3.198 \(\int \frac {\sin ^3(c+d x)}{a+b \sec (c+d x)} \, dx\)

Optimal. Leaf size=89 \[ -\frac {b \cos ^2(c+d x)}{2 a^2 d}+\frac {b \left (a^2-b^2\right ) \log (a \cos (c+d x)+b)}{a^4 d}-\frac {\left (a^2-b^2\right ) \cos (c+d x)}{a^3 d}+\frac {\cos ^3(c+d x)}{3 a d} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 89, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {3872, 2837, 12, 772} \[ -\frac {\left (a^2-b^2\right ) \cos (c+d x)}{a^3 d}+\frac {b \left (a^2-b^2\right ) \log (a \cos (c+d x)+b)}{a^4 d}-\frac {b \cos ^2(c+d x)}{2 a^2 d}+\frac {\cos ^3(c+d x)}{3 a d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(((a^2 - b^2)*Cos[c + d*x])/(a^3*d)) - (b*Cos[c + d*x]^2)/(2*a^2*d) + Cos[c + d*x]^3/(3*a*d) + (b*(a^2 - b^2)
*Log[b + a*Cos[c + d*x]])/(a^4*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 772

Int[((d_.) + (e_.)*(x_))^(m_.)*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_.), x_Symbol] :> Int[ExpandIntegr
and[(d + e*x)^m*(f + g*x)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m}, x] && IGtQ[p, 0]

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 3872

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_.), x_Symbol] :> Int[((g*C
os[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 ^3(c+d x)}{a+b \sec (c+d x)} \, dx &=-\int \frac {\cos (c+d x) \sin ^3(c+d x)}{-b-a \cos (c+d x)} \, dx\\ &=\frac {\operatorname {Subst}\left (\int \frac {x \left (a^2-x^2\right )}{a (-b+x)} \, dx,x,-a \cos (c+d x)\right )}{a^3 d}\\ &=\frac {\operatorname {Subst}\left (\int \frac {x \left (a^2-x^2\right )}{-b+x} \, dx,x,-a \cos (c+d x)\right )}{a^4 d}\\ &=\frac {\operatorname {Subst}\left (\int \left (a^2 \left (1-\frac {b^2}{a^2}\right )+\frac {-a^2 b+b^3}{b-x}-b x-x^2\right ) \, dx,x,-a \cos (c+d x)\right )}{a^4 d}\\ &=-\frac {\left (a^2-b^2\right ) \cos (c+d x)}{a^3 d}-\frac {b \cos ^2(c+d x)}{2 a^2 d}+\frac {\cos ^3(c+d x)}{3 a d}+\frac {b \left (a^2-b^2\right ) \log (b+a \cos (c+d x))}{a^4 d}\\ \end {align*}

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Mathematica [A]  time = 0.21, size = 89, normalized size = 1.00 \[ \frac {\left (12 a b^2-9 a^3\right ) \cos (c+d x)+a^3 \cos (3 (c+d x))-3 a^2 b \cos (2 (c+d x))+12 a^2 b \log (a \cos (c+d x)+b)-12 b^3 \log (a \cos (c+d x)+b)}{12 a^4 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-9*a^3 + 12*a*b^2)*Cos[c + d*x] - 3*a^2*b*Cos[2*(c + d*x)] + a^3*Cos[3*(c + d*x)] + 12*a^2*b*Log[b + a*Cos[c
+ d*x]] - 12*b^3*Log[b + a*Cos[c + d*x]])/(12*a^4*d)

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fricas [A]  time = 0.56, size = 78, normalized size = 0.88 \[ \frac {2 \, a^{3} \cos \left (d x + c\right )^{3} - 3 \, a^{2} b \cos \left (d x + c\right )^{2} - 6 \, {\left (a^{3} - a b^{2}\right )} \cos \left (d x + c\right ) + 6 \, {\left (a^{2} b - b^{3}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{6 \, a^{4} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.23, size = 102, normalized size = 1.15 \[ \frac {{\left (a^{2} b - b^{3}\right )} \log \left ({\left | -a \cos \left (d x + c\right ) - b \right |}\right )}{a^{4} d} + \frac {2 \, a^{2} d^{2} \cos \left (d x + c\right )^{3} - 3 \, a b d^{2} \cos \left (d x + c\right )^{2} - 6 \, a^{2} d^{2} \cos \left (d x + c\right ) + 6 \, b^{2} d^{2} \cos \left (d x + c\right )}{6 \, a^{3} d^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.40, size = 106, normalized size = 1.19 \[ \frac {\cos ^{3}\left (d x +c \right )}{3 d a}-\frac {b \left (\cos ^{2}\left (d x +c \right )\right )}{2 a^{2} d}-\frac {\cos \left (d x +c \right )}{d a}+\frac {\cos \left (d x +c \right ) b^{2}}{d \,a^{3}}+\frac {b \ln \left (b +a \cos \left (d x +c \right )\right )}{a^{2} d}-\frac {b^{3} \ln \left (b +a \cos \left (d x +c \right )\right )}{d \,a^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.58, size = 80, normalized size = 0.90 \[ \frac {\frac {2 \, a^{2} \cos \left (d x + c\right )^{3} - 3 \, a b \cos \left (d x + c\right )^{2} - 6 \, {\left (a^{2} - b^{2}\right )} \cos \left (d x + c\right )}{a^{3}} + \frac {6 \, {\left (a^{2} b - b^{3}\right )} \log \left (a \cos \left (d x + c\right ) + b\right )}{a^{4}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 1.02, size = 79, normalized size = 0.89 \[ -\frac {\cos \left (c+d\,x\right )\,\left (\frac {1}{a}-\frac {b^2}{a^3}\right )-\frac {{\cos \left (c+d\,x\right )}^3}{3\,a}+\frac {b\,{\cos \left (c+d\,x\right )}^2}{2\,a^2}-\frac {\ln \left (b+a\,\cos \left (c+d\,x\right )\right )\,\left (a^2\,b-b^3\right )}{a^4}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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