3.1451 \(\int (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx\)

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

[Out]

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

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Rubi [A]  time = 0.138516, antiderivative size = 94, normalized size of antiderivative = 1., number of steps used = 11, number of rules used = 9, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.429, Rules used = {2722, 3473, 8, 2590, 14, 2591, 288, 321, 203} \[ \frac{a^2 \tan (c+d x)}{d}+a^2 (-x)+\frac{2 a b \cos (c+d x)}{d}+\frac{2 a b \sec (c+d x)}{d}+\frac{3 b^2 \tan (c+d x)}{2 d}-\frac{b^2 \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac{3 b^2 x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2722

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((g_.)*tan[(e_.) + (f_.)*(x_)])^(p_.), x_Symbol] :> Int[Expan
dIntegrand[(g*Tan[e + f*x])^p, (a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, p}, x] && NeQ[a^2 - b^2
, 0] && IGtQ[m, 0]

Rule 3473

Int[((b_.)*tan[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*(b*Tan[c + d*x])^(n - 1))/(d*(n - 1)), x] - Dis
t[b^2, Int[(b*Tan[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, 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 (a+b \sin (c+d x))^2 \tan ^2(c+d x) \, dx &=\int \left (a^2 \tan ^2(c+d x)+2 a b \sin (c+d x) \tan ^2(c+d x)+b^2 \sin ^2(c+d x) \tan ^2(c+d x)\right ) \, dx\\ &=a^2 \int \tan ^2(c+d x) \, dx+(2 a b) \int \sin (c+d x) \tan ^2(c+d x) \, dx+b^2 \int \sin ^2(c+d x) \tan ^2(c+d x) \, dx\\ &=\frac{a^2 \tan (c+d x)}{d}-a^2 \int 1 \, dx-\frac{(2 a b) \operatorname{Subst}\left (\int \frac{1-x^2}{x^2} \, dx,x,\cos (c+d x)\right )}{d}+\frac{b^2 \operatorname{Subst}\left (\int \frac{x^4}{\left (1+x^2\right )^2} \, dx,x,\tan (c+d x)\right )}{d}\\ &=-a^2 x+\frac{a^2 \tan (c+d x)}{d}-\frac{b^2 \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac{(2 a b) \operatorname{Subst}\left (\int \left (-1+\frac{1}{x^2}\right ) \, dx,x,\cos (c+d x)\right )}{d}+\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{x^2}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-a^2 x+\frac{2 a b \cos (c+d x)}{d}+\frac{2 a b \sec (c+d x)}{d}+\frac{a^2 \tan (c+d x)}{d}+\frac{3 b^2 \tan (c+d x)}{2 d}-\frac{b^2 \sin ^2(c+d x) \tan (c+d x)}{2 d}-\frac{\left (3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+x^2} \, dx,x,\tan (c+d x)\right )}{2 d}\\ &=-a^2 x-\frac{3 b^2 x}{2}+\frac{2 a b \cos (c+d x)}{d}+\frac{2 a b \sec (c+d x)}{d}+\frac{a^2 \tan (c+d x)}{d}+\frac{3 b^2 \tan (c+d x)}{2 d}-\frac{b^2 \sin ^2(c+d x) \tan (c+d x)}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.415523, size = 77, normalized size = 0.82 \[ \frac{-4 \left (2 a^2+3 b^2\right ) (c+d x)+\left (8 a^2+9 b^2\right ) \tan (c+d x)+b \sec (c+d x) (8 a \cos (2 (c+d x))+24 a+b \sin (3 (c+d x)))}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*(2*a^2 + 3*b^2)*(c + d*x) + b*Sec[c + d*x]*(24*a + 8*a*Cos[2*(c + d*x)] + b*Sin[3*(c + d*x)]) + (8*a^2 + 9
*b^2)*Tan[c + d*x])/(8*d)

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Maple [A]  time = 0.048, size = 116, normalized size = 1.2 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ( \tan \left ( dx+c \right ) -dx-c \right ) +2\,ab \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}^{2} \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)^2*(a+b*sin(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.49602, size = 112, normalized size = 1.19 \begin{align*} -\frac{2 \,{\left (d x + c - \tan \left (d x + c\right )\right )} a^{2} +{\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} - 4 \, a b{\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)^2*(a+b*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.63604, size = 190, normalized size = 2.02 \begin{align*} -\frac{{\left (2 \, a^{2} + 3 \, b^{2}\right )} d x \cos \left (d x + c\right ) - 4 \, a b \cos \left (d x + c\right )^{2} - 4 \, a b -{\left (b^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} + 2 \, b^{2}\right )} \sin \left (d x + c\right )}{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)^2*(a+b*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.2057, size = 185, normalized size = 1.97 \begin{align*} -\frac{{\left (2 \, a^{2} + 3 \, b^{2}\right )}{\left (d x + c\right )} + \frac{4 \,{\left (a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, a b\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1} + \frac{2 \,{\left (b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, a b\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)^2*(a+b*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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