3.4.9 \(\int \frac {\sec (c+d x)}{a+b \sin (c+d x)} \, dx\) [309]

Optimal. Leaf size=75 \[ -\frac {\log (1-\sin (c+d x))}{2 (a+b) d}+\frac {\log (1+\sin (c+d x))}{2 (a-b) d}-\frac {b \log (a+b \sin (c+d x))}{\left (a^2-b^2\right ) d} \]

[Out]

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

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Rubi [A]
time = 0.06, antiderivative size = 75, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {2747, 720, 31, 647} \begin {gather*} -\frac {b \log (a+b \sin (c+d x))}{d \left (a^2-b^2\right )}-\frac {\log (1-\sin (c+d x))}{2 d (a+b)}+\frac {\log (\sin (c+d x)+1)}{2 d (a-b)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 647

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[(-a)*c, 2]}, Dist[e/2 + c*(d/(2*q)),
Int[1/(-q + c*x), x], x] + Dist[e/2 - c*(d/(2*q)), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[(-a)*c]

Rule 720

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

Rule 2747

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]

Rubi steps

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

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Mathematica [A]
time = 0.04, size = 64, normalized size = 0.85 \begin {gather*} \frac {(-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 (a-b) (a+b) d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-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*(a - b)*(a +
b)*d)

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Maple [A]
time = 0.09, size = 71, normalized size = 0.95

method	result	size
derivativedivides	\(\frac {\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}-\frac {b \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a -b \right ) \left (a +b \right )}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}}{d}\)	\(71\)
default	\(\frac {\frac {\ln \left (1+\sin \left (d x +c \right )\right )}{2 a -2 b}-\frac {b \ln \left (a +b \sin \left (d x +c \right )\right )}{\left (a -b \right ) \left (a +b \right )}-\frac {\ln \left (\sin \left (d x +c \right )-1\right )}{2 a +2 b}}{d}\)	\(71\)
norman	\(\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )+1\right )}{d \left (a -b \right )}-\frac {\ln \left (\tan \left (\frac {d x}{2}+\frac {c}{2}\right )-1\right )}{d \left (a +b \right )}-\frac {b \ln \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+2 b \tan \left (\frac {d x}{2}+\frac {c}{2}\right )+a \right )}{d \left (a^{2}-b^{2}\right )}\)	\(92\)
risch	\(\frac {i x}{a +b}+\frac {i c}{d \left (a +b \right )}-\frac {i x}{a -b}-\frac {i c}{d \left (a -b \right )}+\frac {2 i b x}{a^{2}-b^{2}}+\frac {2 i b c}{d \left (a^{2}-b^{2}\right )}-\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}-i\right )}{d \left (a +b \right )}+\frac {\ln \left ({\mathrm e}^{i \left (d x +c \right )}+i\right )}{d \left (a -b \right )}-\frac {b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 i a \,{\mathrm e}^{i \left (d x +c \right )}}{b}-1\right )}{d \left (a^{2}-b^{2}\right )}\)	\(175\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)/(a+b*sin(d*x+c)),x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.30, size = 64, normalized size = 0.85 \begin {gather*} -\frac {\frac {2 \, b \log \left (b \sin \left (d x + c\right ) + a\right )}{a^{2} - b^{2}} - \frac {\log \left (\sin \left (d x + c\right ) + 1\right )}{a - b} + \frac {\log \left (\sin \left (d x + c\right ) - 1\right )}{a + b}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 0.36, size = 62, normalized size = 0.83 \begin {gather*} -\frac {2 \, b \log \left (b \sin \left (d x + c\right ) + a\right ) - {\left (a + b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left (a - b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, {\left (a^{2} - b^{2}\right )} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec {\left (c + d x \right )}}{a + b \sin {\left (c + d x \right )}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 4.14, size = 71, normalized size = 0.95 \begin {gather*} -\frac {\frac {2 \, b^{2} \log \left ({\left | b \sin \left (d x + c\right ) + a \right |}\right )}{a^{2} b - b^{3}} - \frac {\log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right )}{a - b} + \frac {\log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{a + b}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)/(a+b*sin(d*x+c)),x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 0.20, size = 69, normalized size = 0.92 \begin {gather*} \frac {\ln \left (\sin \left (c+d\,x\right )+1\right )}{2\,d\,\left (a-b\right )}-\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )}{2\,d\,\left (a+b\right )}-\frac {b\,\ln \left (a+b\,\sin \left (c+d\,x\right )\right )}{d\,\left (a^2-b^2\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)*(a + b*sin(c + d*x))),x)

[Out]

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

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