\(\int \frac {\sin ^7(c+d x)}{a-a \sin ^2(c+d x)} \, dx\) [35]

Optimal result

Integrand size = 24, antiderivative size = 62 \[ \int \frac {\sin ^7(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {3 \cos (c+d x)}{a d}-\frac {\cos ^3(c+d x)}{a d}+\frac {\cos ^5(c+d x)}{5 a d}+\frac {\sec (c+d x)}{a d} \]

[Out]

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

Rubi [A] (verified)

Time = 0.06 (sec) , antiderivative size = 62, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.125, Rules used = {3254, 2670, 276} \[ \int \frac {\sin ^7(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\cos ^5(c+d x)}{5 a d}-\frac {\cos ^3(c+d x)}{a d}+\frac {3 \cos (c+d x)}{a d}+\frac {\sec (c+d x)}{a d} \]

[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 276

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]

Rule 2670

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 3254

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\int \sin ^5(c+d x) \tan ^2(c+d x) \, dx}{a} \\ & = -\frac {\text {Subst}\left (\int \frac {\left (1-x^2\right )^3}{x^2} \, dx,x,\cos (c+d x)\right )}{a d} \\ & = -\frac {\text {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] (verified)

Time = 0.14 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.94 \[ \int \frac {\sin ^7(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\frac {19 \cos (c+d x)}{8 d}-\frac {3 \cos (3 (c+d x))}{16 d}+\frac {\cos (5 (c+d x))}{80 d}+\frac {\sec (c+d x)}{d}}{a} \]

[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

Maple [A] (verified)

Time = 0.72 (sec) , antiderivative size = 45, normalized size of antiderivative = 0.73

method	result	size
derivativedivides	\(\frac {\frac {\left (\cos ^{5}\left (d x +c \right )\right )}{5}-\left (\cos ^{3}\left (d x +c \right )\right )+3 \cos \left (d x +c \right )+\frac {1}{\cos \left (d x +c \right )}}{d a}\)	\(45\)
default	\(\frac {\frac {\left (\cos ^{5}\left (d x +c \right )\right )}{5}-\left (\cos ^{3}\left (d x +c \right )\right )+3 \cos \left (d x +c \right )+\frac {1}{\cos \left (d x +c \right )}}{d a}\)	\(45\)
parallelrisch	\(\frac {350+175 \cos \left (2 d x +2 c \right )-14 \cos \left (4 d x +4 c \right )+512 \cos \left (d x +c \right )+\cos \left (6 d x +6 c \right )}{160 a d \cos \left (d x +c \right )}\)	\(58\)
risch	\(\frac {19 \,{\mathrm e}^{i \left (d x +c \right )}}{16 a d}+\frac {19 \,{\mathrm e}^{-i \left (d x +c \right )}}{16 a d}+\frac {2 \,{\mathrm e}^{i \left (d x +c \right )}}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}+\frac {\cos \left (5 d x +5 c \right )}{80 a d}-\frac {3 \cos \left (3 d x +3 c \right )}{16 a d}\)	\(100\)
norman	\(\frac {-\frac {32 \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{d a}-\frac {32}{5 a d}-\frac {192 \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}-\frac {448 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}-\frac {448 \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 d a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{7} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(117\)

[In]

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

[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))

Fricas [A] (verification not implemented)

none

Time = 0.27 (sec) , antiderivative size = 46, normalized size of antiderivative = 0.74 \[ \int \frac {\sin ^7(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\cos \left (d x + c\right )^{6} - 5 \, \cos \left (d x + c\right )^{4} + 15 \, \cos \left (d x + c\right )^{2} + 5}{5 \, a d \cos \left (d x + c\right )} \]

[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))

Sympy [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 314 vs. \(2 (46) = 92\).

Time = 10.46 (sec) , antiderivative size = 314, normalized size of antiderivative = 5.06 \[ \int \frac {\sin ^7(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\begin {cases} - \frac {160 \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{5 a d \tan ^{12}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 20 a d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 25 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 20 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 5 a d} - \frac {128 \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )}}{5 a d \tan ^{12}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 20 a d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 25 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 20 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 5 a d} - \frac {32}{5 a d \tan ^{12}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 20 a d \tan ^{10}{\left (\frac {c}{2} + \frac {d x}{2} \right )} + 25 a d \tan ^{8}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 25 a d \tan ^{4}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 20 a d \tan ^{2}{\left (\frac {c}{2} + \frac {d x}{2} \right )} - 5 a d} & \text {for}\: d \neq 0 \\\frac {x \sin ^{7}{\left (c \right )}}{- a \sin ^{2}{\left (c \right )} + a} & \text {otherwise} \end {cases} \]

[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))

Maxima [A] (verification not implemented)

none

Time = 0.26 (sec) , antiderivative size = 50, normalized size of antiderivative = 0.81 \[ \int \frac {\sin ^7(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\frac {\cos \left (d x + c\right )^{5} - 5 \, \cos \left (d x + c\right )^{3} + 15 \, \cos \left (d x + c\right )}{a} + \frac {5}{a \cos \left (d x + c\right )}}{5 \, d} \]

[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

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 149 vs. \(2 (60) = 120\).

Time = 0.31 (sec) , antiderivative size = 149, normalized size of antiderivative = 2.40 \[ \int \frac {\sin ^7(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {2 \, {\left (\frac {5}{a {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} + 1\right )}} + \frac {\frac {50 \, {\left (\cos \left (d x + c\right ) - 1\right )}}{\cos \left (d x + c\right ) + 1} - \frac {80 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {30 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} - \frac {5 \, {\left (\cos \left (d x + c\right ) - 1\right )}^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - 11}{a {\left (\frac {\cos \left (d x + c\right ) - 1}{\cos \left (d x + c\right ) + 1} - 1\right )}^{5}}\right )}}{5 \, d} \]

[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

Mupad [B] (verification not implemented)

Time = 0.09 (sec) , antiderivative size = 54, normalized size of antiderivative = 0.87 \[ \int \frac {\sin ^7(c+d x)}{a-a \sin ^2(c+d x)} \, dx=\frac {\frac {3\,\cos \left (c+d\,x\right )}{a}+\frac {1}{a\,\cos \left (c+d\,x\right )}-\frac {{\cos \left (c+d\,x\right )}^3}{a}+\frac {{\cos \left (c+d\,x\right )}^5}{5\,a}}{d} \]

[In]

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

[Out]

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