3.1444 \(\int \sin (c+d x) (a+b \sin (c+d x)) \tan ^2(c+d x) \, dx\)

Optimal. Leaf size=65 \[ \frac{a \cos (c+d x)}{d}+\frac{a \sec (c+d x)}{d}+\frac{3 b \tan (c+d x)}{2 d}-\frac{b \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac{3 b x}{2} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.107418, antiderivative size = 65, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.28, Rules used = {2838, 2590, 14, 2591, 288, 321, 203} \[ \frac{a \cos (c+d x)}{d}+\frac{a \sec (c+d x)}{d}+\frac{3 b \tan (c+d x)}{2 d}-\frac{b \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac{3 b x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*b*x)/2 + (a*Cos[c + d*x])/d + (a*Sec[c + d*x])/d + (3*b*Tan[c + d*x])/(2*d) - (b*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 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 14

Int[(u_)*((c_.)*(x_))^(m_.), x_Symbol] :> Int[ExpandIntegrand[(c*x)^m*u, x], x] /; FreeQ[{c, m}, x] && SumQ[u]
&&  !LinearQ[u, x] &&  !MatchQ[u, (a_) + (b_.)*(v_) /; FreeQ[{a, b}, x] && InverseFunctionQ[v]]

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

Mathematica [A]  time = 0.185097, size = 63, normalized size = 0.97 \[ \frac{a \cos (c+d x)}{d}+\frac{a \sec (c+d x)}{d}-\frac{3 b (c+d x)}{2 d}+\frac{b \sin (2 (c+d x))}{4 d}+\frac{b \tan (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

________________________________________________________________________________________

Maple [A]  time = 0.04, size = 94, normalized size = 1.5 \begin{align*}{\frac{1}{d} \left ( a \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{\cos \left ( dx+c \right ) }}+ \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) \right ) +b \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 ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*sin(d*x+c)^3*(a+b*sin(d*x+c)),x)

[Out]

1/d*(a*(sin(d*x+c)^4/cos(d*x+c)+(2+sin(d*x+c)^2)*cos(d*x+c))+b*(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))

________________________________________________________________________________________

Maxima [A]  time = 1.4814, size = 84, normalized size = 1.29 \begin{align*} -\frac{{\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 )} b - 2 \, a{\left (\frac{1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^3*(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

Fricas [A]  time = 2.17512, size = 153, normalized size = 2.35 \begin{align*} -\frac{3 \, b d x \cos \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right )^{2} -{\left (b \cos \left (d x + c\right )^{2} + 2 \, b\right )} \sin \left (d x + c\right ) - 2 \, a}{2 \, 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)^3*(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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)**3*(a+b*sin(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________

Giac [A]  time = 1.23264, size = 140, normalized size = 2.15 \begin{align*} -\frac{3 \,{\left (d x + c\right )} b + \frac{4 \,{\left (b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1} + \frac{2 \,{\left (b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, a \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, a\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^3*(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/2*(3*(d*x + c)*b + 4*(b*tan(1/2*d*x + 1/2*c) + a)/(tan(1/2*d*x + 1/2*c)^2 - 1) + 2*(b*tan(1/2*d*x + 1/2*c)^
3 - 2*a*tan(1/2*d*x + 1/2*c)^2 - b*tan(1/2*d*x + 1/2*c) - 2*a)/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d