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

Optimal. Leaf size=104 \[ \frac{b}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac{2 a b \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^2}-\frac{\log (1-\sin (c+d x))}{2 d (a+b)^2}+\frac{\log (\sin (c+d x)+1)}{2 d (a-b)^2} \]

[Out]

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

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Rubi [A]  time = 0.114654, antiderivative size = 104, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.158, Rules used = {2668, 710, 801} \[ \frac{b}{d \left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac{2 a b \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )^2}-\frac{\log (1-\sin (c+d x))}{2 d (a+b)^2}+\frac{\log (\sin (c+d x)+1)}{2 d (a-b)^2} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2668

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^m*(b^2 - x^2)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x] && Integer
Q[(p - 1)/2] && NeQ[a^2 - b^2, 0]

Rule 710

Int[((d_) + (e_.)*(x_))^(m_)/((a_) + (c_.)*(x_)^2), x_Symbol] :> Simp[(e*(d + e*x)^(m + 1))/((m + 1)*(c*d^2 +
a*e^2)), x] + Dist[c/(c*d^2 + a*e^2), Int[((d + e*x)^(m + 1)*(d - e*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d,
e, m}, x] && NeQ[c*d^2 + a*e^2, 0] && LtQ[m, -1]

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

Mathematica [A]  time = 0.205833, size = 102, normalized size = 0.98 \[ \frac{b \left (\frac{1}{\left (a^2-b^2\right ) (a+b \sin (c+d x))}-\frac{\log (1-\sin (c+d x))}{2 b (a+b)^2}+\frac{\log (\sin (c+d x)+1)}{2 b (a-b)^2}-\frac{2 a \log (a+b \sin (c+d x))}{(a-b)^2 (a+b)^2}\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(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 = 0.091, size = 101, normalized size = 1. \begin{align*}{\frac{b}{d \left ( a+b \right ) \left ( a-b \right ) \left ( a+b\sin \left ( dx+c \right ) \right ) }}-2\,{\frac{ab\ln \left ( a+b\sin \left ( dx+c \right ) \right ) }{d \left ( a+b \right ) ^{2} \left ( a-b \right ) ^{2}}}-{\frac{\ln \left ( \sin \left ( dx+c \right ) -1 \right ) }{2\,d \left ( a+b \right ) ^{2}}}+{\frac{\ln \left ( 1+\sin \left ( dx+c \right ) \right ) }{2\, \left ( a-b \right ) ^{2}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))^2,x)

[Out]

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

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Maxima [A]  time = 1.05678, size = 159, normalized size = 1.53 \begin{align*} -\frac{\frac{4 \, a b \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{4} - 2 \, a^{2} b^{2} + b^{4}} - \frac{2 \, b}{a^{3} - a b^{2} +{\left (a^{2} b - b^{3}\right )} \sin \left (d x + c\right )} - \frac{\log \left (\sin \left (d x + c\right ) + 1\right )}{a^{2} - 2 \, a b + b^{2}} + \frac{\log \left (\sin \left (d x + c\right ) - 1\right )}{a^{2} + 2 \, a b + b^{2}}}{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))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 3.20017, size = 443, normalized size = 4.26 \begin{align*} \frac{2 \, a^{2} b - 2 \, b^{3} - 4 \,{\left (a b^{2} \sin \left (d x + c\right ) + a^{2} b\right )} \log \left (b \sin \left (d x + c\right ) + a\right ) +{\left (a^{3} + 2 \, a^{2} b + a b^{2} +{\left (a^{2} b + 2 \, a b^{2} + b^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (a^{3} - 2 \, a^{2} b + a b^{2} +{\left (a^{2} b - 2 \, a b^{2} + 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))^2,x, algorithm="fricas")

[Out]

1/2*(2*a^2*b - 2*b^3 - 4*(a*b^2*sin(d*x + c) + a^2*b)*log(b*sin(d*x + c) + a) + (a^3 + 2*a^2*b + a*b^2 + (a^2*
b + 2*a*b^2 + b^3)*sin(d*x + c))*log(sin(d*x + c) + 1) - (a^3 - 2*a^2*b + a*b^2 + (a^2*b - 2*a*b^2 + 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{\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))**2,x)

[Out]

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

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Giac [A]  time = 1.14376, size = 198, normalized size = 1.9 \begin{align*} -\frac{\frac{4 \, a b^{2} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{4} b - 2 \, a^{2} b^{3} + b^{5}} - \frac{\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a^{2} - 2 \, a b + b^{2}} + \frac{\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a^{2} + 2 \, a b + b^{2}} - \frac{2 \,{\left (2 \, a b^{2} \sin \left (d x + c\right ) + 3 \, a^{2} b - 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))^2,x, algorithm="giac")

[Out]

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