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

Optimal. Leaf size=71 \[ \frac{2 a^2 \cos (c+d x)}{d}+\frac{a^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac{2 a^2 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac{5 a^2 x}{2} \]

[Out]

(-5*a^2*x)/2 + (2*a^2*Cos[c + d*x])/d + (2*a^2*Cos[c + d*x])/(d*(1 - Sin[c + d*x])) + (a^2*Cos[c + d*x]*Sin[c
+ d*x])/(2*d)

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Rubi [A]  time = 0.0857938, antiderivative size = 71, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.238, Rules used = {2709, 2648, 2638, 2635, 8} \[ \frac{2 a^2 \cos (c+d x)}{d}+\frac{a^2 \sin (c+d x) \cos (c+d x)}{2 d}+\frac{2 a^2 \cos (c+d x)}{d (1-\sin (c+d x))}-\frac{5 a^2 x}{2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-5*a^2*x)/2 + (2*a^2*Cos[c + d*x])/d + (2*a^2*Cos[c + d*x])/(d*(1 - Sin[c + d*x])) + (a^2*Cos[c + d*x]*Sin[c
+ d*x])/(2*d)

Rule 2709

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*tan[(e_.) + (f_.)*(x_)]^(p_), x_Symbol] :> Dist[a^p, Int[Expan
dIntegrand[(Sin[e + f*x]^p*(a + b*Sin[e + f*x])^(m - p/2))/(a - b*Sin[e + f*x])^(p/2), x], x], x] /; FreeQ[{a,
b, e, f}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p/2] && (LtQ[p, 0] || GtQ[m - p/2, 0])

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rule 2635

Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> -Simp[(b*Cos[c + d*x]*(b*Sin[c + d*x])^(n - 1))/(d*n),
x] + Dist[(b^2*(n - 1))/n, Int[(b*Sin[c + d*x])^(n - 2), x], x] /; FreeQ[{b, c, d}, x] && GtQ[n, 1] && Integer
Q[2*n]

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rubi steps

\begin{align*} \int (a+a \sin (c+d x))^2 \tan ^2(c+d x) \, dx &=a^2 \int \left (-2-\frac{2}{-1+\sin (c+d x)}-2 \sin (c+d x)-\sin ^2(c+d x)\right ) \, dx\\ &=-2 a^2 x-a^2 \int \sin ^2(c+d x) \, dx-\left (2 a^2\right ) \int \frac{1}{-1+\sin (c+d x)} \, dx-\left (2 a^2\right ) \int \sin (c+d x) \, dx\\ &=-2 a^2 x+\frac{2 a^2 \cos (c+d x)}{d}+\frac{2 a^2 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac{a^2 \cos (c+d x) \sin (c+d x)}{2 d}-\frac{1}{2} a^2 \int 1 \, dx\\ &=-\frac{5 a^2 x}{2}+\frac{2 a^2 \cos (c+d x)}{d}+\frac{2 a^2 \cos (c+d x)}{d (1-\sin (c+d x))}+\frac{a^2 \cos (c+d x) \sin (c+d x)}{2 d}\\ \end{align*}

Mathematica [B]  time = 0.390829, size = 145, normalized size = 2.04 \[ -\frac{a^2 (\sin (c+d x)+1)^2 \left (\cos \left (\frac{1}{2} (c+d x)\right ) (10 (c+d x)-\sin (2 (c+d x))-8 \cos (c+d x))+\sin \left (\frac{1}{2} (c+d x)\right ) (-2 (5 c+5 d x+8)+\sin (2 (c+d x))+8 \cos (c+d x))\right )}{4 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*(1 + Sin[c + d*x])^2*(Cos[(c + d*x)/2]*(10*(c + d*x) - 8*Cos[c + d*x] - Sin[2*(c + d*x)]) + Sin[(c + d*x
)/2]*(-2*(8 + 5*c + 5*d*x) + 8*Cos[c + d*x] + Sin[2*(c + d*x)])))/(4*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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Maple [A]  time = 0.056, size = 117, normalized size = 1.7 \begin{align*}{\frac{1}{d} \left ({a}^{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 ) +2\,{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 ) +{a}^{2} \left ( \tan \left ( dx+c \right ) -dx-c \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+a*sin(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.63799, size = 113, normalized size = 1.59 \begin{align*} -\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 )} a^{2} + 2 \,{\left (d x + c - \tan \left (d x + c\right )\right )} a^{2} - 4 \, a^{2}{\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+a*sin(d*x+c))^2,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))*a^2 + 2*(d*x + c - tan(d*x + c))*a^2
- 4*a^2*(1/cos(d*x + c) + cos(d*x + c)))/d

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Fricas [A]  time = 1.08748, size = 296, normalized size = 4.17 \begin{align*} \frac{a^{2} \cos \left (d x + c\right )^{3} - 5 \, a^{2} d x + 4 \, a^{2} \cos \left (d x + c\right )^{2} + 4 \, a^{2} -{\left (5 \, a^{2} d x - 7 \, a^{2}\right )} \cos \left (d x + c\right ) +{\left (5 \, a^{2} d x + a^{2} \cos \left (d x + c\right )^{2} - 3 \, a^{2} \cos \left (d x + c\right ) + 4 \, a^{2}\right )} \sin \left (d x + c\right )}{2 \,{\left (d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d\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+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

1/2*(a^2*cos(d*x + c)^3 - 5*a^2*d*x + 4*a^2*cos(d*x + c)^2 + 4*a^2 - (5*a^2*d*x - 7*a^2)*cos(d*x + c) + (5*a^2
*d*x + a^2*cos(d*x + c)^2 - 3*a^2*cos(d*x + c) + 4*a^2)*sin(d*x + c))/(d*cos(d*x + c) - d*sin(d*x + c) + d)

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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+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.20161, size = 138, normalized size = 1.94 \begin{align*} -\frac{5 \,{\left (d x + c\right )} a^{2} + \frac{8 \, a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1} + \frac{2 \,{\left (a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 4 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 4 \, a^{2}\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+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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