3.35 \(\int \frac{\sin ^7(c+d x)}{a-a \sin ^2(c+d x)} \, dx\)

Optimal. Leaf size=62 \[ \frac{\cos ^5(c+d x)}{5 a d}-\frac{\cos ^3(c+d x)}{a d}+\frac{3 \cos (c+d x)}{a d}+\frac{\sec (c+d x)}{a d} \]

[Out]

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

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Rubi [A]  time = 0.0870735, antiderivative size = 62, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.125, Rules used = {3175, 2590, 270} \[ \frac{\cos ^5(c+d x)}{5 a d}-\frac{\cos ^3(c+d x)}{a d}+\frac{3 \cos (c+d x)}{a d}+\frac{\sec (c+d x)}{a d} \]

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]^7/(a - a*Sin[c + d*x]^2),x]

[Out]

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

Rule 3175

Int[(u_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> Dist[a^p, Int[ActivateTrig[u*cos[e + f*x
]^(2*p)], x], x] /; FreeQ[{a, b, e, f, p}, x] && EqQ[a + b, 0] && IntegerQ[p]

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 ^7(c+d x)}{a-a \sin ^2(c+d x)} \, dx &=\frac{\int \sin ^5(c+d x) \tan ^2(c+d x) \, dx}{a}\\ &=-\frac{\operatorname{Subst}\left (\int \frac{\left (1-x^2\right )^3}{x^2} \, dx,x,\cos (c+d x)\right )}{a d}\\ &=-\frac{\operatorname{Subst}\left (\int \left (-3+\frac{1}{x^2}+3 x^2-x^4\right ) \, dx,x,\cos (c+d x)\right )}{a d}\\ &=\frac{3 \cos (c+d x)}{a d}-\frac{\cos ^3(c+d x)}{a d}+\frac{\cos ^5(c+d x)}{5 a d}+\frac{\sec (c+d x)}{a d}\\ \end{align*}

Mathematica [A]  time = 0.0636532, size = 58, normalized size = 0.94 \[ \frac{\frac{19 \cos (c+d x)}{8 d}-\frac{3 \cos (3 (c+d x))}{16 d}+\frac{\cos (5 (c+d x))}{80 d}+\frac{\sec (c+d x)}{d}}{a} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]^7/(a - a*Sin[c + d*x]^2),x]

[Out]

((19*Cos[c + d*x])/(8*d) - (3*Cos[3*(c + d*x)])/(16*d) + Cos[5*(c + d*x)]/(80*d) + Sec[c + d*x]/d)/a

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Maple [A]  time = 0.042, size = 45, normalized size = 0.7 \begin{align*}{\frac{1}{da} \left ({\frac{ \left ( \cos \left ( dx+c \right ) \right ) ^{5}}{5}}- \left ( \cos \left ( dx+c \right ) \right ) ^{3}+3\,\cos \left ( dx+c \right ) + \left ( \cos \left ( dx+c \right ) \right ) ^{-1} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^7/(a-sin(d*x+c)^2*a),x)

[Out]

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

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Maxima [A]  time = 0.966741, size = 68, normalized size = 1.1 \begin{align*} \frac{\frac{\cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )}{a} + \frac{5}{a \cos \left (d x + c\right )}}{5 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^7/(a-a*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

1/5*((cos(d*x + c)^5 - 5*cos(d*x + c)^3 + 15*cos(d*x + c))/a + 5/(a*cos(d*x + c)))/d

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Fricas [A]  time = 1.60437, size = 113, normalized size = 1.82 \begin{align*} \frac{\cos \left (d x + c\right )^{6} - 5 \, \cos \left (d x + c\right )^{4} + 15 \, \cos \left (d x + c\right )^{2} + 5}{5 \, a d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^7/(a-a*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

1/5*(cos(d*x + c)^6 - 5*cos(d*x + c)^4 + 15*cos(d*x + c)^2 + 5)/(a*d*cos(d*x + c))

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Sympy [A]  time = 125.581, size = 314, normalized size = 5.06 \begin{align*} \begin{cases} - \frac{160 \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{5 a d \tan ^{12}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 20 a d \tan ^{10}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 25 a d \tan ^{8}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 25 a d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 20 a d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 5 a d} - \frac{128 \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{5 a d \tan ^{12}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 20 a d \tan ^{10}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 25 a d \tan ^{8}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 25 a d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 20 a d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 5 a d} - \frac{32}{5 a d \tan ^{12}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 20 a d \tan ^{10}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 25 a d \tan ^{8}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 25 a d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 20 a d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 5 a d} & \text{for}\: d \neq 0 \\\frac{x \sin ^{7}{\left (c \right )}}{- a \sin ^{2}{\left (c \right )} + a} & \text{otherwise} \end{cases} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)**7/(a-a*sin(d*x+c)**2),x)

[Out]

Piecewise((-160*tan(c/2 + d*x/2)**4/(5*a*d*tan(c/2 + d*x/2)**12 + 20*a*d*tan(c/2 + d*x/2)**10 + 25*a*d*tan(c/2
+ d*x/2)**8 - 25*a*d*tan(c/2 + d*x/2)**4 - 20*a*d*tan(c/2 + d*x/2)**2 - 5*a*d) - 128*tan(c/2 + d*x/2)**2/(5*a
*d*tan(c/2 + d*x/2)**12 + 20*a*d*tan(c/2 + d*x/2)**10 + 25*a*d*tan(c/2 + d*x/2)**8 - 25*a*d*tan(c/2 + d*x/2)**
4 - 20*a*d*tan(c/2 + d*x/2)**2 - 5*a*d) - 32/(5*a*d*tan(c/2 + d*x/2)**12 + 20*a*d*tan(c/2 + d*x/2)**10 + 25*a*
d*tan(c/2 + d*x/2)**8 - 25*a*d*tan(c/2 + d*x/2)**4 - 20*a*d*tan(c/2 + d*x/2)**2 - 5*a*d), Ne(d, 0)), (x*sin(c)
**7/(-a*sin(c)**2 + a), True))

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Giac [B]  time = 1.16396, size = 201, normalized size = 3.24 \begin{align*} \frac{2 \,{\left (\frac{5}{a{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}} + \frac{\frac{50 \,{\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac{80 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{30 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac{5 \,{\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 11}{a{\left (\frac{\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{5}}\right )}}{5 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^7/(a-a*sin(d*x+c)^2),x, algorithm="giac")

[Out]

2/5*(5/(a*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) + 1)) + (50*(cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 80*(cos(d
*x + c) - 1)^2/(cos(d*x + c) + 1)^2 + 30*(cos(d*x + c) - 1)^3/(cos(d*x + c) + 1)^3 - 5*(cos(d*x + c) - 1)^4/(c
os(d*x + c) + 1)^4 - 11)/(a*((cos(d*x + c) - 1)/(cos(d*x + c) + 1) - 1)^5))/d