3.805 \(\int \sec (c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx\)

Optimal. Leaf size=86 \[ -\frac{4 a^2 \cos (c+d x)}{3 d}+\frac{a^4 \sin ^2(c+d x) \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac{2 a^2 \cos (c+d x)}{d (1-\sin (c+d x))}+2 a^2 x \]

[Out]

2*a^2*x - (4*a^2*Cos[c + d*x])/(3*d) - (2*a^2*Cos[c + d*x])/(d*(1 - Sin[c + d*x])) + (a^4*Cos[c + d*x]*Sin[c +
d*x]^2)/(3*d*(a - a*Sin[c + d*x])^2)

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Rubi [A]  time = 0.245919, antiderivative size = 86, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 7, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.259, Rules used = {2869, 2765, 2968, 3023, 12, 2735, 2648} \[ -\frac{4 a^2 \cos (c+d x)}{3 d}+\frac{a^4 \sin ^2(c+d x) \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac{2 a^2 \cos (c+d x)}{d (1-\sin (c+d x))}+2 a^2 x \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]*(a + a*Sin[c + d*x])^2*Tan[c + d*x]^3,x]

[Out]

2*a^2*x - (4*a^2*Cos[c + d*x])/(3*d) - (2*a^2*Cos[c + d*x])/(d*(1 - Sin[c + d*x])) + (a^4*Cos[c + d*x]*Sin[c +
d*x]^2)/(3*d*(a - a*Sin[c + d*x])^2)

Rule 2869

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

Rule 2765

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[((b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^m*(c + d*Sin[e + f*x])^(n - 1))/(a*f*(2*m + 1)), x] + Dist[1/
(a*b*(2*m + 1)), Int[(a + b*Sin[e + f*x])^(m + 1)*(c + d*Sin[e + f*x])^(n - 2)*Simp[b*(c^2*(m + 1) + d^2*(n -
1)) + a*c*d*(m - n + 1) + d*(a*d*(m - n + 1) + b*c*(m + n))*Sin[e + f*x], x], x], x] /; FreeQ[{a, b, c, d, e,
f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && GtQ[n, 1] && (IntegersQ
[2*m, 2*n] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2968

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)])*((c_.) + (d_.)*sin[(
e_.) + (f_.)*(x_)]), x_Symbol] :> Int[(a + b*Sin[e + f*x])^m*(A*c + (B*c + A*d)*Sin[e + f*x] + B*d*Sin[e + f*x
]^2), x] /; FreeQ[{a, b, c, d, e, f, A, B, m}, x] && NeQ[b*c - a*d, 0]

Rule 3023

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (
f_.)*(x_)]^2), x_Symbol] :> -Simp[(C*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(m + 2)), x] + Dist[1/(b*
(m + 2)), Int[(a + b*Sin[e + f*x])^m*Simp[A*b*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Sin[e + f*x], x], x]
, x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 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]

