3.1533 \(\int \sec ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=59 \[ \frac{(a A-b B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{\sec ^2(c+d x) ((a A+b B) \sin (c+d x)+a B+A b)}{2 d} \]

[Out]

((a*A - b*B)*ArcTanh[Sin[c + d*x]])/(2*d) + (Sec[c + d*x]^2*(A*b + a*B + (a*A + b*B)*Sin[c + d*x]))/(2*d)

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Rubi [A]  time = 0.0757744, antiderivative size = 59, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.103, Rules used = {2837, 778, 206} \[ \frac{(a A-b B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{\sec ^2(c+d x) ((a A+b B) \sin (c+d x)+a B+A b)}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + b*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]

[Out]

((a*A - b*B)*ArcTanh[Sin[c + d*x]])/(2*d) + (Sec[c + d*x]^2*(A*b + a*B + (a*A + b*B)*Sin[c + d*x]))/(2*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 778

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*(e*f + d*g) -
(c*d*f - a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)),
Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1]

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rubi steps

\begin{align*} \int \sec ^3(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx &=\frac{b^3 \operatorname{Subst}\left (\int \frac{(a+x) \left (A+\frac{B x}{b}\right )}{\left (b^2-x^2\right )^2} \, dx,x,b \sin (c+d x)\right )}{d}\\ &=\frac{\sec ^2(c+d x) (A b+a B+(a A+b B) \sin (c+d x))}{2 d}+\frac{(b (a A-b B)) \operatorname{Subst}\left (\int \frac{1}{b^2-x^2} \, dx,x,b \sin (c+d x)\right )}{2 d}\\ &=\frac{(a A-b B) \tanh ^{-1}(\sin (c+d x))}{2 d}+\frac{\sec ^2(c+d x) (A b+a B+(a A+b B) \sin (c+d x))}{2 d}\\ \end{align*}

Mathematica [A]  time = 0.209759, size = 54, normalized size = 0.92 \[ \frac{(a A-b B) \tanh ^{-1}(\sin (c+d x))+\sec ^2(c+d x) ((a A+b B) \sin (c+d x)+a B+A b)}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a + b*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]

[Out]

((a*A - b*B)*ArcTanh[Sin[c + d*x]] + Sec[c + d*x]^2*(A*b + a*B + (a*A + b*B)*Sin[c + d*x]))/(2*d)

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Maple [B]  time = 0.077, size = 129, normalized size = 2.2 \begin{align*}{\frac{aA\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{2\,d}}+{\frac{aA\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}}+{\frac{aB}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{Ab}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{B \left ( \sin \left ( dx+c \right ) \right ) ^{3}b}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{Bb\sin \left ( dx+c \right ) }{2\,d}}-{\frac{Bb\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{2\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

1/2/d*a*A*sec(d*x+c)*tan(d*x+c)+1/2/d*a*A*ln(sec(d*x+c)+tan(d*x+c))+1/2/d*a*B/cos(d*x+c)^2+1/2/d*A*b/cos(d*x+c
)^2+1/2/d*B*b*sin(d*x+c)^3/cos(d*x+c)^2+1/2*b*B*sin(d*x+c)/d-1/2/d*B*b*ln(sec(d*x+c)+tan(d*x+c))

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Maxima [A]  time = 0.989002, size = 105, normalized size = 1.78 \begin{align*} \frac{{\left (A a - B b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (A a - B b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left (B a + A b +{\left (A a + B b\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 1.43386, size = 232, normalized size = 3.93 \begin{align*} \frac{{\left (A a - B b\right )} \cos \left (d x + c\right )^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (A a - B b\right )} \cos \left (d x + c\right )^{2} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, B a + 2 \, A b + 2 \,{\left (A a + B b\right )} \sin \left (d x + c\right )}{4 \, d \cos \left (d x + c\right )^{2}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.30888, size = 113, normalized size = 1.92 \begin{align*} \frac{{\left (A a - B b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) -{\left (A a - B b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (A a \sin \left (d x + c\right ) + B b \sin \left (d x + c\right ) + B a + A b\right )}}{\sin \left (d x + c\right )^{2} - 1}}{4 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+b*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

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