3.993 \(\int \sec ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=105 \[ \frac {a^6 (A+B)}{6 d (a-a \sin (c+d x))^3}+\frac {a^5 (A-B)}{8 d (a-a \sin (c+d x))^2}+\frac {a^4 (A-B)}{8 d (a-a \sin (c+d x))}+\frac {a^3 (A-B) \tanh ^{-1}(\sin (c+d x))}{8 d} \]

[Out]

1/8*a^3*(A-B)*arctanh(sin(d*x+c))/d+1/6*a^6*(A+B)/d/(a-a*sin(d*x+c))^3+1/8*a^5*(A-B)/d/(a-a*sin(d*x+c))^2+1/8*
a^4*(A-B)/d/(a-a*sin(d*x+c))

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Rubi [A]  time = 0.14, antiderivative size = 105, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 31, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.097, Rules used = {2836, 77, 206} \[ \frac {a^6 (A+B)}{6 d (a-a \sin (c+d x))^3}+\frac {a^5 (A-B)}{8 d (a-a \sin (c+d x))^2}+\frac {a^4 (A-B)}{8 d (a-a \sin (c+d x))}+\frac {a^3 (A-B) \tanh ^{-1}(\sin (c+d x))}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^7*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

(a^3*(A - B)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^6*(A + B))/(6*d*(a - a*Sin[c + d*x])^3) + (a^5*(A - B))/(8*d*(a
- a*Sin[c + d*x])^2) + (a^4*(A - B))/(8*d*(a - a*Sin[c + d*x]))

Rule 77

Int[((a_.) + (b_.)*(x_))*((c_) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Int[ExpandIntegran
d[(a + b*x)*(c + d*x)^n*(e + f*x)^p, x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a*d, 0] && ((ILtQ[
n, 0] && ILtQ[p, 0]) || EqQ[p, 1] || (IGtQ[p, 0] && ( !IntegerQ[n] || LeQ[9*p + 5*(n + 2), 0] || GeQ[n + p + 1
, 0] || (GeQ[n + p + 2, 0] && RationalQ[a, b, c, d, e, f]))))

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

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 ^7(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx &=\frac {a^7 \operatorname {Subst}\left (\int \frac {A+\frac {B x}{a}}{(a-x)^4 (a+x)} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^7 \operatorname {Subst}\left (\int \left (\frac {A+B}{2 a (a-x)^4}+\frac {A-B}{4 a^2 (a-x)^3}+\frac {A-B}{8 a^3 (a-x)^2}+\frac {A-B}{8 a^3 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^6 (A+B)}{6 d (a-a \sin (c+d x))^3}+\frac {a^5 (A-B)}{8 d (a-a \sin (c+d x))^2}+\frac {a^4 (A-B)}{8 d (a-a \sin (c+d x))}+\frac {\left (a^4 (A-B)\right ) \operatorname {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{8 d}\\ &=\frac {a^3 (A-B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^6 (A+B)}{6 d (a-a \sin (c+d x))^3}+\frac {a^5 (A-B)}{8 d (a-a \sin (c+d x))^2}+\frac {a^4 (A-B)}{8 d (a-a \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.26, size = 95, normalized size = 0.90 \[ \frac {a^3 \left (-3 (A-B) \sin ^2(c+d x)+9 (A-B) \sin (c+d x)-3 (A-B) \tanh ^{-1}(\sin (c+d x)) \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^6+2 (B-5 A)\right )}{24 d (\sin (c+d x)-1)^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^7*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

(a^3*(2*(-5*A + B) - 3*(A - B)*ArcTanh[Sin[c + d*x]]*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^6 + 9*(A - B)*Sin[c
+ d*x] - 3*(A - B)*Sin[c + d*x]^2))/(24*d*(-1 + Sin[c + d*x])^3)

