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]

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

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Rubi [A]  time = 0.11, antiderivative size = 65, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, 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 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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 0.19, 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

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fricas [A]  time = 0.78, size = 61, normalized size = 0.94 \[ -\frac {3 \, b d x \cos \left (d x + c\right ) - 2 \, a \cos \left (d x + c\right )^{2} - {\left (b \cos \left (d x + c\right )^{2} + 2 \, b\right )} \sin \left (d x + c\right ) - 2 \, a}{2 \, d \cos \left (d x + c\right )} \]

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)
)

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giac [A]  time = 0.18, size = 104, normalized size = 1.60 \[ -\frac {3 \, {\left (d x + c\right )} b + \frac {4 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + a\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1} + \frac {2 \, {\left (b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

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

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maple [A]  time = 0.40, size = 94, normalized size = 1.45 \[ \frac {a \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+b \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )}{d} \]

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))

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maxima [A]  time = 0.42, size = 62, normalized size = 0.95 \[ -\frac {{\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} b - 2 \, a {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{2 \, d} \]

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

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mupad [B]  time = 15.97, size = 98, normalized size = 1.51 \[ -\frac {3\,b\,x}{2}-\frac {3\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+2\,b\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+4\,a\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+3\,b\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+4\,a}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2-1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (3*b*x)/2 - (4*a + 3*b*tan(c/2 + (d*x)/2) + 4*a*tan(c/2 + (d*x)/2)^2 + 2*b*tan(c/2 + (d*x)/2)^3 + 3*b*tan(c/
2 + (d*x)/2)^5)/(d*(tan(c/2 + (d*x)/2)^2 - 1)*(tan(c/2 + (d*x)/2)^2 + 1)^2)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sin {\left (c + d x \right )}\right ) \sin ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx \]

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]

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

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