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

Optimal. Leaf size=49 \[ \frac{3 \tan (c+d x)}{2 a d}-\frac{\sin ^2(c+d x) \tan (c+d x)}{2 a d}-\frac{3 x}{2 a} \]

[Out]

(-3*x)/(2*a) + (3*Tan[c + d*x])/(2*a*d) - (Sin[c + d*x]^2*Tan[c + d*x])/(2*a*d)

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Rubi [A]  time = 0.0815679, antiderivative size = 49, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 24, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.208, Rules used = {3175, 2591, 288, 321, 203} \[ \frac{3 \tan (c+d x)}{2 a d}-\frac{\sin ^2(c+d x) \tan (c+d x)}{2 a d}-\frac{3 x}{2 a} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*x)/(2*a) + (3*Tan[c + d*x])/(2*a*d) - (Sin[c + d*x]^2*Tan[c + d*x])/(2*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 2591

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> With[{ff = FreeFactors[Ta
n[e + f*x], x]}, Dist[(b*ff)/f, Subst[Int[(ff*x)^(m + n)/(b^2 + ff^2*x^2)^(m/2 + 1), x], x, (b*Tan[e + f*x])/f
f], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/2]

Rule 288

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^
n)^(p + 1))/(b*n*(p + 1)), x] - Dist[(c^n*(m - n + 1))/(b*n*(p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^(p + 1), x
], x] /; FreeQ[{a, b, c}, x] && IGtQ[n, 0] && LtQ[p, -1] && GtQ[m + 1, n] &&  !ILtQ[(m + n*(p + 1) + 1)/n, 0]
&& IntBinomialQ[a, b, c, n, m, p, x]

Rule 321

Int[((c_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c^(n - 1)*(c*x)^(m - n + 1)*(a + b*x^n
)^(p + 1))/(b*(m + n*p + 1)), x] - Dist[(a*c^n*(m - n + 1))/(b*(m + n*p + 1)), Int[(c*x)^(m - n)*(a + b*x^n)^p
, x], x] /; FreeQ[{a, b, c, p}, x] && IGtQ[n, 0] && GtQ[m, n - 1] && NeQ[m + n*p + 1, 0] && IntBinomialQ[a, b,
c, n, m, p, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.126989, size = 34, normalized size = 0.69 \[ \frac{-6 (c+d x)+\sin (2 (c+d x))+4 \tan (c+d x)}{4 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*(c + d*x) + Sin[2*(c + d*x)] + 4*Tan[c + d*x])/(4*a*d)

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Maple [A]  time = 0.041, size = 56, normalized size = 1.1 \begin{align*}{\frac{\tan \left ( dx+c \right ) }{da}}+{\frac{\tan \left ( dx+c \right ) }{2\,da \left ( \left ( \tan \left ( dx+c \right ) \right ) ^{2}+1 \right ) }}-{\frac{3\,\arctan \left ( \tan \left ( dx+c \right ) \right ) }{2\,da}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)^4/(a-sin(d*x+c)^2*a),x)

[Out]

tan(d*x+c)/d/a+1/2/d/a*tan(d*x+c)/(tan(d*x+c)^2+1)-3/2/d/a*arctan(tan(d*x+c))

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Maxima [A]  time = 1.43283, size = 66, normalized size = 1.35 \begin{align*} -\frac{\frac{3 \,{\left (d x + c\right )}}{a} - \frac{\tan \left (d x + c\right )}{a \tan \left (d x + c\right )^{2} + a} - \frac{2 \, \tan \left (d x + c\right )}{a}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4/(a-a*sin(d*x+c)^2),x, algorithm="maxima")

[Out]

-1/2*(3*(d*x + c)/a - tan(d*x + c)/(a*tan(d*x + c)^2 + a) - 2*tan(d*x + c)/a)/d

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Fricas [A]  time = 1.59544, size = 111, normalized size = 2.27 \begin{align*} -\frac{3 \, d x \cos \left (d x + c\right ) -{\left (\cos \left (d x + c\right )^{2} + 2\right )} \sin \left (d x + c\right )}{2 \, 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)^4/(a-a*sin(d*x+c)^2),x, algorithm="fricas")

