3.821 \(\int \frac{\sin ^2(c+d x) \tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx\)

Optimal. Leaf size=117 \[ -\frac{\cos (c+d x)}{a d}+\frac{\tan ^5(c+d x)}{5 a d}-\frac{\tan ^3(c+d x)}{3 a d}+\frac{\tan (c+d x)}{a d}-\frac{\sec ^5(c+d x)}{5 a d}+\frac{\sec ^3(c+d x)}{a d}-\frac{3 \sec (c+d x)}{a d}-\frac{x}{a} \]

[Out]

-(x/a) - Cos[c + d*x]/(a*d) - (3*Sec[c + d*x])/(a*d) + Sec[c + d*x]^3/(a*d) - Sec[c + d*x]^5/(5*a*d) + Tan[c +
d*x]/(a*d) - Tan[c + d*x]^3/(3*a*d) + Tan[c + d*x]^5/(5*a*d)

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Rubi [A]  time = 0.158352, antiderivative size = 117, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {2839, 3473, 8, 2590, 270} \[ -\frac{\cos (c+d x)}{a d}+\frac{\tan ^5(c+d x)}{5 a d}-\frac{\tan ^3(c+d x)}{3 a d}+\frac{\tan (c+d x)}{a d}-\frac{\sec ^5(c+d x)}{5 a d}+\frac{\sec ^3(c+d x)}{a d}-\frac{3 \sec (c+d x)}{a d}-\frac{x}{a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(x/a) - Cos[c + d*x]/(a*d) - (3*Sec[c + d*x])/(a*d) + Sec[c + d*x]^3/(a*d) - Sec[c + d*x]^5/(5*a*d) + Tan[c +
d*x]/(a*d) - Tan[c + d*x]^3/(3*a*d) + Tan[c + d*x]^5/(5*a*d)

Rule 2839

Int[((cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.))/((a_) + (b_.)*sin[(e_.) + (f_
.)*(x_)]), x_Symbol] :> Dist[g^2/a, Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^n, x], x] - Dist[g^2/(b*d),
Int[(g*Cos[e + f*x])^(p - 2)*(d*Sin[e + f*x])^(n + 1), x], x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2
- b^2, 0]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

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

Rule 2590

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*tan[(e_.) + (f_.)*(x_)]^(n_.), x_Symbol] :> -Dist[f^(-1), Subst[Int[(1 - x^2
)^((m + n - 1)/2)/x^n, x], x, Cos[e + f*x]], x] /; FreeQ[{e, f}, x] && IntegersQ[m, n, (m + n - 1)/2]

Rule 270

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*(a + b*x^n)^p,
x], x] /; FreeQ[{a, b, c, m, n}, x] && IGtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{\sin ^2(c+d x) \tan ^4(c+d x)}{a+a \sin (c+d x)} \, dx &=\frac{\int \tan ^6(c+d x) \, dx}{a}-\frac{\int \sin (c+d x) \tan ^6(c+d x) \, dx}{a}\\ &=\frac{\tan ^5(c+d x)}{5 a d}-\frac{\int \tan ^4(c+d x) \, dx}{a}+\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^3}{x^6} \, dx,x,\cos (c+d x)\right )}{a d}\\ &=-\frac{\tan ^3(c+d x)}{3 a d}+\frac{\tan ^5(c+d x)}{5 a d}+\frac{\int \tan ^2(c+d x) \, dx}{a}+\frac{\operatorname{Subst}\left (\int \left (-1+\frac{1}{x^6}-\frac{3}{x^4}+\frac{3}{x^2}\right ) \, dx,x,\cos (c+d x)\right )}{a d}\\ &=-\frac{\cos (c+d x)}{a d}-\frac{3 \sec (c+d x)}{a d}+\frac{\sec ^3(c+d x)}{a d}-\frac{\sec ^5(c+d x)}{5 a d}+\frac{\tan (c+d x)}{a d}-\frac{\tan ^3(c+d x)}{3 a d}+\frac{\tan ^5(c+d x)}{5 a d}-\frac{\int 1 \, dx}{a}\\ &=-\frac{x}{a}-\frac{\cos (c+d x)}{a d}-\frac{3 \sec (c+d x)}{a d}+\frac{\sec ^3(c+d x)}{a d}-\frac{\sec ^5(c+d x)}{5 a d}+\frac{\tan (c+d x)}{a d}-\frac{\tan ^3(c+d x)}{3 a d}+\frac{\tan ^5(c+d x)}{5 a d}\\ \end{align*}