Rubi steps

\begin{align*} \int \sec (c+d x) (a+a \sin (c+d x))^2 \tan ^3(c+d x) \, dx &=a^4 \int \frac{\sin ^3(c+d x)}{(a-a \sin (c+d x))^2} \, dx\\ &=\frac{a^4 \cos (c+d x) \sin ^2(c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac{1}{3} a^2 \int \frac{\sin (c+d x) (-2 a-4 a \sin (c+d x))}{a-a \sin (c+d x)} \, dx\\ &=\frac{a^4 \cos (c+d x) \sin ^2(c+d x)}{3 d (a-a \sin (c+d x))^2}+\frac{1}{3} a^2 \int \frac{-2 a \sin (c+d x)-4 a \sin ^2(c+d x)}{a-a \sin (c+d x)} \, dx\\ &=-\frac{4 a^2 \cos (c+d x)}{3 d}+\frac{a^4 \cos (c+d x) \sin ^2(c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac{1}{3} a \int \frac{6 a^2 \sin (c+d x)}{a-a \sin (c+d x)} \, dx\\ &=-\frac{4 a^2 \cos (c+d x)}{3 d}+\frac{a^4 \cos (c+d x) \sin ^2(c+d x)}{3 d (a-a \sin (c+d x))^2}-\left (2 a^3\right ) \int \frac{\sin (c+d x)}{a-a \sin (c+d x)} \, dx\\ &=2 a^2 x-\frac{4 a^2 \cos (c+d x)}{3 d}+\frac{a^4 \cos (c+d x) \sin ^2(c+d x)}{3 d (a-a \sin (c+d x))^2}-\left (2 a^3\right ) \int \frac{1}{a-a \sin (c+d x)} \, dx\\ &=2 a^2 x-\frac{4 a^2 \cos (c+d x)}{3 d}+\frac{a^4 \cos (c+d x) \sin ^2(c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac{2 a^3 \cos (c+d x)}{d (a-a \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 1.0341, size = 131, normalized size = 1.52 \[ \frac{a^2 (\sin (c+d x)+1)^2 \left (-3 \cos (c+d x)+\frac{2 \sin \left (\frac{1}{2} (c+d x)\right ) (8 \sin (c+d x)-7)}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{1}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+6 c+6 d x\right )}{3 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]*(a + a*Sin[c + d*x])^2*Tan[c + d*x]^3,x]

[Out]

(a^2*(1 + Sin[c + d*x])^2*(6*c + 6*d*x - 3*Cos[c + d*x] + (Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^(-2) + (2*Sin[
(c + d*x)/2]*(-7 + 8*Sin[c + d*x]))/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3))/(3*d*(Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2])^4)

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Maple [A]  time = 0.073, size = 162, normalized size = 1.9 \begin{align*}{\frac{1}{d} \left ({a}^{2} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{6}}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\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 ) +2\,{a}^{2} \left ( 1/3\, \left ( \tan \left ( dx+c \right ) \right ) ^{3}-\tan \left ( dx+c \right ) +dx+c \right ) +{a}^{2} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{3\,\cos \left ( dx+c \right ) }}-{\frac{ \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) }{3}} \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.64283, size = 128, normalized size = 1.49 \begin{align*} \frac{2 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{2} - a^{2}{\left (\frac{6 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} + 3 \, \cos \left (d x + c\right )\right )} - \frac{{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{2}}{\cos \left (d x + c\right )^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.38732, size = 400, normalized size = 4.65 \begin{align*} -\frac{3 \, a^{2} \cos \left (d x + c\right )^{3} + 12 \, a^{2} d x -{\left (6 \, a^{2} d x + 11 \, a^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} +{\left (6 \, a^{2} d x - 13 \, a^{2}\right )} \cos \left (d x + c\right ) -{\left (12 \, a^{2} d x - 3 \, a^{2} \cos \left (d x + c\right )^{2} - a^{2} + 2 \,{\left (3 \, a^{2} d x - 7 \, a^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \,{\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) +{\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/3*(3*a^2*cos(d*x + c)^3 + 12*a^2*d*x - (6*a^2*d*x + 11*a^2)*cos(d*x + c)^2 + a^2 + (6*a^2*d*x - 13*a^2)*cos
(d*x + c) - (12*a^2*d*x - 3*a^2*cos(d*x + c)^2 - a^2 + 2*(3*a^2*d*x - 7*a^2)*cos(d*x + c))*sin(d*x + c))/(d*co
s(d*x + c)^2 - d*cos(d*x + c) + (d*cos(d*x + c) + 2*d)*sin(d*x + c) - 2*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)**4*sin(d*x+c)**3*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.2729, size = 116, normalized size = 1.35 \begin{align*} \frac{2 \,{\left (3 \,{\left (d x + c\right )} a^{2} - \frac{3 \, a^{2}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1} + \frac{6 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 15 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 7 \, a^{2}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}}\right )}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^3*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

2/3*(3*(d*x + c)*a^2 - 3*a^2/(tan(1/2*d*x + 1/2*c)^2 + 1) + (6*a^2*tan(1/2*d*x + 1/2*c)^2 - 15*a^2*tan(1/2*d*x
+ 1/2*c) + 7*a^2)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d