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

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Rubi [A]  time = 0.10, antiderivative size = 60, normalized size of antiderivative = 1.00, 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 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]))))

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]

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*}

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Mathematica [A]  time = 0.08, 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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fricas [A]  time = 0.79, size = 54, normalized size = 0.90 \[ \frac {B a^{2} \cos \left (d x + c\right )^{2} - 4 \, {\left (A + B\right )} a^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (A + 2 \, B\right )} a^{2} \sin \left (d x + c\right )}{2 \, d} \]

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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giac [B]  time = 0.20, size = 220, normalized size = 3.67 \[ \frac {2 \, {\left (A a^{2} + B a^{2}\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 4 \, {\left (A a^{2} + B a^{2}\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {3 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 3 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 2 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 6 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 8 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, A a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, B a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 3 \, A a^{2} + 3 \, B a^{2}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{d} \]

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

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maple [B]  time = 0.37, size = 127, normalized size = 2.12 \[ -\frac {a^{2} A \sin \left (d x +c \right )}{d}+\frac {2 a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-\frac {B \,a^{2} \left (\sin ^{2}\left (d x +c \right )\right )}{2 d}-\frac {2 B \,a^{2} \ln \left (\cos \left (d x +c \right )\right )}{d}-\frac {2 a^{2} A \ln \left (\cos \left (d x +c \right )\right )}{d}-\frac {2 B \,a^{2} \sin \left (d x +c \right )}{d}+\frac {2 B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d} \]

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 = 0.31, size = 52, normalized size = 0.87 \[ -\frac {B a^{2} \sin \left (d x + c\right )^{2} + 4 \, {\left (A + B\right )} a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) + 2 \, {\left (A + 2 \, B\right )} a^{2} \sin \left (d x + c\right )}{2 \, d} \]

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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mupad [B]  time = 0.08, size = 63, normalized size = 1.05 \[ -\frac {\sin \left (c+d\,x\right )\,\left (a^2\,\left (A+B\right )+B\,a^2\right )+\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (2\,A\,a^2+2\,B\,a^2\right )+\frac {B\,a^2\,{\sin \left (c+d\,x\right )}^2}{2}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{2} \left (\int A \sec {\left (c + d x \right )}\, dx + \int 2 A \sin {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int A \sin ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \sin {\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int 2 B \sin ^{2}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx + \int B \sin ^{3}{\left (c + d x \right )} \sec {\left (c + d x \right )}\, dx\right ) \]

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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