Mathematica [A]  time = 0.674661, size = 224, normalized size = 1.91 \[ -\frac{216 \sin (c+d x)+240 c \sin (2 (c+d x))+240 d x \sin (2 (c+d x))-618 \sin (2 (c+d x))+532 \sin (3 (c+d x))+120 c \sin (4 (c+d x))+120 d x \sin (4 (c+d x))-309 \sin (4 (c+d x))+60 \sin (5 (c+d x))+18 (40 c+40 d x-103) \cos (c+d x)+1568 \cos (2 (c+d x))+240 c \cos (3 (c+d x))+240 d x \cos (3 (c+d x))-618 \cos (3 (c+d x))+304 \cos (4 (c+d x))+1200}{960 a d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(1200 + 18*(-103 + 40*c + 40*d*x)*Cos[c + d*x] + 1568*Cos[2*(c + d*x)] - 618*Cos[3*(c + d*x)] + 240*c*Cos[3*(
c + d*x)] + 240*d*x*Cos[3*(c + d*x)] + 304*Cos[4*(c + d*x)] + 216*Sin[c + d*x] - 618*Sin[2*(c + d*x)] + 240*c*
Sin[2*(c + d*x)] + 240*d*x*Sin[2*(c + d*x)] + 532*Sin[3*(c + d*x)] - 309*Sin[4*(c + d*x)] + 120*c*Sin[4*(c + d
*x)] + 120*d*x*Sin[4*(c + d*x)] + 60*Sin[5*(c + d*x)])/(960*a*d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3*(Cos[(
c + d*x)/2] + Sin[(c + d*x)/2])^5)

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Maple [A]  time = 0.096, size = 210, normalized size = 1.8 \begin{align*} -{\frac{1}{6\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-3}}-{\frac{1}{4\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-2}}+{\frac{7}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) -1 \right ) ^{-1}}-2\,{\frac{1}{da \left ( 1+ \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) ^{2} \right ) }}-2\,{\frac{\arctan \left ( \tan \left ( 1/2\,dx+c/2 \right ) \right ) }{da}}-{\frac{2}{5\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-5}}+{\frac{1}{da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-4}}+{\frac{1}{3\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-3}}-{\frac{3}{2\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-2}}-{\frac{23}{8\,da} \left ( \tan \left ({\frac{dx}{2}}+{\frac{c}{2}} \right ) +1 \right ) ^{-1}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*sin(d*x+c)^6/(a+a*sin(d*x+c)),x)

[Out]

-1/6/d/a/(tan(1/2*d*x+1/2*c)-1)^3-1/4/d/a/(tan(1/2*d*x+1/2*c)-1)^2+7/8/d/a/(tan(1/2*d*x+1/2*c)-1)-2/a/d/(1+tan
(1/2*d*x+1/2*c)^2)-2/a/d*arctan(tan(1/2*d*x+1/2*c))-2/5/d/a/(tan(1/2*d*x+1/2*c)+1)^5+1/d/a/(tan(1/2*d*x+1/2*c)
+1)^4+1/3/d/a/(tan(1/2*d*x+1/2*c)+1)^3-3/2/d/a/(tan(1/2*d*x+1/2*c)+1)^2-23/8/a/d/(tan(1/2*d*x+1/2*c)+1)

