3.1.43 \(\int \frac {\sin ^4(c+d x)}{a-a \sin ^2(c+d x)} \, dx\) [43]

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

[Out]

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

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Rubi [A]
time = 0.05, antiderivative size = 49, normalized size of antiderivative = 1.00, 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 = {3254, 2671, 294, 327, 209} \begin {gather*} \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} \end {gather*}

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 209

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

Rule 294

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 327

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 2671

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]/ff
)], x]] /; FreeQ[{b, e, f, n}, x] && IntegerQ[m/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*} \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 {\text {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 \text {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 \text {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*}

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Mathematica [A]
time = 0.10, size = 34, normalized size = 0.69 \begin {gather*} \frac {-6 (c+d x)+\sin (2 (c+d x))+4 \tan (c+d x)}{4 a d} \end {gather*}

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 = 0.18, size = 44, normalized size = 0.90

method	result	size
derivativedivides	\(\frac {\tan \left (d x +c \right )+\frac {\tan \left (d x +c \right )}{2 \left (\tan ^{2}\left (d x +c \right )\right )+2}-\frac {3 \arctan \left (\tan \left (d x +c \right )\right )}{2}}{d a}\)	\(44\)
default	\(\frac {\tan \left (d x +c \right )+\frac {\tan \left (d x +c \right )}{2 \left (\tan ^{2}\left (d x +c \right )\right )+2}-\frac {3 \arctan \left (\tan \left (d x +c \right )\right )}{2}}{d a}\)	\(44\)
risch	\(-\frac {3 x}{2 a}-\frac {i {\mathrm e}^{2 i \left (d x +c \right )}}{8 a d}+\frac {i {\mathrm e}^{-2 i \left (d x +c \right )}}{8 a d}+\frac {2 i}{d a \left ({\mathrm e}^{2 i \left (d x +c \right )}+1\right )}\)	\(66\)
norman	\(\frac {\frac {3 x}{2 a}-\frac {3 \tan \left (\frac {d x}{2}+\frac {c}{2}\right )}{a d}-\frac {8 \left (\tan ^{3}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {10 \left (\tan ^{5}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {8 \left (\tan ^{7}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}-\frac {3 \left (\tan ^{9}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a d}+\frac {9 x \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}+\frac {3 x \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {3 x \left (\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a}-\frac {9 x \left (\tan ^{8}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}-\frac {3 x \left (\tan ^{10}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )^{4} \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}\)	\(217\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]
time = 0.51, size = 49, normalized size = 1.00 \begin {gather*} -\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 {gather*}

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 = 0.39, size = 45, normalized size = 0.92 \begin {gather*} -\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 {gather*}

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 [B] Leaf count of result is larger than twice the leaf count of optimal. 502 vs. \(2 (39) = 78\).
time = 2.89, size = 502, normalized size = 10.24 \begin {gather*} \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 {gather*}

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 = 0.44, size = 50, normalized size = 1.02 \begin {gather*} -\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 {gather*}

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

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Mupad [B]
time = 13.53, size = 45, normalized size = 0.92 \begin {gather*} \frac {\mathrm {tan}\left (c+d\,x\right )}{2\,d\,\left (a\,{\mathrm {tan}\left (c+d\,x\right )}^2+a\right )}-\frac {3\,x}{2\,a}+\frac {\mathrm {tan}\left (c+d\,x\right )}{a\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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