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

Optimal. Leaf size=60 \[ -\frac{2 a^2 \tan (c+d x)}{3 d}-\frac{2 a^2 \sec (c+d x)}{3 d}+\frac{\sec ^3(c+d x) (a \sin (c+d x)+a)^2}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0864827, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {2855, 2669, 3767, 8} \[ -\frac{2 a^2 \tan (c+d x)}{3 d}-\frac{2 a^2 \sec (c+d x)}{3 d}+\frac{\sec ^3(c+d x) (a \sin (c+d x)+a)^2}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2855

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

Rule 2669

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

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

Rubi steps

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

Mathematica [A]  time = 0.282257, size = 72, normalized size = 1.2 \[ \frac{a^2 \left (-3 \sin \left (\frac{1}{2} (c+d x)\right )+3 \cos \left (\frac{1}{2} (c+d x)\right )-2 \cos \left (\frac{3}{2} (c+d x)\right )\right )}{3 d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.059, size = 99, normalized size = 1.7 \begin{align*}{\frac{1}{d} \left ({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 ) +{\frac{2\,{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{{a}^{2}}{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)*(a+a*sin(d*x+c))^2,x)

[Out]

1/d*(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))+2/3*a^2*s
in(d*x+c)^3/cos(d*x+c)^3+1/3*a^2/cos(d*x+c)^3)

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Maxima [A]  time = 1.12491, size = 76, normalized size = 1.27 \begin{align*} \frac{2 \, a^{2} \tan \left (d x + c\right )^{3} - \frac{{\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{2}}{\cos \left (d x + c\right )^{3}} + \frac{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)*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.40958, size = 236, normalized size = 3.93 \begin{align*} \frac{2 \, a^{2} \cos \left (d x + c\right )^{2} + a^{2} \cos \left (d x + c\right ) - a^{2} -{\left (2 \, a^{2} \cos \left (d x + c\right ) + a^{2}\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)*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.27201, size = 51, normalized size = 0.85 \begin{align*} -\frac{2 \,{\left (3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - a^{2}\right )}}{3 \, d{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1\right )}^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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