3.992 \(\int \sec ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=43 \[ \frac {a^3 (a A+a B \sin (c+d x))^2}{2 d (A+B) (a-a \sin (c+d x))^2} \]

[Out]

1/2*a^3*(a*A+a*B*sin(d*x+c))^2/(A+B)/d/(a-a*sin(d*x+c))^2

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Rubi [A]  time = 0.08, antiderivative size = 43, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.065, Rules used = {2836, 37} \[ \frac {a^3 (a A+a B \sin (c+d x))^2}{2 d (A+B) (a-a \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

(a^3*(a*A + a*B*Sin[c + d*x])^2)/(2*(A + B)*d*(a - a*Sin[c + d*x])^2)

Rule 37

Int[((a_.) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[((a + b*x)^(m + 1)*(c + d*x)^(n +
1))/((b*c - a*d)*(m + 1)), x] /; FreeQ[{a, b, c, d, m, n}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n + 2, 0] && NeQ
[m, -1]

Rule 2836

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 + (p - 1)/2)*(a - x)^((p - 1)/2)*(c + (d*x)/b
)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, c, d, m, n}, x] && IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2,
0]

Rubi steps

\begin {align*} \int \sec ^5(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx &=\frac {a^5 \operatorname {Subst}\left (\int \frac {A+\frac {B x}{a}}{(a-x)^3} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^3 (a A+a B \sin (c+d x))^2}{2 (A+B) d (a-a \sin (c+d x))^2}\\ \end {align*}

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Mathematica [A]  time = 0.05, size = 37, normalized size = 0.86 \[ \frac {a^3 (A+B \sin (c+d x))^2}{2 d (A+B) (\sin (c+d x)-1)^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

(a^3*(A + B*Sin[c + d*x])^2)/(2*(A + B)*d*(-1 + Sin[c + d*x])^2)

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fricas [A]  time = 0.75, size = 49, normalized size = 1.14 \[ -\frac {2 \, B a^{3} \sin \left (d x + c\right ) + {\left (A - B\right )} a^{3}}{2 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/2*(2*B*a^3*sin(d*x + c) + (A - B)*a^3)/(d*cos(d*x + c)^2 + 2*d*sin(d*x + c) - 2*d)

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giac [A]  time = 0.26, size = 82, normalized size = 1.91 \[ \frac {2 \, {\left (A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{d {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

2*(A*a^3*tan(1/2*d*x + 1/2*c)^3 - A*a^3*tan(1/2*d*x + 1/2*c)^2 + B*a^3*tan(1/2*d*x + 1/2*c)^2 + A*a^3*tan(1/2*
d*x + 1/2*c))/(d*(tan(1/2*d*x + 1/2*c) - 1)^4)

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maple [B]  time = 0.58, size = 312, normalized size = 7.26 \[ \frac {a^{3} A \left (\sin ^{4}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {B \,a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}-\frac {B \,a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}-\frac {B \,a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d}+\frac {3 a^{3} A \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {3 a^{3} A \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}+\frac {3 a^{3} A \sin \left (d x +c \right )}{8 d}+\frac {3 B \,a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {3 a^{3} A}{4 d \cos \left (d x +c \right )^{4}}+\frac {3 B \,a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {3 B \,a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}+\frac {a^{3} A \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 a^{3} A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {B \,a^{3}}{4 d \cos \left (d x +c \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x)

[Out]

1/4/d*a^3*A*sin(d*x+c)^4/cos(d*x+c)^4+1/4/d*B*a^3*sin(d*x+c)^5/cos(d*x+c)^4-1/8/d*B*a^3*sin(d*x+c)^5/cos(d*x+c
)^2-1/8/d*B*a^3*sin(d*x+c)^3+3/4/d*a^3*A*sin(d*x+c)^3/cos(d*x+c)^4+3/8/d*a^3*A*sin(d*x+c)^3/cos(d*x+c)^2+3/8/d
*a^3*A*sin(d*x+c)+3/4/d*B*a^3*sin(d*x+c)^4/cos(d*x+c)^4+3/4/d*a^3*A/cos(d*x+c)^4+3/4/d*B*a^3*sin(d*x+c)^3/cos(
d*x+c)^4+3/8/d*B*a^3*sin(d*x+c)^3/cos(d*x+c)^2+1/4/d*a^3*A*tan(d*x+c)*sec(d*x+c)^3+3/8/d*a^3*A*sec(d*x+c)*tan(
d*x+c)+1/4/d*B*a^3/cos(d*x+c)^4

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maxima [A]  time = 0.30, size = 47, normalized size = 1.09 \[ \frac {2 \, B a^{3} \sin \left (d x + c\right ) + {\left (A - B\right )} a^{3}}{2 \, {\left (\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1\right )} d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/2*(2*B*a^3*sin(d*x + c) + (A - B)*a^3)/((sin(d*x + c)^2 - 2*sin(d*x + c) + 1)*d)

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mupad [B]  time = 9.15, size = 36, normalized size = 0.84 \[ \frac {\frac {a^3\,\left (A-B\right )}{2}+B\,a^3\,\sin \left (c+d\,x\right )}{d\,{\left (\sin \left (c+d\,x\right )-1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x))^3)/cos(c + d*x)^5,x)

[Out]

((a^3*(A - B))/2 + B*a^3*sin(c + d*x))/(d*(sin(c + d*x) - 1)^2)

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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(sec(d*x+c)**5*(a+a*sin(d*x+c))**3*(A+B*sin(d*x+c)),x)

[Out]

Timed out

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