3.1.96 \(\int \frac {\sin (c+d x)}{(a+a \sec (c+d x))^3} \, dx\) [96]

Optimal. Leaf size=75 \[ -\frac {\cos (c+d x)}{a^3 d}-\frac {1}{2 a d (a+a \cos (c+d x))^2}+\frac {3}{d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {3 \log (1+\cos (c+d x))}{a^3 d} \]

[Out]

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

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Rubi [A]
time = 0.09, antiderivative size = 75, 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 {\cos (c+d x)}{a^3 d}+\frac {3}{d \left (a^3 \cos (c+d x)+a^3\right )}+\frac {3 \log (\cos (c+d x)+1)}{a^3 d}-\frac {1}{2 a d (a \cos (c+d x)+a)^2} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Sin[c + d*x]/(a + a*Sec[c + d*x])^3,x]

[Out]

-(Cos[c + d*x]/(a^3*d)) - 1/(2*a*d*(a + a*Cos[c + d*x])^2) + 3/(d*(a^3 + a^3*Cos[c + d*x])) + (3*Log[1 + 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+a \sec (c+d x))^3} \, dx &=-\int \frac {\cos ^3(c+d x) \sin (c+d x)}{(-a-a \cos (c+d x))^3} \, dx\\ &=\frac {\text {Subst}\left (\int \frac {x^3}{a^3 (-a+x)^3} \, dx,x,-a \cos (c+d x)\right )}{a d}\\ &=\frac {\text {Subst}\left (\int \frac {x^3}{(-a+x)^3} \, dx,x,-a \cos (c+d x)\right )}{a^4 d}\\ &=\frac {\text {Subst}\left (\int \left (1-\frac {a^3}{(a-x)^3}+\frac {3 a^2}{(a-x)^2}-\frac {3 a}{a-x}\right ) \, dx,x,-a \cos (c+d x)\right )}{a^4 d}\\ &=-\frac {\cos (c+d x)}{a^3 d}-\frac {1}{2 a d (a+a \cos (c+d x))^2}+\frac {3}{d \left (a^3+a^3 \cos (c+d x)\right )}+\frac {3 \log (1+\cos (c+d x))}{a^3 d}\\ \end {align*}

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Mathematica [A]
time = 0.21, size = 103, normalized size = 1.37 \begin {gather*} \frac {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \left (21-2 \cos (3 (c+d x))+72 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )+\cos (2 (c+d x)) \left (-5+24 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )+\cos (c+d x) \left (22+96 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )\right )\right )\right )}{4 a^3 d (1+\cos (c+d x))^3} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[Sin[c + d*x]/(a + a*Sec[c + d*x])^3,x]

[Out]

(Cos[(c + d*x)/2]^2*(21 - 2*Cos[3*(c + d*x)] + 72*Log[Cos[(c + d*x)/2]] + Cos[2*(c + d*x)]*(-5 + 24*Log[Cos[(c
+ d*x)/2]]) + Cos[c + d*x]*(22 + 96*Log[Cos[(c + d*x)/2]])))/(4*a^3*d*(1 + Cos[c + d*x])^3)

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Maple [A]
time = 0.06, size = 63, normalized size = 0.84

method	result	size
derivativedivides	\(\frac {-\frac {1}{\sec \left (d x +c \right )}-3 \ln \left (\sec \left (d x +c \right )\right )-\frac {1}{2 \left (1+\sec \left (d x +c \right )\right )^{2}}-\frac {2}{1+\sec \left (d x +c \right )}+3 \ln \left (1+\sec \left (d x +c \right )\right )}{a^{3} d}\)	\(63\)
default	\(\frac {-\frac {1}{\sec \left (d x +c \right )}-3 \ln \left (\sec \left (d x +c \right )\right )-\frac {1}{2 \left (1+\sec \left (d x +c \right )\right )^{2}}-\frac {2}{1+\sec \left (d x +c \right )}+3 \ln \left (1+\sec \left (d x +c \right )\right )}{a^{3} d}\)	\(63\)
norman	\(\frac {\frac {9 \left (\tan ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{8 a d}-\frac {\tan ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )}{8 a d}-\frac {13}{4 a d}}{\left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) a^{2}}-\frac {3 \ln \left (1+\tan ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{a^{3} d}\)	\(90\)
risch	\(-\frac {3 i x}{a^{3}}-\frac {{\mathrm e}^{i \left (d x +c \right )}}{2 a^{3} d}-\frac {{\mathrm e}^{-i \left (d x +c \right )}}{2 a^{3} d}-\frac {6 i c}{a^{3} d}+\frac {6 \,{\mathrm e}^{3 i \left (d x +c \right )}+10 \,{\mathrm e}^{2 i \left (d x +c \right )}+6 \,{\mathrm e}^{i \left (d x +c \right )}}{a^{3} d \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )^{4}}+\frac {6 \ln \left ({\mathrm e}^{i \left (d x +c \right )}+1\right )}{a^{3} d}\)	\(128\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(d*x+c)/(a+a*sec(d*x+c))^3,x,method=_RETURNVERBOSE)