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fricas [B]  time = 0.72, size = 242, normalized size = 2.30 \[ \frac {6 \, {\left (A - B\right )} a^{3} \cos \left (d x + c\right )^{2} + 18 \, {\left (A - B\right )} a^{3} \sin \left (d x + c\right ) - 2 \, {\left (13 \, A - 5 \, B\right )} a^{3} + 3 \, {\left (3 \, {\left (A - B\right )} a^{3} \cos \left (d x + c\right )^{2} - 4 \, {\left (A - B\right )} a^{3} - {\left ({\left (A - B\right )} a^{3} \cos \left (d x + c\right )^{2} - 4 \, {\left (A - B\right )} a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (3 \, {\left (A - B\right )} a^{3} \cos \left (d x + c\right )^{2} - 4 \, {\left (A - B\right )} a^{3} - {\left ({\left (A - B\right )} a^{3} \cos \left (d x + c\right )^{2} - 4 \, {\left (A - B\right )} a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{48 \, {\left (3 \, d \cos \left (d x + c\right )^{2} - {\left (d \cos \left (d x + c\right )^{2} - 4 \, d\right )} \sin \left (d x + c\right ) - 4 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/48*(6*(A - B)*a^3*cos(d*x + c)^2 + 18*(A - B)*a^3*sin(d*x + c) - 2*(13*A - 5*B)*a^3 + 3*(3*(A - B)*a^3*cos(d
*x + c)^2 - 4*(A - B)*a^3 - ((A - B)*a^3*cos(d*x + c)^2 - 4*(A - B)*a^3)*sin(d*x + c))*log(sin(d*x + c) + 1) -
3*(3*(A - B)*a^3*cos(d*x + c)^2 - 4*(A - B)*a^3 - ((A - B)*a^3*cos(d*x + c)^2 - 4*(A - B)*a^3)*sin(d*x + c))*
log(-sin(d*x + c) + 1))/(3*d*cos(d*x + c)^2 - (d*cos(d*x + c)^2 - 4*d)*sin(d*x + c) - 4*d)

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giac [A]  time = 0.29, size = 158, normalized size = 1.50 \[ \frac {6 \, {\left (A a^{3} - B a^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 6 \, {\left (A a^{3} - B a^{3}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac {11 \, A a^{3} \sin \left (d x + c\right )^{3} - 11 \, B a^{3} \sin \left (d x + c\right )^{3} - 45 \, A a^{3} \sin \left (d x + c\right )^{2} + 45 \, B a^{3} \sin \left (d x + c\right )^{2} + 69 \, A a^{3} \sin \left (d x + c\right ) - 69 \, B a^{3} \sin \left (d x + c\right ) - 51 \, A a^{3} + 19 \, B a^{3}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{3}}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/96*(6*(A*a^3 - B*a^3)*log(abs(sin(d*x + c) + 1)) - 6*(A*a^3 - B*a^3)*log(abs(sin(d*x + c) - 1)) + (11*A*a^3*
sin(d*x + c)^3 - 11*B*a^3*sin(d*x + c)^3 - 45*A*a^3*sin(d*x + c)^2 + 45*B*a^3*sin(d*x + c)^2 + 69*A*a^3*sin(d*
x + c) - 69*B*a^3*sin(d*x + c) - 51*A*a^3 + 19*B*a^3)/(sin(d*x + c) - 1)^3)/d

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maple [B]  time = 0.57, size = 521, normalized size = 4.96 \[ \frac {3 a^{3} A \sin \left (d x +c \right )}{16 d}+\frac {a^{3} A}{2 d \cos \left (d x +c \right )^{6}}+\frac {B \,a^{3}}{6 d \cos \left (d x +c \right )^{6}}-\frac {B \,a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{48 d}+\frac {a^{3} B \sin \left (d x +c \right )}{8 d}-\frac {B \,a^{3} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {a^{3} A \left (\sin ^{4}\left (d x +c \right )\right )}{6 d \cos \left (d x +c \right )^{6}}+\frac {B \,a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{6 d \cos \left (d x +c \right )^{6}}+\frac {B \,a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{24 d \cos \left (d x +c \right )^{4}}-\frac {B \,a^{3} \left (\sin ^{5}\left (d x +c \right )\right )}{48 d \cos \left (d x +c \right )^{2}}+\frac {a^{3} A \left (\sin ^{3}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{6}}+\frac {3 a^{3} A \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{4}}+\frac {3 a^{3} A \left (\sin ^{3}\left (d x +c \right )\right )}{16 d \cos \left (d x +c \right )^{2}}+\frac {B \,a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{6}}+\frac {B \,a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{6}}+\frac {3 B \,a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{4}}+\frac {3 B \,a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{16 d \cos \left (d x +c \right )^{2}}+\frac {a^{3} A \tan \left (d x +c \right ) \left (\sec ^{5}\left (d x +c \right )\right )}{6 d}+\frac {5 a^{3} A \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{24 d}+\frac {5 a^{3} A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}+\frac {a^{3} A \left (\sin ^{4}\left (d x +c \right )\right )}{12 d \cos \left (d x +c \right )^{4}}+\frac {B \,a^{3} \left (\sin ^{4}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {a^{3} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^7*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x)

