3.973 \(\int \sec (c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=60 \[ -\frac{a^2 (A+B) \sin (c+d x)}{d}-\frac{2 a^2 (A+B) \log (1-\sin (c+d x))}{d}-\frac{B (a \sin (c+d x)+a)^2}{2 d} \]

[Out]

(-2*a^2*(A + B)*Log[1 - Sin[c + d*x]])/d - (a^2*(A + B)*Sin[c + d*x])/d - (B*(a + a*Sin[c + d*x])^2)/(2*d)

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Rubi [A]  time = 0.0962114, antiderivative size = 60, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.069, Rules used = {2836, 77} \[ -\frac{a^2 (A+B) \sin (c+d x)}{d}-\frac{2 a^2 (A+B) \log (1-\sin (c+d x))}{d}-\frac{B (a \sin (c+d x)+a)^2}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*(A + B)*Log[1 - Sin[c + d*x]])/d - (a^2*(A + B)*Sin[c + d*x])/d - (B*(a + a*Sin[c + d*x])^2)/(2*d)

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 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rubi steps

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

Mathematica [A]  time = 0.0786707, size = 51, normalized size = 0.85 \[ \frac{a \left (-a (A+2 B) \sin (c+d x)-2 a (A+B) \log (1-\sin (c+d x))-\frac{1}{2} a B \sin ^2(c+d x)\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(-2*a*(A + B)*Log[1 - Sin[c + d*x]] - a*(A + 2*B)*Sin[c + d*x] - (a*B*Sin[c + d*x]^2)/2))/d

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Maple [B]  time = 0.075, size = 127, normalized size = 2.1 \begin{align*} -{\frac{{a}^{2}A\sin \left ( dx+c \right ) }{d}}+2\,{\frac{{a}^{2}A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-{\frac{B{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-2\,{\frac{B{a}^{2}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-2\,{\frac{{a}^{2}A\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-2\,{\frac{B{a}^{2}\sin \left ( dx+c \right ) }{d}}+2\,{\frac{B{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.01459, size = 70, normalized size = 1.17 \begin{align*} -\frac{B a^{2} \sin \left (d x + c\right )^{2} + 4 \,{\left (A + B\right )} a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) + 2 \,{\left (A + 2 \, B\right )} a^{2} \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(B*a^2*sin(d*x + c)^2 + 4*(A + B)*a^2*log(sin(d*x + c) - 1) + 2*(A + 2*B)*a^2*sin(d*x + c))/d

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Fricas [A]  time = 1.84076, size = 135, normalized size = 2.25 \begin{align*} \frac{B a^{2} \cos \left (d x + c\right )^{2} - 4 \,{\left (A + B\right )} a^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (A + 2 \, B\right )} a^{2} \sin \left (d x + c\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(B*a^2*cos(d*x + c)^2 - 4*(A + B)*a^2*log(-sin(d*x + c) + 1) - 2*(A + 2*B)*a^2*sin(d*x + c))/d

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} a^{2} \left (\int A \sec{\left (c + d x \right )}\, dx + \int 2 A \sin{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int A \sin ^{2}{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int B \sin{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int 2 B \sin ^{2}{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx + \int B \sin ^{3}{\left (c + d x \right )} \sec{\left (c + d x \right )}\, dx\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+a*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)

[Out]

a**2*(Integral(A*sec(c + d*x), x) + Integral(2*A*sin(c + d*x)*sec(c + d*x), x) + Integral(A*sin(c + d*x)**2*se
c(c + d*x), x) + Integral(B*sin(c + d*x)*sec(c + d*x), x) + Integral(2*B*sin(c + d*x)**2*sec(c + d*x), x) + In
tegral(B*sin(c + d*x)**3*sec(c + d*x), x))

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Giac [B]  time = 1.34599, size = 297, normalized size = 4.95 \begin{align*} \frac{2 \,{\left (A a^{2} + B a^{2}\right )} \log \left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right ) - 4 \,{\left (A a^{2} + B a^{2}\right )} \log \left ({\left | \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1 \right |}\right ) - \frac{3 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 3 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} + 2 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 6 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 8 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 4 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, A a^{2} + 3 \, B a^{2}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2}}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(2*(A*a^2 + B*a^2)*log(tan(1/2*d*x + 1/2*c)^2 + 1) - 4*(A*a^2 + B*a^2)*log(abs(tan(1/2*d*x + 1/2*c) - 1)) - (3
*A*a^2*tan(1/2*d*x + 1/2*c)^4 + 3*B*a^2*tan(1/2*d*x + 1/2*c)^4 + 2*A*a^2*tan(1/2*d*x + 1/2*c)^3 + 4*B*a^2*tan(
1/2*d*x + 1/2*c)^3 + 6*A*a^2*tan(1/2*d*x + 1/2*c)^2 + 8*B*a^2*tan(1/2*d*x + 1/2*c)^2 + 2*A*a^2*tan(1/2*d*x + 1
/2*c) + 4*B*a^2*tan(1/2*d*x + 1/2*c) + 3*A*a^2 + 3*B*a^2)/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d