3.1450 \(\int \sin (c+d x) (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx\)

Optimal. Leaf size=94 \[ \frac{\left (a^2+2 b^2\right ) \cos (c+d x)}{d}+\frac{\left (a^2+b^2\right ) \sec (c+d x)}{d}+\frac{3 a b \tan (c+d x)}{d}-\frac{a b \sin ^2(c+d x) \tan (c+d x)}{d}-3 a b x-\frac{b^2 \cos ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.18344, antiderivative size = 94, 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 = {2911, 2591, 288, 321, 203, 4357, 448} \[ \frac{\left (a^2+2 b^2\right ) \cos (c+d x)}{d}+\frac{\left (a^2+b^2\right ) \sec (c+d x)}{d}+\frac{3 a b \tan (c+d x)}{d}-\frac{a b \sin ^2(c+d x) \tan (c+d x)}{d}-3 a b x-\frac{b^2 \cos ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2911

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^2, x_Symbol] :> Dist[(2*a*b)/d, Int[(g*Cos[e + f*x])^p*(d*Sin[e + f*x])^(n + 1), x], x] + Int[(g*Cos[e
+ f*x])^p*(d*Sin[e + f*x])^n*(a^2 + b^2*Sin[e + f*x]^2), x] /; FreeQ[{a, b, d, e, f, g, n, p}, x] && NeQ[a^2 -
b^2, 0]

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 4357

Int[(u_)*(F_)[(c_.)*((a_.) + (b_.)*(x_))], x_Symbol] :> With[{d = FreeFactors[Cos[c*(a + b*x)], x]}, -Dist[d/(
b*c), Subst[Int[SubstFor[1, Cos[c*(a + b*x)]/d, u, x], x], x, Cos[c*(a + b*x)]/d], x] /; FunctionOfQ[Cos[c*(a
+ b*x)]/d, u, x]] /; FreeQ[{a, b, c}, x] && (EqQ[F, Sin] || EqQ[F, sin])

Rule 448

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> Int[ExpandI
ntegrand[(e*x)^m*(a + b*x^n)^p*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, e, m, n}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[p, 0] && IGtQ[q, 0]

Rubi steps

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

Mathematica [A]  time = 0.426966, size = 104, normalized size = 1.11 \[ \frac{\sec (c+d x) \left (-24 \cos (c+d x) \left (a^2+3 a b (c+d x)+b^2\right )+4 \left (3 a^2+5 b^2\right ) \cos (2 (c+d x))+36 a^2+54 a b \sin (c+d x)+6 a b \sin (3 (c+d x))-b^2 \cos (4 (c+d x))+45 b^2\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sec[c + d*x]*(36*a^2 + 45*b^2 - 24*(a^2 + b^2 + 3*a*b*(c + d*x))*Cos[c + d*x] + 4*(3*a^2 + 5*b^2)*Cos[2*(c +
d*x)] - b^2*Cos[4*(c + d*x)] + 54*a*b*Sin[c + d*x] + 6*a*b*Sin[3*(c + d*x)]))/(24*d)

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Maple [A]  time = 0.05, size = 147, normalized size = 1.6 \begin{align*}{\frac{1}{d} \left ({a}^{2} \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 ) +2\,ab \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{\cos \left ( dx+c \right ) }}+ \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{3}+3/2\,\sin \left ( dx+c \right ) \right ) \cos \left ( dx+c \right ) -3/2\,dx-3/2\,c \right ) +{b}^{2} \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)^3*(a+b*sin(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.54784, size = 131, normalized size = 1.39 \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 b +{\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^{2} - 3 \, a^{2}{\left (\frac{1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )}}{3 \, 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))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.66871, size = 221, normalized size = 2.35 \begin{align*} -\frac{b^{2} \cos \left (d x + c\right )^{4} + 9 \, a b d x \cos \left (d x + c\right ) - 3 \,{\left (a^{2} + 2 \, b^{2}\right )} \cos \left (d x + c\right )^{2} - 3 \, a^{2} - 3 \, b^{2} - 3 \,{\left (a b \cos \left (d x + c\right )^{2} + 2 \, a b\right )} \sin \left (d x + c\right )}{3 \, 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))^2,x, algorithm="fricas")

[Out]

-1/3*(b^2*cos(d*x + c)^4 + 9*a*b*d*x*cos(d*x + c) - 3*(a^2 + 2*b^2)*cos(d*x + c)^2 - 3*a^2 - 3*b^2 - 3*(a*b*co
s(d*x + c)^2 + 2*a*b)*sin(d*x + c))/(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)**3*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.19564, size = 232, normalized size = 2.47 \begin{align*} -\frac{9 \,{\left (d x + c\right )} a b + \frac{6 \,{\left (2 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + a^{2} + b^{2}\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1} + \frac{2 \,{\left (3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} - 3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 3 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 6 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 12 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 3 \, a^{2} - 5 \, b^{2}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{3}}}{3 \, 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))^2,x, algorithm="giac")

[Out]

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