[Out]

3/16/d*a^3*A*sin(d*x+c)+1/2/d*a^3*A/cos(d*x+c)^6+1/6/d*B*a^3/cos(d*x+c)^6-1/48/d*B*a^3*sin(d*x+c)^3+1/8*a^3*B*
sin(d*x+c)/d-1/8/d*B*a^3*ln(sec(d*x+c)+tan(d*x+c))+1/6/d*a^3*A*sin(d*x+c)^4/cos(d*x+c)^6+1/6/d*B*a^3*sin(d*x+c
)^5/cos(d*x+c)^6+1/24/d*B*a^3*sin(d*x+c)^5/cos(d*x+c)^4-1/48/d*B*a^3*sin(d*x+c)^5/cos(d*x+c)^2+1/2/d*a^3*A*sin
(d*x+c)^3/cos(d*x+c)^6+3/8/d*a^3*A*sin(d*x+c)^3/cos(d*x+c)^4+3/16/d*a^3*A*sin(d*x+c)^3/cos(d*x+c)^2+1/2/d*B*a^
3*sin(d*x+c)^4/cos(d*x+c)^6+1/2/d*B*a^3*sin(d*x+c)^3/cos(d*x+c)^6+3/8/d*B*a^3*sin(d*x+c)^3/cos(d*x+c)^4+3/16/d
*B*a^3*sin(d*x+c)^3/cos(d*x+c)^2+1/6/d*a^3*A*tan(d*x+c)*sec(d*x+c)^5+5/24/d*a^3*A*tan(d*x+c)*sec(d*x+c)^3+5/16
/d*a^3*A*sec(d*x+c)*tan(d*x+c)+1/12/d*a^3*A*sin(d*x+c)^4/cos(d*x+c)^4+1/4/d*B*a^3*sin(d*x+c)^4/cos(d*x+c)^4+1/
8/d*a^3*A*ln(sec(d*x+c)+tan(d*x+c))

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maxima [A]  time = 0.31, size = 123, normalized size = 1.17 \[ \frac {3 \, {\left (A - B\right )} a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (A - B\right )} a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (A - B\right )} a^{3} \sin \left (d x + c\right )^{2} - 9 \, {\left (A - B\right )} a^{3} \sin \left (d x + c\right ) + 2 \, {\left (5 \, A - B\right )} a^{3}\right )}}{\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right ) - 1}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/48*(3*(A - B)*a^3*log(sin(d*x + c) + 1) - 3*(A - B)*a^3*log(sin(d*x + c) - 1) - 2*(3*(A - B)*a^3*sin(d*x + c
)^2 - 9*(A - B)*a^3*sin(d*x + c) + 2*(5*A - B)*a^3)/(sin(d*x + c)^3 - 3*sin(d*x + c)^2 + 3*sin(d*x + c) - 1))/
d

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mupad [B]  time = 9.14, size = 112, normalized size = 1.07 \[ \frac {a^3\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (A-B\right )}{8\,d}-\frac {{\sin \left (c+d\,x\right )}^2\,\left (\frac {A\,a^3}{8}-\frac {B\,a^3}{8}\right )+\frac {5\,A\,a^3}{12}-\frac {B\,a^3}{12}-\sin \left (c+d\,x\right )\,\left (\frac {3\,A\,a^3}{8}-\frac {3\,B\,a^3}{8}\right )}{d\,\left ({\sin \left (c+d\,x\right )}^3-3\,{\sin \left (c+d\,x\right )}^2+3\,\sin \left (c+d\,x\right )-1\right )} \]

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)^7,x)

[Out]

(a^3*atanh(sin(c + d*x))*(A - B))/(8*d) - (sin(c + d*x)^2*((A*a^3)/8 - (B*a^3)/8) + (5*A*a^3)/12 - (B*a^3)/12
- sin(c + d*x)*((3*A*a^3)/8 - (3*B*a^3)/8))/(d*(3*sin(c + d*x) - 3*sin(c + d*x)^2 + sin(c + d*x)^3 - 1))

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

[Out]

Timed out

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