3.957 \(\int \sec (c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=34 \[ -\frac {a (A+B) \log (1-\sin (c+d x))}{d}-\frac {a B \sin (c+d x)}{d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.07, antiderivative size = 34, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2836, 43} \[ -\frac {a (A+B) \log (1-\sin (c+d x))}{d}-\frac {a B \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 43

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

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

________________________________________________________________________________________

Mathematica [A]  time = 0.03, size = 68, normalized size = 2.00 \[ \frac {a A \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a A \log (\cos (c+d x))}{d}-\frac {a B \sin (c+d x)}{d}+\frac {a B \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a B \log (\cos (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*A*ArcTanh[Sin[c + d*x]])/d + (a*B*ArcTanh[Sin[c + d*x]])/d - (a*A*Log[Cos[c + d*x]])/d - (a*B*Log[Cos[c + d
*x]])/d - (a*B*Sin[c + d*x])/d

________________________________________________________________________________________

fricas [A]  time = 0.84, size = 31, normalized size = 0.91 \[ -\frac {{\left (A + B\right )} a \log \left (-\sin \left (d x + c\right ) + 1\right ) + B a \sin \left (d x + c\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

giac [B]  time = 0.17, size = 114, normalized size = 3.35 \[ \frac {{\left (A a + B a\right )} \log \left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right ) - 2 \, {\left (A a + B a\right )} \log \left ({\left | \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) - \frac {A a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 2 \, B a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + A a + B a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maple [A]  time = 0.36, size = 47, normalized size = 1.38 \[ -\frac {a \ln \left (\sin \left (d x +c \right )-1\right ) A}{d}-\frac {a B \sin \left (d x +c \right )}{d}-\frac {a \ln \left (\sin \left (d x +c \right )-1\right ) B}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.36, size = 29, normalized size = 0.85 \[ -\frac {{\left (A + B\right )} a \log \left (\sin \left (d x + c\right ) - 1\right ) + B a \sin \left (d x + c\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.07, size = 35, normalized size = 1.03 \[ -\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (A\,a+B\,a\right )}{d}-\frac {B\,a\,\sin \left (c+d\,x\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________