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Maxima [B]  time = 1.58587, size = 540, normalized size = 4.62 \begin{align*} -\frac{2 \,{\left (\frac{\frac{81 \, \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{78 \, \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{172 \, \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{26 \, \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{22 \, \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{70 \, \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{20 \, \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac{30 \, \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac{15 \, \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} + 48}{a + \frac{2 \, a \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{a \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{4 \, a \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{2 \, a \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{2 \, a \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{4 \, a \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} + \frac{a \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}} - \frac{2 \, a \sin \left (d x + c\right )^{9}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{9}} - \frac{a \sin \left (d x + c\right )^{10}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{10}}} + \frac{15 \, \arctan \left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1}\right )}{a}\right )}}{15 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

-2/15*((81*sin(d*x + c)/(cos(d*x + c) + 1) - 78*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 - 172*sin(d*x + c)^3/(cos(
d*x + c) + 1)^3 - 26*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 22*sin(d*x + c)^5/(cos(d*x + c) + 1)^5 + 70*sin(d*x
+ c)^6/(cos(d*x + c) + 1)^6 + 20*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 30*sin(d*x + c)^8/(cos(d*x + c) + 1)^8
- 15*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 + 48)/(a + 2*a*sin(d*x + c)/(cos(d*x + c) + 1) - a*sin(d*x + c)^2/(c
os(d*x + c) + 1)^2 - 4*a*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 - 2*a*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 + 2*a*s
in(d*x + c)^6/(cos(d*x + c) + 1)^6 + 4*a*sin(d*x + c)^7/(cos(d*x + c) + 1)^7 + a*sin(d*x + c)^8/(cos(d*x + c)
+ 1)^8 - 2*a*sin(d*x + c)^9/(cos(d*x + c) + 1)^9 - a*sin(d*x + c)^10/(cos(d*x + c) + 1)^10) + 15*arctan(sin(d*
x + c)/(cos(d*x + c) + 1))/a)/d

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Fricas [A]  time = 1.5073, size = 288, normalized size = 2.46 \begin{align*} -\frac{15 \, d x \cos \left (d x + c\right )^{3} + 38 \, \cos \left (d x + c\right )^{4} + 11 \, \cos \left (d x + c\right )^{2} +{\left (15 \, d x \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )^{4} + 22 \, \cos \left (d x + c\right )^{2} - 4\right )} \sin \left (d x + c\right ) - 1}{15 \,{\left (a d \cos \left (d x + c\right )^{3} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{3}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/15*(15*d*x*cos(d*x + c)^3 + 38*cos(d*x + c)^4 + 11*cos(d*x + c)^2 + (15*d*x*cos(d*x + c)^3 + 15*cos(d*x + c
)^4 + 22*cos(d*x + c)^2 - 4)*sin(d*x + c) - 1)/(a*d*cos(d*x + c)^3*sin(d*x + c) + a*d*cos(d*x + c)^3)

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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(sec(d*x+c)**4*sin(d*x+c)**6/(a+a*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.29895, size = 201, normalized size = 1.72 \begin{align*} -\frac{\frac{120 \,{\left (d x + c\right )}}{a} + \frac{240}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )} a} - \frac{5 \,{\left (21 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 48 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 23\right )}}{a{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}} + \frac{345 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 1560 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 2570 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1720 \, \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 413}{a{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 1\right )}^{5}}}{120 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^6/(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/120*(120*(d*x + c)/a + 240/((tan(1/2*d*x + 1/2*c)^2 + 1)*a) - 5*(21*tan(1/2*d*x + 1/2*c)^2 - 48*tan(1/2*d*x
+ 1/2*c) + 23)/(a*(tan(1/2*d*x + 1/2*c) - 1)^3) + (345*tan(1/2*d*x + 1/2*c)^4 + 1560*tan(1/2*d*x + 1/2*c)^3 +
2570*tan(1/2*d*x + 1/2*c)^2 + 1720*tan(1/2*d*x + 1/2*c) + 413)/(a*(tan(1/2*d*x + 1/2*c) + 1)^5))/d