3.1548 \(\int \frac{\sec (c+d x) (A+B \sin (c+d x))}{a+b \sin (c+d x)} \, dx\)

Optimal. Leaf size=90 \[ -\frac{(A b-a B) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )}-\frac{(A+B) \log (1-\sin (c+d x))}{2 d (a+b)}+\frac{(A-B) \log (\sin (c+d x)+1)}{2 d (a-b)} \]

[Out]

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

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Rubi [A]  time = 0.14837, antiderivative size = 90, normalized size of antiderivative = 1., 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 = {2837, 801} \[ -\frac{(A b-a B) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )}-\frac{(A+B) \log (1-\sin (c+d x))}{2 d (a+b)}+\frac{(A-B) \log (\sin (c+d x)+1)}{2 d (a-b)} \]

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)*Log[1 - Sin[c + d*x]])/(2*(a + b)*d) + ((A - B)*Log[1 + Sin[c + d*x]])/(2*(a - b)*d) - ((A*b - a*B)*
Log[a + b*Sin[c + d*x]])/((a^2 - b^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 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

Rubi steps

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

Mathematica [A]  time = 0.177182, size = 99, normalized size = 1.1 \[ \frac{\frac{(a B-A b) \log (a+b \sin (c+d x))+(a+b) (A-B) \log \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )}{a-b}-(A+B) \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d (a+b)} \]

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

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Maple [A]  time = 0.076, size = 156, normalized size = 1.7 \begin{align*} -{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ) Ab}{d \left ( a+b \right ) \left ( a-b \right ) }}+{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ) aB}{d \left ( a+b \right ) \left ( a-b \right ) }}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) A}{d \left ( 2\,a+2\,b \right ) }}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) B}{d \left ( 2\,a+2\,b \right ) }}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) A}{d \left ( 2\,a-2\,b \right ) }}-{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) B}{d \left ( 2\,a-2\,b \right ) }} \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)/(a-b)*ln(a+b*sin(d*x+c))*A*b+1/d/(a+b)/(a-b)*ln(a+b*sin(d*x+c))*a*B-1/d/(2*a+2*b)*ln(sin(d*x+c)-1)*
A-1/d/(2*a+2*b)*ln(sin(d*x+c)-1)*B+1/d/(2*a-2*b)*ln(1+sin(d*x+c))*A-1/d/(2*a-2*b)*ln(1+sin(d*x+c))*B

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Maxima [A]  time = 0.966279, size = 107, normalized size = 1.19 \begin{align*} \frac{\frac{2 \,{\left (B a - A b\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{2} - b^{2}} + \frac{{\left (A - B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} - \frac{{\left (A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a + b}}{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*a - A*b)*log(b*sin(d*x + c) + a)/(a^2 - b^2) + (A - B)*log(sin(d*x + c) + 1)/(a - b) - (A + B)*log(s
in(d*x + c) - 1)/(a + b))/d

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Fricas [A]  time = 1.87924, size = 213, normalized size = 2.37 \begin{align*} \frac{2 \,{\left (B a - A b\right )} \log \left (b \sin \left (d x + c\right ) + a\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 \,{\left (a^{2} - b^{2}\right )} 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*a - A*b)*log(b*sin(d*x + c) + a) + ((A - B)*a + (A - B)*b)*log(sin(d*x + c) + 1) - ((A + B)*a - (A +
B)*b)*log(-sin(d*x + c) + 1))/((a^2 - b^2)*d)

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

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Giac [A]  time = 1.1869, size = 117, normalized size = 1.3 \begin{align*} \frac{\frac{2 \,{\left (B a b - A b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{2} b - b^{3}} + \frac{{\left (A - B\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a - b} - \frac{{\left (A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a + b}}{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*a*b - A*b^2)*log(abs(b*sin(d*x + c) + a))/(a^2*b - b^3) + (A - B)*log(abs(sin(d*x + c) + 1))/(a - b)
- (A + B)*log(abs(sin(d*x + c) - 1))/(a + b))/d