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

Optimal. Leaf size=89 \[ -\frac {a^2 \cos ^3(c+d x)}{3 d}+\frac {3 a^2 \cos (c+d x)}{d}+\frac {3 a^2 \tan (c+d x)}{d}+\frac {2 a^2 \sec (c+d x)}{d}-\frac {a^2 \sin ^2(c+d x) \tan (c+d x)}{d}-3 a^2 x \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2873, 2590, 14, 2591, 288, 321, 203, 270} \[ -\frac {a^2 \cos ^3(c+d x)}{3 d}+\frac {3 a^2 \cos (c+d x)}{d}+\frac {3 a^2 \tan (c+d x)}{d}+\frac {2 a^2 \sec (c+d x)}{d}-\frac {a^2 \sin ^2(c+d x) \tan (c+d x)}{d}-3 a^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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 2873

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Int[ExpandTrig[(g*cos[e + f*x])^p, (d*sin[e + f*x])^n*(a + b*sin[e + f*x])^m, x], x]
/; FreeQ[{a, b, d, e, f, g, n, p}, x] && EqQ[a^2 - b^2, 0] && IGtQ[m, 0]

Rubi steps

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

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Mathematica [A]  time = 0.52, size = 161, normalized size = 1.81 \[ -\frac {a^2 (\sin (c+d x)+1)^2 \left (\cos \left (\frac {1}{2} (c+d x)\right ) (-6 \sin (2 (c+d x))-33 \cos (c+d x)+\cos (3 (c+d x))+36 c+36 d x)-\sin \left (\frac {1}{2} (c+d x)\right ) (-6 \sin (2 (c+d x))-33 \cos (c+d x)+\cos (3 (c+d x))+36 c+36 d x+48)\right )}{12 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/12*(a^2*(1 + Sin[c + d*x])^2*(Cos[(c + d*x)/2]*(36*c + 36*d*x - 33*Cos[c + d*x] + Cos[3*(c + d*x)] - 6*Sin[
2*(c + d*x)]) - Sin[(c + d*x)/2]*(48 + 36*c + 36*d*x - 33*Cos[c + d*x] + Cos[3*(c + d*x)] - 6*Sin[2*(c + d*x)]
)))/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)

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fricas [A]  time = 0.43, size = 152, normalized size = 1.71 \[ -\frac {a^{2} \cos \left (d x + c\right )^{4} - 2 \, a^{2} \cos \left (d x + c\right )^{3} + 9 \, a^{2} d x - 9 \, a^{2} \cos \left (d x + c\right )^{2} - 6 \, a^{2} + 3 \, {\left (3 \, a^{2} d x - 4 \, a^{2}\right )} \cos \left (d x + c\right ) - {\left (a^{2} \cos \left (d x + c\right )^{3} + 9 \, a^{2} d x + 3 \, a^{2} \cos \left (d x + c\right )^{2} - 6 \, a^{2} \cos \left (d x + c\right ) + 6 \, a^{2}\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 119, normalized size = 1.34 \[ -\frac {9 \, {\left (d x + c\right )} a^{2} + \frac {12 \, a^{2}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1} + \frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 18 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 8 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3*(9*(d*x + c)*a^2 + 12*a^2/(tan(1/2*d*x + 1/2*c) - 1) + 2*(3*a^2*tan(1/2*d*x + 1/2*c)^5 - 6*a^2*tan(1/2*d*
x + 1/2*c)^4 - 18*a^2*tan(1/2*d*x + 1/2*c)^2 - 3*a^2*tan(1/2*d*x + 1/2*c) - 8*a^2)/(tan(1/2*d*x + 1/2*c)^2 + 1
)^3)/d

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maple [A]  time = 0.50, size = 148, normalized size = 1.66 \[ \frac {a^{2} \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+2 a^{2} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.62, size = 98, normalized size = 1.10 \[ -\frac {{\left (\cos \left (d x + c\right )^{3} - \frac {3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} a^{2} + 3 \, {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a^{2} - 3 \, a^{2} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 14.71, size = 288, normalized size = 3.24 \[ -3\,a^2\,x-\frac {3\,a^2\,\left (c+d\,x\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^2\,\left (c+d\,x\right )-\frac {a^2\,\left (9\,c+9\,d\,x-10\right )}{3}\right )-\frac {a^2\,\left (9\,c+9\,d\,x-28\right )}{3}+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6\,\left (3\,a^2\,\left (c+d\,x\right )-\frac {a^2\,\left (9\,c+9\,d\,x-18\right )}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5\,\left (9\,a^2\,\left (c+d\,x\right )-\frac {a^2\,\left (27\,c+27\,d\,x-18\right )}{3}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (9\,a^2\,\left (c+d\,x\right )-\frac {a^2\,\left (27\,c+27\,d\,x-36\right )}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (9\,a^2\,\left (c+d\,x\right )-\frac {a^2\,\left (27\,c+27\,d\,x-48\right )}{3}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (9\,a^2\,\left (c+d\,x\right )-\frac {a^2\,\left (27\,c+27\,d\,x-66\right )}{3}\right )}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- 3*a^2*x - (3*a^2*(c + d*x) - tan(c/2 + (d*x)/2)*(3*a^2*(c + d*x) - (a^2*(9*c + 9*d*x - 10))/3) - (a^2*(9*c +
9*d*x - 28))/3 + tan(c/2 + (d*x)/2)^6*(3*a^2*(c + d*x) - (a^2*(9*c + 9*d*x - 18))/3) - tan(c/2 + (d*x)/2)^5*(
9*a^2*(c + d*x) - (a^2*(27*c + 27*d*x - 18))/3) - tan(c/2 + (d*x)/2)^3*(9*a^2*(c + d*x) - (a^2*(27*c + 27*d*x
- 36))/3) + tan(c/2 + (d*x)/2)^4*(9*a^2*(c + d*x) - (a^2*(27*c + 27*d*x - 48))/3) + tan(c/2 + (d*x)/2)^2*(9*a^
2*(c + d*x) - (a^2*(27*c + 27*d*x - 66))/3))/(d*(tan(c/2 + (d*x)/2) - 1)*(tan(c/2 + (d*x)/2)^2 + 1)^3)

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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