-\frac{4 a^4 \cos ^3(c+d x)}{3 d}+\frac{8 a^4 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac{31 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{95 a^4 x}{8}+\frac{12 a^4 \cos (c+d x)}{d}-\frac{4 a^4 \cos ^3(c+d x)}{3 d}+\frac{8 a^4 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac{31 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 = 1.12581, size = 125, normalized size = 1.11 \[ \frac{(a \sin (c+d x)+a)^4 \left (-1140 (c+d x)+192 \sin (2 (c+d x))-3 \sin (4 (c+d x))+1056 \cos (c+d x)-32 \cos (3 (c+d x))+\frac{1536 \sin \left (\frac{1}{2} (c+d x)\right )}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}\right )}{96 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^8} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^4*Tan[c + d*x]^2,x]

[Out]

((a + a*Sin[c + d*x])^4*(-1140*(c + d*x) + 1056*Cos[c + d*x] - 32*Cos[3*(c + d*x)] + (1536*Sin[(c + d*x)/2])/(
Cos[(c + d*x)/2] - Sin[(c + d*x)/2]) + 192*Sin[2*(c + d*x)] - 3*Sin[4*(c + d*x)]))/(96*d*(Cos[(c + d*x)/2] + S
in[(c + d*x)/2])^8)

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Maple [B]  time = 0.06, size = 231, normalized size = 2. \begin{align*}{\frac{1}{d} \left ({a}^{4} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{7}}{\cos \left ( dx+c \right ) }}+ \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{5}+{\frac{5\, \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4}}+{\frac{15\,\sin \left ( dx+c \right ) }{8}} \right ) \cos \left ( dx+c \right ) -{\frac{15\,dx}{8}}-{\frac{15\,c}{8}} \right ) +4\,{a}^{4} \left ({\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 ) +6\,{a}^{4} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{\cos \left ( dx+c \right ) }}+ \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{3}+3/2\,\sin \left ( dx+c \right ) \right ) \cos \left ( dx+c \right ) -3/2\,dx-3/2\,c \right ) +4\,{a}^{4} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{\cos \left ( dx+c \right ) }}+ \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) \right ) +{a}^{4} \left ( \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)^2,x)

[Out]

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

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Maxima [A]  time = 1.58886, size = 244, normalized size = 2.16 \begin{align*} -\frac{32 \,{\left (\cos \left (d x + c\right )^{3} - \frac{3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} a^{4} + 3 \,{\left (15 \, d x + 15 \, c - \frac{9 \, \tan \left (d x + c\right )^{3} + 7 \, \tan \left (d x + c\right )}{\tan \left (d x + c\right )^{4} + 2 \, \tan \left (d x + c\right )^{2} + 1} - 8 \, \tan \left (d x + c\right )\right )} a^{4} + 72 \,{\left (3 \, d x + 3 \, c - \frac{\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a^{4} + 24 \,{\left (d x + c - \tan \left (d x + c\right )\right )} a^{4} - 96 \, a^{4}{\left (\frac{1}{\cos \left (d x + c\right )} + \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)^2,x, algorithm="maxima")

[Out]

-1/24*(32*(cos(d*x + c)^3 - 3/cos(d*x + c) - 6*cos(d*x + c))*a^4 + 3*(15*d*x + 15*c - (9*tan(d*x + c)^3 + 7*ta
n(d*x + c))/(tan(d*x + c)^4 + 2*tan(d*x + c)^2 + 1) - 8*tan(d*x + c))*a^4 + 72*(3*d*x + 3*c - tan(d*x + c)/(ta
n(d*x + c)^2 + 1) - 2*tan(d*x + c))*a^4 + 24*(d*x + c - tan(d*x + c))*a^4 - 96*a^4*(1/cos(d*x + c) + cos(d*x +
c)))/d

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Fricas [A]  time = 1.5892, size = 456, normalized size = 4.04 \begin{align*} -\frac{6 \, a^{4} \cos \left (d x + c\right )^{5} + 32 \, a^{4} \cos \left (d x + c\right )^{4} - 73 \, a^{4} \cos \left (d x + c\right )^{3} + 285 \, a^{4} d x - 288 \, a^{4} \cos \left (d x + c\right )^{2} - 192 \, a^{4} + 3 \,{\left (95 \, a^{4} d x - 127 \, a^{4}\right )} \cos \left (d x + c\right ) +{\left (6 \, a^{4} \cos \left (d x + c\right )^{4} - 26 \, a^{4} \cos \left (d x + c\right )^{3} - 285 \, a^{4} d x - 99 \, a^{4} \cos \left (d x + c\right )^{2} + 189 \, a^{4} \cos \left (d x + c\right ) - 192 \, a^{4}\right )} \sin \left (d x + c\right )}{24 \,{\left (d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + 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)^2,x, algorithm="fricas")

[Out]

-1/24*(6*a^4*cos(d*x + c)^5 + 32*a^4*cos(d*x + c)^4 - 73*a^4*cos(d*x + c)^3 + 285*a^4*d*x - 288*a^4*cos(d*x +
c)^2 - 192*a^4 + 3*(95*a^4*d*x - 127*a^4)*cos(d*x + c) + (6*a^4*cos(d*x + c)^4 - 26*a^4*cos(d*x + c)^3 - 285*a
^4*d*x - 99*a^4*cos(d*x + c)^2 + 189*a^4*cos(d*x + c) - 192*a^4)*sin(d*x + c))/(d*cos(d*x + c) - d*sin(d*x + c
) + 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)**2,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)^2,x, algorithm="giac")

[Out]

Timed out