3.1556 \(\int \frac{\sec (c+d x) (A+B \sin (c+d x))}{(a+b \sin (c+d x))^2} \, dx\)

Optimal. Leaf size=135 \[ \frac{A b-a B}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac{\left (a^2 (-B)+2 a A b-b^2 B\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^2}-\frac{(A+B) \log (1-\sin (c+d x))}{2 d (a+b)^2}+\frac{(A-B) \log (\sin (c+d x)+1)}{2 d (a-b)^2} \]

[Out]

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

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Rubi [A]  time = 0.193716, antiderivative size = 135, 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} \[ \frac{A b-a B}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac{\left (a^2 (-B)+2 a A b-b^2 B\right ) \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^2}-\frac{(A+B) \log (1-\sin (c+d x))}{2 d (a+b)^2}+\frac{(A-B) \log (\sin (c+d x)+1)}{2 d (a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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

Mathematica [A]  time = 1.27208, size = 178, normalized size = 1.32 \[ \frac{b \left (A-\frac{a B}{b}\right ) \left (\frac{1}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac{\log (1-\sin (c+d x))}{2 b (a+b)^2}+\frac{\log (\sin (c+d x)+1)}{2 b (a-b)^2}-\frac{2 a \log (a+b \sin (c+d x))}{(a-b)^2 (a+b)^2}\right )-\frac{B ((b-a) \log (1-\sin (c+d x))+(a+b) \log (\sin (c+d x)+1)-2 b \log (a+b \sin (c+d x)))}{2 b (b-a) (a+b)}}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [A]  time = 0.135, size = 240, normalized size = 1.8 \begin{align*}{\frac{Ab}{d \left ( a+b \right ) \left ( a-b \right ) \left ( a+b\sin \left ( dx+c \right ) \right ) }}-{\frac{aB}{d \left ( a+b \right ) \left ( a-b \right ) \left ( a+b\sin \left ( dx+c \right ) \right ) }}-2\,{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ) Aab}{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}}}+{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ) B{a}^{2}}{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}}}+{\frac{\ln \left ( a+b\sin \left ( dx+c \right ) \right ) B{b}^{2}}{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}}}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) A}{2\,d \left ( a+b \right ) ^{2}}}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) B}{2\,d \left ( a+b \right ) ^{2}}}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) A}{2\,d \left ( a-b \right ) ^{2}}}-{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) B}{2\,d \left ( a-b \right ) ^{2}}} \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))^2,x)

[Out]

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

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Maxima [A]  time = 0.99445, size = 198, normalized size = 1.47 \begin{align*} \frac{\frac{2 \,{\left (B a^{2} - 2 \, A a b + B b^{2}\right )} \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} + \frac{{\left (A - B\right )} \log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} - \frac{{\left (A + B\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}} - \frac{2 \,{\left (B a - A b\right )}}{a^{3} - a b^{2} +{\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\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))^2,x, algorithm="maxima")

[Out]

1/2*(2*(B*a^2 - 2*A*a*b + B*b^2)*log(b*sin(d*x + c) + a)/(a^4 - 2*a^2*b^2 + b^4) + (A - B)*log(sin(d*x + c) +
1)/(a^2 - 2*a*b + b^2) - (A + B)*log(sin(d*x + c) - 1)/(a^2 + 2*a*b + b^2) - 2*(B*a - A*b)/(a^3 - a*b^2 + (a^2
*b - b^3)*sin(d*x + c)))/d

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Fricas [B]  time = 3.37979, size = 671, normalized size = 4.97 \begin{align*} -\frac{2 \, B a^{3} - 2 \, A a^{2} b - 2 \, B a b^{2} + 2 \, A b^{3} - 2 \,{\left (B a^{3} - 2 \, A a^{2} b + B a b^{2} +{\left (B a^{2} b - 2 \, A a b^{2} + B b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) -{\left ({\left (A - B\right )} a^{3} + 2 \,{\left (A - B\right )} a^{2} b +{\left (A - B\right )} a b^{2} +{\left ({\left (A - B\right )} a^{2} b + 2 \,{\left (A - B\right )} a b^{2} +{\left (A - B\right )} b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left ({\left (A + B\right )} a^{3} - 2 \,{\left (A + B\right )} a^{2} b +{\left (A + B\right )} a b^{2} +{\left ({\left (A + B\right )} a^{2} b - 2 \,{\left (A + B\right )} a b^{2} +{\left (A + B\right )} b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \,{\left ({\left (a^{4} b - 2 \, a^{2} b^{3} + b^{5}\right )} d \sin \left (d x + c\right ) +{\left (a^{5} - 2 \, a^{3} b^{2} + a b^{4}\right )} d\right )}} \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))^2,x, algorithm="fricas")

[Out]

-1/2*(2*B*a^3 - 2*A*a^2*b - 2*B*a*b^2 + 2*A*b^3 - 2*(B*a^3 - 2*A*a^2*b + B*a*b^2 + (B*a^2*b - 2*A*a*b^2 + B*b^
3)*sin(d*x + c))*log(b*sin(d*x + c) + a) - ((A - B)*a^3 + 2*(A - B)*a^2*b + (A - B)*a*b^2 + ((A - B)*a^2*b + 2
*(A - B)*a*b^2 + (A - B)*b^3)*sin(d*x + c))*log(sin(d*x + c) + 1) + ((A + B)*a^3 - 2*(A + B)*a^2*b + (A + B)*a
*b^2 + ((A + B)*a^2*b - 2*(A + B)*a*b^2 + (A + B)*b^3)*sin(d*x + c))*log(-sin(d*x + c) + 1))/((a^4*b - 2*a^2*b
^3 + b^5)*d*sin(d*x + c) + (a^5 - 2*a^3*b^2 + a*b^4)*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 )}}{\left (a + b \sin{\left (c + d x \right )}\right )^{2}}\, 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))**2,x)

[Out]

Integral((A + B*sin(c + d*x))*sec(c + d*x)/(a + b*sin(c + d*x))**2, x)

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Giac [A]  time = 1.33868, size = 277, normalized size = 2.05 \begin{align*} \frac{\frac{2 \,{\left (B a^{2} b - 2 \, A a b^{2} + B b^{3}\right )} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} - \frac{{\left (A + B\right )} \log \left ({\left | -\sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} + 2 \, a b + b^{2}} + \frac{{\left (A - B\right )} \log \left ({\left | -\sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} - \frac{2 \,{\left (B a^{2} b \sin \left (d x + c\right ) - 2 \, A a b^{2} \sin \left (d x + c\right ) + B b^{3} \sin \left (d x + c\right ) + 2 \, B a^{3} - 3 \, A a^{2} b + A b^{3}\right )}}{{\left (a^{4} - 2 \, a^{2} b^{2} + b^{4}\right )}{\left (b \sin \left (d x + c\right ) + a\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))^2,x, algorithm="giac")

[Out]

1/2*(2*(B*a^2*b - 2*A*a*b^2 + B*b^3)*log(abs(b*sin(d*x + c) + a))/(a^4*b - 2*a^2*b^3 + b^5) - (A + B)*log(abs(
-sin(d*x + c) + 1))/(a^2 + 2*a*b + b^2) + (A - B)*log(abs(-sin(d*x + c) - 1))/(a^2 - 2*a*b + b^2) - 2*(B*a^2*b
*sin(d*x + c) - 2*A*a*b^2*sin(d*x + c) + B*b^3*sin(d*x + c) + 2*B*a^3 - 3*A*a^2*b + A*b^3)/((a^4 - 2*a^2*b^2 +
b^4)*(b*sin(d*x + c) + a)))/d