3.1443 \(\int \sin ^2(c+d x) (a+b \sin (c+d x)) \tan ^2(c+d x) \, dx\)

Optimal. Leaf size=82 \[ \frac{3 a \tan (c+d x)}{2 d}-\frac{a \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac{3 a x}{2}-\frac{b \cos ^3(c+d x)}{3 d}+\frac{2 b \cos (c+d x)}{d}+\frac{b \sec (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.13668, antiderivative size = 82, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {2838, 2591, 288, 321, 203, 2590, 270} \[ \frac{3 a \tan (c+d x)}{2 d}-\frac{a \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac{3 a x}{2}-\frac{b \cos ^3(c+d x)}{3 d}+\frac{2 b \cos (c+d x)}{d}+\frac{b \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2838

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

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])

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

Mathematica [A]  time = 0.408716, size = 82, normalized size = 1. \[ -\frac{3 a (c+d x)}{2 d}+\frac{a \sin (2 (c+d x))}{4 d}+\frac{a \tan (c+d x)}{d}+\frac{7 b \cos (c+d x)}{4 d}-\frac{b \cos (3 (c+d x))}{12 d}+\frac{b \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*(c + d*x))/(2*d) + (7*b*Cos[c + d*x])/(4*d) - (b*Cos[3*(c + d*x)])/(12*d) + (b*Sec[c + d*x])/d + (a*Sin[
2*(c + d*x)])/(4*d) + (a*Tan[c + d*x])/d

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Maple [A]  time = 0.043, size = 104, normalized size = 1.3 \begin{align*}{\frac{1}{d} \left ( a \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{\cos \left ( dx+c \right ) }}+ \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\sin \left ( dx+c \right ) }{2}} \right ) \cos \left ( dx+c \right ) -{\frac{3\,dx}{2}}-{\frac{3\,c}{2}} \right ) +b \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{\cos \left ( dx+c \right ) }}+ \left ({\frac{8}{3}}+ \left ( \sin \left ( dx+c \right ) \right ) ^{4}+{\frac{4\, \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) \cos \left ( dx+c \right ) \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*sin(d*x+c)^4*(a+b*sin(d*x+c)),x)

[Out]

1/d*(a*(sin(d*x+c)^5/cos(d*x+c)+(sin(d*x+c)^3+3/2*sin(d*x+c))*cos(d*x+c)-3/2*d*x-3/2*c)+b*(sin(d*x+c)^6/cos(d*
x+c)+(8/3+sin(d*x+c)^4+4/3*sin(d*x+c)^2)*cos(d*x+c)))

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Maxima [A]  time = 1.53752, size = 101, normalized size = 1.23 \begin{align*} -\frac{3 \,{\left (3 \, d x + 3 \, c - \frac{\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a + 2 \,{\left (\cos \left (d x + c\right )^{3} - \frac{3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} b}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*(3*d*x + 3*c - tan(d*x + c)/(tan(d*x + c)^2 + 1) - 2*tan(d*x + c))*a + 2*(cos(d*x + c)^3 - 3/cos(d*x +
c) - 6*cos(d*x + c))*b)/d

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Fricas [A]  time = 2.14495, size = 185, normalized size = 2.26 \begin{align*} -\frac{2 \, b \cos \left (d x + c\right )^{4} + 9 \, a d x \cos \left (d x + c\right ) - 12 \, b \cos \left (d x + c\right )^{2} - 3 \,{\left (a \cos \left (d x + c\right )^{2} + 2 \, a\right )} \sin \left (d x + c\right ) - 6 \, b}{6 \, d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^4*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/6*(2*b*cos(d*x + c)^4 + 9*a*d*x*cos(d*x + c) - 12*b*cos(d*x + c)^2 - 3*(a*cos(d*x + c)^2 + 2*a)*sin(d*x + c
) - 6*b)/(d*cos(d*x + c))

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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)**2*sin(d*x+c)**4*(a+b*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.19943, size = 161, normalized size = 1.96 \begin{align*} -\frac{9 \,{\left (d x + c\right )} a + \frac{12 \,{\left (a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1} + \frac{2 \,{\left (3 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 6 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 24 \, b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 10 \, b\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(9*(d*x + c)*a + 12*(a*tan(1/2*d*x + 1/2*c) + b)/(tan(1/2*d*x + 1/2*c)^2 - 1) + 2*(3*a*tan(1/2*d*x + 1/2*
c)^5 - 6*b*tan(1/2*d*x + 1/2*c)^4 - 24*b*tan(1/2*d*x + 1/2*c)^2 - 3*a*tan(1/2*d*x + 1/2*c) - 10*b)/(tan(1/2*d*
x + 1/2*c)^2 + 1)^3)/d