[Out]

-1/2*(3*d*x*cos(d*x + c) - (cos(d*x + c)^2 + 2)*sin(d*x + c))/(a*d*cos(d*x + c))

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Sympy [A]  time = 39.793, size = 502, normalized size = 10.24 \begin{align*} \begin{cases} - \frac{3 d x \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2 a d \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 2 a d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 2 a d} - \frac{3 d x \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2 a d \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 2 a d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 2 a d} + \frac{3 d x \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2 a d \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 2 a d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 2 a d} + \frac{3 d x}{2 a d \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 2 a d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 2 a d} - \frac{6 \tan ^{5}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2 a d \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 2 a d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 2 a d} - \frac{4 \tan ^{3}{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2 a d \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 2 a d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 2 a d} - \frac{6 \tan{\left (\frac{c}{2} + \frac{d x}{2} \right )}}{2 a d \tan ^{6}{\left (\frac{c}{2} + \frac{d x}{2} \right )} + 2 a d \tan ^{4}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 2 a d \tan ^{2}{\left (\frac{c}{2} + \frac{d x}{2} \right )} - 2 a d} & \text{for}\: d \neq 0 \\\frac{x \sin ^{4}{\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)**4/(a-a*sin(d*x+c)**2),x)

[Out]

Piecewise((-3*d*x*tan(c/2 + d*x/2)**6/(2*a*d*tan(c/2 + d*x/2)**6 + 2*a*d*tan(c/2 + d*x/2)**4 - 2*a*d*tan(c/2 +
d*x/2)**2 - 2*a*d) - 3*d*x*tan(c/2 + d*x/2)**4/(2*a*d*tan(c/2 + d*x/2)**6 + 2*a*d*tan(c/2 + d*x/2)**4 - 2*a*d
*tan(c/2 + d*x/2)**2 - 2*a*d) + 3*d*x*tan(c/2 + d*x/2)**2/(2*a*d*tan(c/2 + d*x/2)**6 + 2*a*d*tan(c/2 + d*x/2)*
*4 - 2*a*d*tan(c/2 + d*x/2)**2 - 2*a*d) + 3*d*x/(2*a*d*tan(c/2 + d*x/2)**6 + 2*a*d*tan(c/2 + d*x/2)**4 - 2*a*d
*tan(c/2 + d*x/2)**2 - 2*a*d) - 6*tan(c/2 + d*x/2)**5/(2*a*d*tan(c/2 + d*x/2)**6 + 2*a*d*tan(c/2 + d*x/2)**4 -
2*a*d*tan(c/2 + d*x/2)**2 - 2*a*d) - 4*tan(c/2 + d*x/2)**3/(2*a*d*tan(c/2 + d*x/2)**6 + 2*a*d*tan(c/2 + d*x/2
)**4 - 2*a*d*tan(c/2 + d*x/2)**2 - 2*a*d) - 6*tan(c/2 + d*x/2)/(2*a*d*tan(c/2 + d*x/2)**6 + 2*a*d*tan(c/2 + d*
x/2)**4 - 2*a*d*tan(c/2 + d*x/2)**2 - 2*a*d), Ne(d, 0)), (x*sin(c)**4/(-a*sin(c)**2 + a), True))

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Giac [A]  time = 1.16521, size = 68, normalized size = 1.39 \begin{align*} -\frac{\frac{3 \,{\left (d x + c\right )}}{a} - \frac{2 \, \tan \left (d x + c\right )}{a} - \frac{\tan \left (d x + c\right )}{{\left (\tan \left (d x + c\right )^{2} + 1\right )} a}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)^4/(a-a*sin(d*x+c)^2),x, algorithm="giac")

[Out]

-1/2*(3*(d*x + c)/a - 2*tan(d*x + c)/a - tan(d*x + c)/((tan(d*x + c)^2 + 1)*a))/d