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

Optimal. Leaf size=77 \[ -\frac{3 a^3 \cos (c+d x)}{d}-\frac{2 a^5 \cos ^3(c+d x)}{d (a-a \sin (c+d x))^2}+3 a^3 x+\frac{\sec ^3(c+d x) (a \sin (c+d x)+a)^3}{3 d} \]

[Out]

3*a^3*x - (3*a^3*Cos[c + d*x])/d - (2*a^5*Cos[c + d*x]^3)/(d*(a - a*Sin[c + d*x])^2) + (Sec[c + d*x]^3*(a + a*
Sin[c + d*x])^3)/(3*d)

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Rubi [A]  time = 0.218969, antiderivative size = 77, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.172, Rules used = {2871, 2670, 2680, 2682, 8} \[ -\frac{3 a^3 \cos (c+d x)}{d}-\frac{2 a^5 \cos ^3(c+d x)}{d (a-a \sin (c+d x))^2}+3 a^3 x+\frac{\sec ^3(c+d x) (a \sin (c+d x)+a)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

3*a^3*x - (3*a^3*Cos[c + d*x])/d - (2*a^5*Cos[c + d*x]^3)/(d*(a - a*Sin[c + d*x])^2) + (Sec[c + d*x]^3*(a + a*
Sin[c + d*x])^3)/(3*d)

Rule 2871

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*sin[(e_.) + (f_.)*(x_)]^2*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)
, x_Symbol] :> Simp[(b*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*m), x] - Dist[1/g^2, Int[(g*Cos
[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0] && E
qQ[m + p + 1, 0]

Rule 2670

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Dist[(a/g)^
(2*m), Int[(g*Cos[e + f*x])^(2*m + p)/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 -
b^2, 0] && IntegerQ[m] && LtQ[p, -1] && GeQ[2*m + p, 0]

Rule 2680

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rule 2682

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

Rule 8

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

Rubi steps

\begin{align*} \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx &=\frac{\sec ^3(c+d x) (a+a \sin (c+d x))^3}{3 d}-\int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \, dx\\ &=\frac{\sec ^3(c+d x) (a+a \sin (c+d x))^3}{3 d}-a^6 \int \frac{\cos ^4(c+d x)}{(a-a \sin (c+d x))^3} \, dx\\ &=-\frac{2 a^5 \cos ^3(c+d x)}{d (a-a \sin (c+d x))^2}+\frac{\sec ^3(c+d x) (a+a \sin (c+d x))^3}{3 d}+\left (3 a^4\right ) \int \frac{\cos ^2(c+d x)}{a-a \sin (c+d x)} \, dx\\ &=-\frac{3 a^3 \cos (c+d x)}{d}-\frac{2 a^5 \cos ^3(c+d x)}{d (a-a \sin (c+d x))^2}+\frac{\sec ^3(c+d x) (a+a \sin (c+d x))^3}{3 d}+\left (3 a^3\right ) \int 1 \, dx\\ &=3 a^3 x-\frac{3 a^3 \cos (c+d x)}{d}-\frac{2 a^5 \cos ^3(c+d x)}{d (a-a \sin (c+d x))^2}+\frac{\sec ^3(c+d x) (a+a \sin (c+d x))^3}{3 d}\\ \end{align*}

Mathematica [A]  time = 1.25296, size = 133, normalized size = 1.73 \[ \frac{a^3 (\sin (c+d x)+1)^3 \left (-3 \cos (c+d x)+\frac{2 \sin \left (\frac{1}{2} (c+d x)\right ) (13 \sin (c+d x)-11)}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{2}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}+9 c+9 d x\right )}{3 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(1 + Sin[c + d*x])^3*(9*c + 9*d*x - 3*Cos[c + d*x] + 2/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (2*Sin[(
c + d*x)/2]*(-11 + 13*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])^6)

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Maple [B]  time = 0.08, size = 184, normalized size = 2.4 \begin{align*}{\frac{1}{d} \left ({a}^{3} \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 ) +3\,{a}^{3} \left ( 1/3\, \left ( \tan \left ( dx+c \right ) \right ) ^{3}-\tan \left ( dx+c \right ) +dx+c \right ) +3\,{a}^{3} \left ( 1/3\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{ \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-1/3\,{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{\cos \left ( dx+c \right ) }}-1/3\, \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) \right ) +{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.68552, size = 144, normalized size = 1.87 \begin{align*} \frac{a^{3} \tan \left (d x + c\right )^{3} + 3 \,{\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{3} - a^{3}{\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{3 \,{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{3}}{\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)^2*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 1.37299, size = 404, normalized size = 5.25 \begin{align*} -\frac{3 \, a^{3} \cos \left (d x + c\right )^{3} + 18 \, a^{3} d x + 2 \, a^{3} -{\left (9 \, a^{3} d x + 16 \, a^{3}\right )} \cos \left (d x + c\right )^{2} +{\left (9 \, a^{3} d x - 17 \, a^{3}\right )} \cos \left (d x + c\right ) -{\left (18 \, a^{3} d x - 3 \, a^{3} \cos \left (d x + c\right )^{2} - 2 \, a^{3} +{\left (9 \, a^{3} d x - 19 \, a^{3}\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)^2*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/3*(3*a^3*cos(d*x + c)^3 + 18*a^3*d*x + 2*a^3 - (9*a^3*d*x + 16*a^3)*cos(d*x + c)^2 + (9*a^3*d*x - 17*a^3)*c
os(d*x + c) - (18*a^3*d*x - 3*a^3*cos(d*x + c)^2 - 2*a^3 + (9*a^3*d*x - 19*a^3)*cos(d*x + c))*sin(d*x + c))/(d
*cos(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)**2*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.29811, size = 117, normalized size = 1.52 \begin{align*} \frac{9 \,{\left (d x + c\right )} a^{3} - \frac{6 \, a^{3}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1} + \frac{2 \,{\left (9 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 24 \, a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 11 \, a^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\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)^2*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/3*(9*(d*x + c)*a^3 - 6*a^3/(tan(1/2*d*x + 1/2*c)^2 + 1) + 2*(9*a^3*tan(1/2*d*x + 1/2*c)^2 - 24*a^3*tan(1/2*d
*x + 1/2*c) + 11*a^3)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d