3.976 \(\int \sec ^7(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=132 \[ \frac {a^5 (A+B)}{12 d (a-a \sin (c+d x))^3}+\frac {a^4 A}{8 d (a-a \sin (c+d x))^2}+\frac {a^3 (3 A-B)}{16 d (a-a \sin (c+d x))}-\frac {a^3 (A-B)}{16 d (a \sin (c+d x)+a)}+\frac {a^2 (2 A-B) \tanh ^{-1}(\sin (c+d x))}{8 d} \]

[Out]

1/8*a^2*(2*A-B)*arctanh(sin(d*x+c))/d+1/12*a^5*(A+B)/d/(a-a*sin(d*x+c))^3+1/8*a^4*A/d/(a-a*sin(d*x+c))^2+1/16*
a^3*(3*A-B)/d/(a-a*sin(d*x+c))-1/16*a^3*(A-B)/d/(a+a*sin(d*x+c))

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Rubi [A]  time = 0.16, antiderivative size = 132, 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} \[ \frac {a^5 (A+B)}{12 d (a-a \sin (c+d x))^3}+\frac {a^3 (3 A-B)}{16 d (a-a \sin (c+d x))}-\frac {a^3 (A-B)}{16 d (a \sin (c+d x)+a)}+\frac {a^2 (2 A-B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^4 A}{8 d (a-a \sin (c+d x))^2} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^7*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(a^2*(2*A - B)*ArcTanh[Sin[c + d*x]])/(8*d) + (a^5*(A + B))/(12*d*(a - a*Sin[c + d*x])^3) + (a^4*A)/(8*d*(a -
a*Sin[c + d*x])^2) + (a^3*(3*A - B))/(16*d*(a - a*Sin[c + d*x])) - (a^3*(A - B))/(16*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))^2 (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)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^7 \operatorname {Subst}\left (\int \left (\frac {A+B}{4 a^2 (a-x)^4}+\frac {A}{4 a^3 (a-x)^3}+\frac {3 A-B}{16 a^4 (a-x)^2}+\frac {A-B}{16 a^4 (a+x)^2}+\frac {2 A-B}{8 a^4 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^5 (A+B)}{12 d (a-a \sin (c+d x))^3}+\frac {a^4 A}{8 d (a-a \sin (c+d x))^2}+\frac {a^3 (3 A-B)}{16 d (a-a \sin (c+d x))}-\frac {a^3 (A-B)}{16 d (a+a \sin (c+d x))}+\frac {\left (a^3 (2 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^2 (2 A-B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^5 (A+B)}{12 d (a-a \sin (c+d x))^3}+\frac {a^4 A}{8 d (a-a \sin (c+d x))^2}+\frac {a^3 (3 A-B)}{16 d (a-a \sin (c+d x))}-\frac {a^3 (A-B)}{16 d (a+a \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.71, size = 90, normalized size = 0.68 \[ \frac {a^2 \left (\frac {3 B-9 A}{\sin (c+d x)-1}-\frac {3 (A-B)}{\sin (c+d x)+1}-\frac {4 (A+B)}{(\sin (c+d x)-1)^3}+6 (2 A-B) \tanh ^{-1}(\sin (c+d x))+\frac {6 A}{(\sin (c+d x)-1)^2}\right )}{48 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^7*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

(a^2*(6*(2*A - B)*ArcTanh[Sin[c + d*x]] - (4*(A + B))/(-1 + Sin[c + d*x])^3 + (6*A)/(-1 + Sin[c + d*x])^2 + (-
9*A + 3*B)/(-1 + Sin[c + d*x]) - (3*(A - B))/(1 + Sin[c + d*x])))/(48*d)

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fricas [B]  time = 0.78, size = 271, normalized size = 2.05 \[ -\frac {12 \, {\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{2} - 8 \, {\left (A - 2 \, B\right )} a^{2} - 3 \, {\left ({\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{4} + 2 \, {\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, {\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left ({\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{4} + 2 \, {\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, {\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (3 \, {\left (2 \, A - B\right )} a^{2} \cos \left (d x + c\right )^{2} - 4 \, {\left (2 \, A - B\right )} a^{2}\right )} \sin \left (d x + c\right )}{48 \, {\left (d \cos \left (d x + c\right )^{4} + 2 \, d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/48*(12*(2*A - B)*a^2*cos(d*x + c)^2 - 8*(A - 2*B)*a^2 - 3*((2*A - B)*a^2*cos(d*x + c)^4 + 2*(2*A - B)*a^2*c
os(d*x + c)^2*sin(d*x + c) - 2*(2*A - B)*a^2*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + 3*((2*A - B)*a^2*cos(d*x
+ c)^4 + 2*(2*A - B)*a^2*cos(d*x + c)^2*sin(d*x + c) - 2*(2*A - B)*a^2*cos(d*x + c)^2)*log(-sin(d*x + c) + 1)
- 2*(3*(2*A - B)*a^2*cos(d*x + c)^2 - 4*(2*A - B)*a^2)*sin(d*x + c))/(d*cos(d*x + c)^4 + 2*d*cos(d*x + c)^2*si
n(d*x + c) - 2*d*cos(d*x + c)^2)

