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]

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

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Rubi [A]  time = 0.15, antiderivative size = 90, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [A]  time = 0.19, size = 99, normalized size = 1.10 \[ \frac {\frac {(a B-A b) \log (a+b \sin (c+d x))+(a+b) (A-B) \log \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )}{a-b}-(A+B) \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d (a+b)} \]

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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fricas [A]  time = 0.52, size = 88, normalized size = 0.98 \[ \frac {2 \, {\left (B a - A b\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) + {\left ({\left (A - B\right )} a + {\left (A - B\right )} b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (A + B\right )} a - {\left (A + B\right )} b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{2} - b^{2}\right )} d} \]

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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giac [A]  time = 0.20, size = 87, normalized size = 0.97 \[ \frac {\frac {2 \, {\left (B a b - A b^{2}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{2} b - b^{3}} + \frac {{\left (A - B\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a - b} - \frac {{\left (A + B\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a + b}}{2 \, d} \]

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

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maple [A]  time = 0.44, size = 156, normalized size = 1.73 \[ -\frac {\ln \left (\sin \left (d x +c \right )-1\right ) A}{d \left (2 a +2 b \right )}-\frac {\ln \left (\sin \left (d x +c \right )-1\right ) B}{d \left (2 a +2 b \right )}-\frac {\ln \left (a +b \sin \left (d x +c \right )\right ) A b}{d \left (a +b \right ) \left (a -b \right )}+\frac {\ln \left (a +b \sin \left (d x +c \right )\right ) a B}{d \left (a +b \right ) \left (a -b \right )}+\frac {\ln \left (1+\sin \left (d x +c \right )\right ) A}{d \left (2 a -2 b \right )}-\frac {\ln \left (1+\sin \left (d x +c \right )\right ) B}{d \left (2 a -2 b \right )} \]

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/(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/(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(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 = 0.56, size = 79, normalized size = 0.88 \[ \frac {\frac {2 \, {\left (B a - A b\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{2} - b^{2}} + \frac {{\left (A - B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} - \frac {{\left (A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a + b}}{2 \, d} \]

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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mupad [B]  time = 0.31, size = 89, normalized size = 0.99 \[ \frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (\frac {A}{2}-\frac {B}{2}\right )}{d\,\left (a-b\right )}-\frac {\ln \left (a+b\,\sin \left (c+d\,x\right )\right )\,\left (A\,b-B\,a\right )}{d\,\left (a^2-b^2\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (\frac {A}{2}+\frac {B}{2}\right )}{d\,\left (a+b\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

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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