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

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Rubi [A]  time = 0.0695984, antiderivative size = 34, normalized size of antiderivative = 1., 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 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 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])

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 = 0.0337236, size = 68, normalized size = 2. \[ \frac{a A \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a A \log (\cos (c+d x))}{d}-\frac{a B \sin (c+d x)}{d}+\frac{a B \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a B \log (\cos (c+d x))}{d} \]

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

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Maple [A]  time = 0.065, size = 47, normalized size = 1.4 \begin{align*} -{\frac{a\ln \left ( \sin \left ( dx+c \right ) -1 \right ) A}{d}}-{\frac{aB\sin \left ( dx+c \right ) }{d}}-{\frac{a\ln \left ( \sin \left ( dx+c \right ) -1 \right ) B}{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))*(A+B*sin(d*x+c)),x)

[Out]

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

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Maxima [A]  time = 1.01055, size = 39, normalized size = 1.15 \begin{align*} -\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} \end{align*}

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

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Fricas [A]  time = 1.77425, size = 78, normalized size = 2.29 \begin{align*} -\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} \end{align*}

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

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} 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 ) \end{align*}

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))

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Giac [B]  time = 1.37615, size = 154, normalized size = 4.53 \begin{align*} \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} \end{align*}

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