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

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

[Out]

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

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Rubi [A]  time = 0.13, antiderivative size = 82, normalized size of antiderivative = 1.00, 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} \[ -\frac {a \cos ^3(c+d x)}{3 d}+\frac {2 a \cos (c+d x)}{d}+\frac {3 a \tan (c+d x)}{2 d}+\frac {a \sec (c+d x)}{d}-\frac {a \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac {3 a x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-3*a*x)/2 + (2*a*Cos[c + d*x])/d - (a*Cos[c + d*x]^3)/(3*d) + (a*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 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 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 ^2(c+d x) (a+a \sin (c+d x)) \tan ^2(c+d x) \, dx &=a \int \sin ^2(c+d x) \tan ^2(c+d x) \, dx+a \int \sin ^3(c+d x) \tan ^2(c+d x) \, dx\\ &=-\frac {a \operatorname {Subst}\left (\int \frac {\left (1-x^2\right )^2}{x^2} \, dx,x,\cos (c+d x)\right )}{d}+\frac {a \operatorname {Subst}\left (\int \frac {x^4}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-\frac {a \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac {a \operatorname {Subst}\left (\int \left (-2+\frac {1}{x^2}+x^2\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac {(3 a) \operatorname {Subst}\left (\int \frac {x^2}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=\frac {2 a \cos (c+d x)}{d}-\frac {a \cos ^3(c+d x)}{3 d}+\frac {a \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 a \cos (c+d x)}{d}-\frac {a \cos ^3(c+d x)}{3 d}+\frac {a \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*}

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Mathematica [A]  time = 0.31, size = 82, normalized size = 1.00 \[ -\frac {3 a (c+d x)}{2 d}+\frac {a \sin (2 (c+d x))}{4 d}+\frac {7 a \cos (c+d x)}{4 d}-\frac {a \cos (3 (c+d x))}{12 d}+\frac {a \tan (c+d x)}{d}+\frac {a \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.45, size = 130, normalized size = 1.59 \[ -\frac {2 \, a \cos \left (d x + c\right )^{4} - a \cos \left (d x + c\right )^{3} + 9 \, a d x - 12 \, a \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, a d x - 5 \, a\right )} \cos \left (d x + c\right ) - {\left (2 \, a \cos \left (d x + c\right )^{3} + 9 \, a d x + 3 \, a \cos \left (d x + c\right )^{2} - 9 \, a \cos \left (d x + c\right ) + 6 \, a\right )} \sin \left (d x + c\right ) - 6 \, a}{6 \, {\left (d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(2*a*cos(d*x + c)^4 - a*cos(d*x + c)^3 + 9*a*d*x - 12*a*cos(d*x + c)^2 + 3*(3*a*d*x - 5*a)*cos(d*x + c) -
(2*a*cos(d*x + c)^3 + 9*a*d*x + 3*a*cos(d*x + c)^2 - 9*a*cos(d*x + c) + 6*a)*sin(d*x + c) - 6*a)/(d*cos(d*x +
c) - d*sin(d*x + c) + d)

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giac [A]  time = 0.18, size = 105, normalized size = 1.28 \[ -\frac {9 \, {\left (d x + c\right )} a + \frac {12 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1} + \frac {2 \, {\left (3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} - 6 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 24 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 3 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 10 \, a\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)^4*(a+a*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) - 1) + 2*(3*a*tan(1/2*d*x + 1/2*c)^5 - 6*a*tan(1/2*d*x + 1/2*
c)^4 - 24*a*tan(1/2*d*x + 1/2*c)^2 - 3*a*tan(1/2*d*x + 1/2*c) - 10*a)/(tan(1/2*d*x + 1/2*c)^2 + 1)^3)/d

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maple [A]  time = 0.40, size = 104, normalized size = 1.27 \[ \frac {a \left (\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+a \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a*(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))+a*(sin(d*x+c)^5/cos(d*x+c)+(si
n(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 = 0.44, size = 75, normalized size = 0.91 \[ -\frac {2 \, {\left (\cos \left (d x + c\right )^{3} - \frac {3}{\cos \left (d x + c\right )} - 6 \, \cos \left (d x + c\right )\right )} a + 3 \, {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 14.82, size = 257, normalized size = 3.13 \[ \frac {\left (\frac {a\,\left (9\,c+9\,d\,x-18\right )}{6}-\frac {3\,a\,\left (c+d\,x\right )}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^6+\left (\frac {9\,a\,\left (c+d\,x\right )}{2}-\frac {a\,\left (27\,c+27\,d\,x-18\right )}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5+\left (\frac {a\,\left (27\,c+27\,d\,x-48\right )}{6}-\frac {9\,a\,\left (c+d\,x\right )}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+\left (\frac {9\,a\,\left (c+d\,x\right )}{2}-\frac {a\,\left (27\,c+27\,d\,x-48\right )}{6}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3+\left (\frac {a\,\left (27\,c+27\,d\,x-78\right )}{6}-\frac {9\,a\,\left (c+d\,x\right )}{2}\right )\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\left (\frac {3\,a\,\left (c+d\,x\right )}{2}-\frac {a\,\left (9\,c+9\,d\,x-14\right )}{6}\right )\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )+\frac {a\,\left (9\,c+9\,d\,x-32\right )}{6}-\frac {3\,a\,\left (c+d\,x\right )}{2}}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^3}-\frac {3\,a\,x}{2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a*(9*c + 9*d*x - 32))/6 - tan(c/2 + (d*x)/2)*((a*(9*c + 9*d*x - 14))/6 - (3*a*(c + d*x))/2) - (3*a*(c + d*x)
)/2 + tan(c/2 + (d*x)/2)^6*((a*(9*c + 9*d*x - 18))/6 - (3*a*(c + d*x))/2) - tan(c/2 + (d*x)/2)^5*((a*(27*c + 2
7*d*x - 18))/6 - (9*a*(c + d*x))/2) - tan(c/2 + (d*x)/2)^3*((a*(27*c + 27*d*x - 48))/6 - (9*a*(c + d*x))/2) +
tan(c/2 + (d*x)/2)^4*((a*(27*c + 27*d*x - 48))/6 - (9*a*(c + d*x))/2) + tan(c/2 + (d*x)/2)^2*((a*(27*c + 27*d*
x - 78))/6 - (9*a*(c + d*x))/2))/(d*(tan(c/2 + (d*x)/2) - 1)*(tan(c/2 + (d*x)/2)^2 + 1)^3) - (3*a*x)/2

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

[Out]

Timed out

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