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

Optimal. Leaf size=64 \[ \frac{a^4}{4 d (a-a \sin (c+d x))^2}-\frac{3 a^3}{4 d (a-a \sin (c+d x))}+\frac{a^2 \tanh ^{-1}(\sin (c+d x))}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.11848, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {2836, 12, 88, 206} \[ \frac{a^4}{4 d (a-a \sin (c+d x))^2}-\frac{3 a^3}{4 d (a-a \sin (c+d x))}+\frac{a^2 \tanh ^{-1}(\sin (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2836

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

Rule 12

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

Rule 88

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandI
ntegrand[(a + b*x)^m*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, p}, x] && IntegersQ[m, n] &&
(IntegerQ[p] || (GtQ[m, 0] && GeQ[n, -1]))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

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

Mathematica [A]  time = 0.10705, size = 39, normalized size = 0.61 \[ \frac{a^2 \left (\frac{3 \sin (c+d x)-2}{(\sin (c+d x)-1)^2}+\tanh ^{-1}(\sin (c+d x))\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.075, size = 174, normalized size = 2.7 \begin{align*}{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}{a}^{2}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}-{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}{a}^{2}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}-{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d}}-{\frac{{a}^{2}\sin \left ( dx+c \right ) }{4\,d}}+{\frac{{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}{a}^{2}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*sin(d*x+c)^2*(a+a*sin(d*x+c))^2,x)

[Out]

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

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Maxima [A]  time = 1.08203, size = 97, normalized size = 1.52 \begin{align*} \frac{a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) + \frac{2 \,{\left (3 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{8 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/8*(a^2*log(sin(d*x + c) + 1) - a^2*log(sin(d*x + c) - 1) + 2*(3*a^2*sin(d*x + c) - 2*a^2)/(sin(d*x + c)^2 -
2*sin(d*x + c) + 1))/d

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Fricas [B]  time = 1.38907, size = 308, normalized size = 4.81 \begin{align*} -\frac{6 \, a^{2} \sin \left (d x + c\right ) - 4 \, a^{2} -{\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{8 \,{\left (d \cos \left (d x + c\right )^{2} + 2 \, d \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)^5*sin(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

-1/8*(6*a^2*sin(d*x + c) - 4*a^2 - (a^2*cos(d*x + c)^2 + 2*a^2*sin(d*x + c) - 2*a^2)*log(sin(d*x + c) + 1) + (
a^2*cos(d*x + c)^2 + 2*a^2*sin(d*x + c) - 2*a^2)*log(-sin(d*x + c) + 1))/(d*cos(d*x + c)^2 + 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)**5*sin(d*x+c)**2*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.26176, size = 104, normalized size = 1.62 \begin{align*} \frac{2 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 2 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac{3 \, a^{2} \sin \left (d x + c\right )^{2} + 6 \, a^{2} \sin \left (d x + c\right ) - 5 \, a^{2}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*sin(d*x+c)^2*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/16*(2*a^2*log(abs(sin(d*x + c) + 1)) - 2*a^2*log(abs(sin(d*x + c) - 1)) + (3*a^2*sin(d*x + c)^2 + 6*a^2*sin(
d*x + c) - 5*a^2)/(sin(d*x + c) - 1)^2)/d