3.1532 \(\int \sec (c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=64 \[ -\frac{(a+b) (A+B) \log (1-\sin (c+d x))}{2 d}+\frac{(a-b) (A-B) \log (\sin (c+d x)+1)}{2 d}-\frac{b B \sin (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.108234, antiderivative size = 64, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.148, Rules used = {2837, 774, 633, 31} \[ -\frac{(a+b) (A+B) \log (1-\sin (c+d x))}{2 d}+\frac{(a-b) (A-B) \log (\sin (c+d x)+1)}{2 d}-\frac{b B \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2837

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*(c + (d*x)/b)^n*(b^2 - x^2)^((p - 1)/2), x], x
, b*Sin[e + f*x]], x] /; FreeQ[{a, b, c, d, e, f, m, n}, x] && IntegerQ[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 774

Int[(((d_.) + (e_.)*(x_))*((f_) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*g*x)/c, x] + Dist[1
/c, Int[(c*d*f - a*e*g + c*(e*f + d*g)*x)/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rubi steps

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

Mathematica [A]  time = 0.0294744, size = 68, normalized size = 1.06 \[ \frac{a A \tanh ^{-1}(\sin (c+d x))}{d}-\frac{a B \log (\cos (c+d x))}{d}-\frac{A b \log (\cos (c+d x))}{d}-\frac{b B \sin (c+d x)}{d}+\frac{b B \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.063, size = 83, normalized size = 1.3 \begin{align*} -{\frac{Ab\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{aA\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-{\frac{Bb\sin \left ( dx+c \right ) }{d}}-{\frac{aB\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}+{\frac{Bb\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+b*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

-1/d*A*b*ln(cos(d*x+c))+1/d*a*A*ln(sec(d*x+c)+tan(d*x+c))-b*B*sin(d*x+c)/d-1/d*a*B*ln(cos(d*x+c))+1/d*B*b*ln(s
ec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 0.981998, size = 86, normalized size = 1.34 \begin{align*} -\frac{2 \, B b \sin \left (d x + c\right ) -{\left ({\left (A - B\right )} a -{\left (A - B\right )} b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left ({\left (A + B\right )} a +{\left (A + B\right )} b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.50181, size = 170, normalized size = 2.66 \begin{align*} -\frac{2 \, B b \sin \left (d x + c\right ) -{\left ({\left (A - B\right )} a -{\left (A - B\right )} b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left ({\left (A + B\right )} a +{\left (A + B\right )} b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]  time = 1.28562, size = 90, normalized size = 1.41 \begin{align*} -\frac{2 \, B b \sin \left (d x + c\right ) -{\left (A a - B a - A b + B b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) +{\left (A a + B a + A b + B b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{2 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*B*b*sin(d*x + c) - (A*a - B*a - A*b + B*b)*log(abs(sin(d*x + c) + 1)) + (A*a + B*a + A*b + B*b)*log(ab
s(sin(d*x + c) - 1)))/d