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giac [A]  time = 0.27, size = 209, normalized size = 1.58 \[ \frac {6 \, {\left (2 \, A a^{2} - B a^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 6 \, {\left (2 \, A a^{2} - B a^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {6 \, {\left (2 \, A a^{2} \sin \left (d x + c\right ) - B a^{2} \sin \left (d x + c\right ) + 3 \, A a^{2} - 2 \, B a^{2}\right )}}{\sin \left (d x + c\right ) + 1} + \frac {22 \, A a^{2} \sin \left (d x + c\right )^{3} - 11 \, B a^{2} \sin \left (d x + c\right )^{3} - 84 \, A a^{2} \sin \left (d x + c\right )^{2} + 39 \, B a^{2} \sin \left (d x + c\right )^{2} + 114 \, A a^{2} \sin \left (d x + c\right ) - 45 \, B a^{2} \sin \left (d x + c\right ) - 60 \, A a^{2} + 9 \, B a^{2}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{3}}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/96*(6*(2*A*a^2 - B*a^2)*log(abs(sin(d*x + c) + 1)) - 6*(2*A*a^2 - B*a^2)*log(abs(sin(d*x + c) - 1)) - 6*(2*A
*a^2*sin(d*x + c) - B*a^2*sin(d*x + c) + 3*A*a^2 - 2*B*a^2)/(sin(d*x + c) + 1) + (22*A*a^2*sin(d*x + c)^3 - 11
*B*a^2*sin(d*x + c)^3 - 84*A*a^2*sin(d*x + c)^2 + 39*B*a^2*sin(d*x + c)^2 + 114*A*a^2*sin(d*x + c) - 45*B*a^2*
sin(d*x + c) - 60*A*a^2 + 9*B*a^2)/(sin(d*x + c) - 1)^3)/d

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maple [B]  time = 0.73, size = 379, normalized size = 2.87 \[ \frac {a^{2} A \left (\sin ^{3}\left (d x +c \right )\right )}{6 d \cos \left (d x +c \right )^{6}}+\frac {a^{2} A \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{4}}+\frac {a^{2} A \left (\sin ^{3}\left (d x +c \right )\right )}{16 d \cos \left (d x +c \right )^{2}}+\frac {a^{2} A \sin \left (d x +c \right )}{16 d}+\frac {a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {B \,a^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{6 d \cos \left (d x +c \right )^{6}}+\frac {B \,a^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{12 d \cos \left (d x +c \right )^{4}}+\frac {a^{2} A}{3 d \cos \left (d x +c \right )^{6}}+\frac {B \,a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 d \cos \left (d x +c \right )^{6}}+\frac {B \,a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {B \,a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}+\frac {B \,a^{2} \sin \left (d x +c \right )}{8 d}-\frac {B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d}+\frac {a^{2} A \tan \left (d x +c \right ) \left (\sec ^{5}\left (d x +c \right )\right )}{6 d}+\frac {5 a^{2} A \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{24 d}+\frac {5 a^{2} A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}+\frac {B \,a^{2}}{6 d \cos \left (d x +c \right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^7*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.43, size = 148, normalized size = 1.12 \[ \frac {3 \, {\left (2 \, A - B\right )} a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (2 \, A - B\right )} a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (2 \, A - B\right )} a^{2} \sin \left (d x + c\right )^{3} - 6 \, {\left (2 \, A - B\right )} a^{2} \sin \left (d x + c\right )^{2} + {\left (2 \, A - B\right )} a^{2} \sin \left (d x + c\right ) + 2 \, {\left (4 \, A + B\right )} a^{2}\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{3} + 2 \, \sin \left (d x + c\right ) - 1}}{48 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 9.22, size = 136, normalized size = 1.03 \[ \frac {a^2\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (2\,A-B\right )}{8\,d}-\frac {{\sin \left (c+d\,x\right )}^3\,\left (\frac {A\,a^2}{4}-\frac {B\,a^2}{8}\right )-{\sin \left (c+d\,x\right )}^2\,\left (\frac {A\,a^2}{2}-\frac {B\,a^2}{4}\right )+\frac {A\,a^2}{3}+\frac {B\,a^2}{12}+\sin \left (c+d\,x\right )\,\left (\frac {A\,a^2}{12}-\frac {B\,a^2}{24}\right )}{d\,\left ({\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^3+2\,\sin \left (c+d\,x\right )-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*sin(c + d*x))*(a + a*sin(c + d*x))^2)/cos(c + d*x)^7,x)

[Out]

(a^2*atanh(sin(c + d*x))*(2*A - B))/(8*d) - (sin(c + d*x)^3*((A*a^2)/4 - (B*a^2)/8) - sin(c + d*x)^2*((A*a^2)/
2 - (B*a^2)/4) + (A*a^2)/3 + (B*a^2)/12 + sin(c + d*x)*((A*a^2)/12 - (B*a^2)/24))/(d*(2*sin(c + d*x) - 2*sin(c
+ d*x)^3 + sin(c + d*x)^4 - 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))**2*(A+B*sin(d*x+c)),x)

[Out]

Timed out

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