3.3.12 \(\int \frac {\sin (c+d x)}{(a+b \sec (c+d x))^2} \, dx\) [212]

Optimal. Leaf size=57 \[ -\frac {\cos (c+d x)}{a^2 d}+\frac {b^2}{a^3 d (b+a \cos (c+d x))}+\frac {2 b \log (b+a \cos (c+d x))}{a^3 d} \]

[Out]

-cos(d*x+c)/a^2/d+b^2/a^3/d/(b+a*cos(d*x+c))+2*b*ln(b+a*cos(d*x+c))/a^3/d

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Rubi [A]
time = 0.08, antiderivative size = 57, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.210, Rules used = {3957, 2912, 12, 45} \begin {gather*} \frac {b^2}{a^3 d (a \cos (c+d x)+b)}+\frac {2 b \log (a \cos (c+d x)+b)}{a^3 d}-\frac {\cos (c+d x)}{a^2 d} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-(Cos[c + d*x]/(a^2*d)) + b^2/(a^3*d*(b + a*Cos[c + d*x])) + (2*b*Log[b + a*Cos[c + d*x]])/(a^3*d)

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 45

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d
*x)^n, x], x] /; FreeQ[{a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && IGtQ[m, 0] && ( !IntegerQ[n] || (EqQ[c, 0]
&& LeQ[7*m + 4*n + 4, 0]) || LtQ[9*m + 5*(n + 1), 0] || GtQ[m + n + 2, 0])

Rule 2912

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d/b)*x)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rule 3957

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[(g*Co
s[e + f*x])^p*((b + a*Sin[e + f*x])^m/Sin[e + f*x]^m), x] /; FreeQ[{a, b, e, f, g, p}, x] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {\sin (c+d x)}{(a+b \sec (c+d x))^2} \, dx &=\int \frac {\cos ^2(c+d x) \sin (c+d x)}{(-b-a \cos (c+d x))^2} \, dx\\ &=\frac {\text {Subst}\left (\int \frac {x^2}{a^2 (-b+x)^2} \, dx,x,-a \cos (c+d x)\right )}{a d}\\ &=\frac {\text {Subst}\left (\int \frac {x^2}{(-b+x)^2} \, dx,x,-a \cos (c+d x)\right )}{a^3 d}\\ &=\frac {\text {Subst}\left (\int \left (1+\frac {b^2}{(b-x)^2}-\frac {2 b}{b-x}\right ) \, dx,x,-a \cos (c+d x)\right )}{a^3 d}\\ &=-\frac {\cos (c+d x)}{a^2 d}+\frac {b^2}{a^3 d (b+a \cos (c+d x))}+\frac {2 b \log (b+a \cos (c+d x))}{a^3 d}\\ \end {align*}

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Mathematica [A]
time = 0.11, size = 76, normalized size = 1.33 \begin {gather*} \frac {-a^2 \cos ^2(c+d x)+a b \cos (c+d x) (-1+2 \log (b+a \cos (c+d x)))+b^2 (1+2 \log (b+a \cos (c+d x)))}{a^3 d (b+a \cos (c+d x))} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-(a^2*Cos[c + d*x]^2) + a*b*Cos[c + d*x]*(-1 + 2*Log[b + a*Cos[c + d*x]]) + b^2*(1 + 2*Log[b + a*Cos[c + d*x]
]))/(a^3*d*(b + a*Cos[c + d*x]))

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Maple [A]
time = 0.06, size = 67, normalized size = 1.18

method	result	size
derivativedivides	\(\frac {-\frac {1}{a^{2} \sec \left (d x +c \right )}-\frac {2 b \ln \left (\sec \left (d x +c \right )\right )}{a^{3}}-\frac {b}{a^{2} \left (a +b \sec \left (d x +c \right )\right )}+\frac {2 b \ln \left (a +b \sec \left (d x +c \right )\right )}{a^{3}}}{d}\)	\(67\)
default	\(\frac {-\frac {1}{a^{2} \sec \left (d x +c \right )}-\frac {2 b \ln \left (\sec \left (d x +c \right )\right )}{a^{3}}-\frac {b}{a^{2} \left (a +b \sec \left (d x +c \right )\right )}+\frac {2 b \ln \left (a +b \sec \left (d x +c \right )\right )}{a^{3}}}{d}\)	\(67\)
risch	\(-\frac {2 i b x}{a^{3}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 a^{2} d}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 a^{2} d}-\frac {4 i b c}{a^{3} d}+\frac {2 b^{2} {\mathrm e}^{i \left (d x +c \right )}}{a^{3} d \left (a \,{\mathrm e}^{2 i \left (d x +c \right )}+2 b \,{\mathrm e}^{i \left (d x +c \right )}+a \right )}+\frac {2 b \ln \left ({\mathrm e}^{2 i \left (d x +c \right )}+\frac {2 b \,{\mathrm e}^{i \left (d x +c \right )}}{a}+1\right )}{a^{3} d}\)	\(138\)
norman	\(\frac {\frac {2 a^{2}+4 b a +4 b^{2}}{2 a^{2} d b}-\frac {\left (2 a^{2}-4 b a +4 b^{2}\right ) \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{2 a^{2} d b}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )}-\frac {2 b \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3} d}+\frac {2 b \ln \left (a \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-b \left (\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-a -b \right )}{a^{3} d}\)	\(185\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)/(a+b*sec(d*x+c))^2,x,method=_RETURNVERBOSE)

[Out]

1/d*(-1/a^2/sec(d*x+c)-2/a^3*b*ln(sec(d*x+c))-b/a^2/(a+b*sec(d*x+c))+2/a^3*b*ln(a+b*sec(d*x+c)))

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Maxima [A]
time = 0.27, size = 55, normalized size = 0.96 \begin {gather*} \frac {\frac {b^{2}}{a^{4} \cos \left (d x + c\right ) + a^{3} b} - \frac {\cos \left (d x + c\right )}{a^{2}} + \frac {2 \, b \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{3}}}{d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)/(a+b*sec(d*x+c))^2,x, algorithm="maxima")

[Out]

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

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Fricas [A]
time = 2.69, size = 75, normalized size = 1.32 \begin {gather*} -\frac {a^{2} \cos \left (d x + c\right )^{2} + a b \cos \left (d x + c\right ) - b^{2} - 2 \, {\left (a b \cos \left (d x + c\right ) + b^{2}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{4} d \cos \left (d x + c\right ) + a^{3} b d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)/(a+b*sec(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin {\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{2}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [A]
time = 0.47, size = 61, normalized size = 1.07 \begin {gather*} -\frac {\cos \left (d x + c\right )}{a^{2} d} + \frac {2 \, b \log \left ({\left | -a \cos \left (d x + c\right ) - b \right |}\right )}{a^{3} d} + \frac {b^{2}}{{\left (a \cos \left (d x + c\right ) + b\right )} a^{3} d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)/(a+b*sec(d*x+c))^2,x, algorithm="giac")

[Out]

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

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Mupad [B]
time = 1.02, size = 60, normalized size = 1.05 \begin {gather*} \frac {b^2}{d\,\left (\cos \left (c+d\,x\right )\,a^4+b\,a^3\right )}-\frac {\cos \left (c+d\,x\right )}{a^2\,d}+\frac {2\,b\,\ln \left (b+a\,\cos \left (c+d\,x\right )\right )}{a^3\,d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)/(a + b/cos(c + d*x))^2,x)

[Out]

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

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