[Out]

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

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Maxima [A]
time = 0.26, size = 71, normalized size = 0.95 \begin {gather*} \frac {\frac {6 \, \cos \left (d x + c\right ) + 5}{a^{3} \cos \left (d x + c\right )^{2} + 2 \, a^{3} \cos \left (d x + c\right ) + a^{3}} - \frac {2 \, \cos \left (d x + c\right )}{a^{3}} + \frac {6 \, \log \left (\cos \left (d x + c\right ) + 1\right )}{a^{3}}}{2 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)/(a+a*sec(d*x+c))^3,x, algorithm="maxima")

[Out]

1/2*((6*cos(d*x + c) + 5)/(a^3*cos(d*x + c)^2 + 2*a^3*cos(d*x + c) + a^3) - 2*cos(d*x + c)/a^3 + 6*log(cos(d*x
+ c) + 1)/a^3)/d

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Fricas [A]
time = 3.05, size = 96, normalized size = 1.28 \begin {gather*} -\frac {2 \, \cos \left (d x + c\right )^{3} + 4 \, \cos \left (d x + c\right )^{2} - 6 \, {\left (\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1\right )} \log \left (\frac {1}{2} \, \cos \left (d x + c\right ) + \frac {1}{2}\right ) - 4 \, \cos \left (d x + c\right ) - 5}{2 \, {\left (a^{3} d \cos \left (d x + c\right )^{2} + 2 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)/(a+a*sec(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/2*(2*cos(d*x + c)^3 + 4*cos(d*x + c)^2 - 6*(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)*log(1/2*cos(d*x + c) + 1/2
) - 4*cos(d*x + c) - 5)/(a^3*d*cos(d*x + c)^2 + 2*a^3*d*cos(d*x + c) + a^3*d)

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \frac {\int \frac {\sin {\left (c + d x \right )}}{\sec ^{3}{\left (c + d x \right )} + 3 \sec ^{2}{\left (c + d x \right )} + 3 \sec {\left (c + d x \right )} + 1}\, dx}{a^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)/(a+a*sec(d*x+c))**3,x)

[Out]

Integral(sin(c + d*x)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1), x)/a**3

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Giac [A]
time = 0.49, size = 63, normalized size = 0.84 \begin {gather*} -\frac {\cos \left (d x + c\right )}{a^{3} d} + \frac {3 \, \log \left ({\left | -\cos \left (d x + c\right ) - 1 \right |}\right )}{a^{3} d} + \frac {6 \, \cos \left (d x + c\right ) + 5}{2 \, a^{3} d {\left (\cos \left (d x + c\right ) + 1\right )}^{2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(d*x+c)/(a+a*sec(d*x+c))^3,x, algorithm="giac")

[Out]

-cos(d*x + c)/(a^3*d) + 3*log(abs(-cos(d*x + c) - 1))/(a^3*d) + 1/2*(6*cos(d*x + c) + 5)/(a^3*d*(cos(d*x + c)
+ 1)^2)

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Mupad [B]
time = 0.08, size = 59, normalized size = 0.79 \begin {gather*} \frac {3\,\ln \left (\cos \left (c+d\,x\right )+1\right )}{a^3\,d}-\frac {\cos \left (c+d\,x\right )}{a^3\,d}+\frac {3\,\cos \left (c+d\,x\right )+\frac {5}{2}}{a^3\,d\,{\left (\cos \left (c+d\,x\right )+1\right )}^2} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(c + d*x)/(a + a/cos(c + d*x))^3,x)

[Out]

(3*log(cos(c + d*x) + 1))/(a^3*d) - cos(c + d*x)/(a^3*d) + (3*cos(c + d*x) + 5/2)/(a^3*d*(cos(c + d*x) + 1)^2)

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