3.22 \(\int \sec ^4(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=63 \[ \frac{a^4 \cos (c+d x)}{3 d \left (a^2-a^2 \sin (c+d x)\right )}+\frac{a^4 \cos (c+d x)}{3 d (a-a \sin (c+d x))^2} \]

[Out]

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

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Rubi [A]  time = 0.0691645, antiderivative size = 63, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.143, Rules used = {2670, 2650, 2648} \[ \frac{a^4 \cos (c+d x)}{3 d \left (a^2-a^2 \sin (c+d x)\right )}+\frac{a^4 \cos (c+d x)}{3 d (a-a \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]

[Out]

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

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 2650

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

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

Mathematica [A]  time = 0.0086958, size = 58, normalized size = 0.92 \[ -\frac{a^2 \tan ^3(c+d x)}{3 d}+\frac{2 a^2 \sec ^3(c+d x)}{3 d}+\frac{a^2 \tan (c+d x) \sec ^2(c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + a*Sin[c + d*x])^2,x]

[Out]

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

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Maple [A]  time = 0.065, size = 63, normalized size = 1. \begin{align*}{\frac{1}{d} \left ({\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{2\,{a}^{2}}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}-{a}^{2} \left ( -{\frac{2}{3}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) \tan \left ( dx+c \right ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.953012, size = 70, normalized size = 1.11 \begin{align*} \frac{a^{2} \tan \left (d x + c\right )^{3} +{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} + \frac{2 \, 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*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.67741, size = 235, normalized size = 3.73 \begin{align*} -\frac{a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \cos \left (d x + c\right ) + a^{2} -{\left (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*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

[Out]

Timed out

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Giac [A]  time = 1.2081, size = 73, normalized size = 1.16 \begin{align*} -\frac{2 \,{\left (3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, 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*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

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