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

Optimal. Leaf size=157 \[ \frac {a^4 (A+B)}{24 d (a-a \sin (c+d x))^3}+\frac {a^3 (3 A+B)}{32 d (a-a \sin (c+d x))^2}-\frac {a^3 (A-B)}{32 d (a \sin (c+d x)+a)^2}-\frac {a^2 (2 A-B)}{16 d (a \sin (c+d x)+a)}+\frac {3 a^2 A}{16 d (a-a \sin (c+d x))}+\frac {a (5 A-B) \tanh ^{-1}(\sin (c+d x))}{16 d} \]

[Out]

1/16*a*(5*A-B)*arctanh(sin(d*x+c))/d+1/24*a^4*(A+B)/d/(a-a*sin(d*x+c))^3+1/32*a^3*(3*A+B)/d/(a-a*sin(d*x+c))^2
+3/16*a^2*A/d/(a-a*sin(d*x+c))-1/32*a^3*(A-B)/d/(a+a*sin(d*x+c))^2-1/16*a^2*(2*A-B)/d/(a+a*sin(d*x+c))

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Rubi [A]  time = 0.17, antiderivative size = 157, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2836, 77, 206} \[ \frac {a^4 (A+B)}{24 d (a-a \sin (c+d x))^3}+\frac {a^3 (3 A+B)}{32 d (a-a \sin (c+d x))^2}-\frac {a^3 (A-B)}{32 d (a \sin (c+d x)+a)^2}-\frac {a^2 (2 A-B)}{16 d (a \sin (c+d x)+a)}+\frac {3 a^2 A}{16 d (a-a \sin (c+d x))}+\frac {a (5 A-B) \tanh ^{-1}(\sin (c+d x))}{16 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(5*A - B)*ArcTanh[Sin[c + d*x]])/(16*d) + (a^4*(A + B))/(24*d*(a - a*Sin[c + d*x])^3) + (a^3*(3*A + B))/(32
*d*(a - a*Sin[c + d*x])^2) + (3*a^2*A)/(16*d*(a - a*Sin[c + d*x])) - (a^3*(A - B))/(32*d*(a + a*Sin[c + d*x])^
2) - (a^2*(2*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)) (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)^3} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^7 \operatorname {Subst}\left (\int \left (\frac {A+B}{8 a^3 (a-x)^4}+\frac {3 A+B}{16 a^4 (a-x)^3}+\frac {3 A}{16 a^5 (a-x)^2}+\frac {A-B}{16 a^4 (a+x)^3}+\frac {2 A-B}{16 a^5 (a+x)^2}+\frac {5 A-B}{16 a^5 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^4 (A+B)}{24 d (a-a \sin (c+d x))^3}+\frac {a^3 (3 A+B)}{32 d (a-a \sin (c+d x))^2}+\frac {3 a^2 A}{16 d (a-a \sin (c+d x))}-\frac {a^3 (A-B)}{32 d (a+a \sin (c+d x))^2}-\frac {a^2 (2 A-B)}{16 d (a+a \sin (c+d x))}+\frac {\left (a^2 (5 A-B)\right ) \operatorname {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{16 d}\\ &=\frac {a (5 A-B) \tanh ^{-1}(\sin (c+d x))}{16 d}+\frac {a^4 (A+B)}{24 d (a-a \sin (c+d x))^3}+\frac {a^3 (3 A+B)}{32 d (a-a \sin (c+d x))^2}+\frac {3 a^2 A}{16 d (a-a \sin (c+d x))}-\frac {a^3 (A-B)}{32 d (a+a \sin (c+d x))^2}-\frac {a^2 (2 A-B)}{16 d (a+a \sin (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 1.75, size = 451, normalized size = 2.87 \[ \frac {a (\sin (c+d x)+1) \left (3 i x (5 A-B) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4+\frac {3 (3 A+B) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^4}+\frac {4 (A+B) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^6}-\frac {6 (2 A-B) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}{d}-\frac {6 (5 A-B) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )}{d}+\frac {3 (5 A-B) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4 \log \left (\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2\right )}{d}-\frac {6 i (5 A-B) \tan ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4}{d}+\frac {3 (B-A)}{d}+\frac {18 A \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4}{d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}\right )}{96 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*((3*(-A + B))/d - (6*(2*A - B)*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/d + (3*I)*(5*A - B)*x*(Cos[(c + d*x
)/2] + Sin[(c + d*x)/2])^4 - ((6*I)*(5*A - B)*ArcTan[Tan[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4
)/d - (6*(5*A - B)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)/d + (3*(5
*A - B)*Log[(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)/d + (4*(A + B)*(
Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^6) + (3*(3*A + B)*(Cos[(c + d
*x)/2] + Sin[(c + d*x)/2])^4)/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^4) + (18*A*(Cos[(c + d*x)/2] + Sin[(c +
d*x)/2])^4)/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2))*(1 + Sin[c + d*x]))/(96*(Cos[(c + d*x)/2] + Sin[(c +
d*x)/2])^6)

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fricas [A]  time = 0.73, size = 222, normalized size = 1.41 \[ -\frac {6 \, {\left (5 \, A - B\right )} a \cos \left (d x + c\right )^{4} - 2 \, {\left (5 \, A - B\right )} a \cos \left (d x + c\right )^{2} - 4 \, {\left (A - 5 \, B\right )} a - 3 \, {\left ({\left (5 \, A - B\right )} a \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - {\left (5 \, A - B\right )} a \cos \left (d x + c\right )^{4}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left ({\left (5 \, A - B\right )} a \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - {\left (5 \, A - B\right )} a \cos \left (d x + c\right )^{4}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 2 \, {\left (3 \, {\left (5 \, A - B\right )} a \cos \left (d x + c\right )^{2} + 2 \, {\left (5 \, A - B\right )} a\right )} \sin \left (d x + c\right )}{96 \, {\left (d \cos \left (d x + c\right )^{4} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{4}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/96*(6*(5*A - B)*a*cos(d*x + c)^4 - 2*(5*A - B)*a*cos(d*x + c)^2 - 4*(A - 5*B)*a - 3*((5*A - B)*a*cos(d*x +
c)^4*sin(d*x + c) - (5*A - B)*a*cos(d*x + c)^4)*log(sin(d*x + c) + 1) + 3*((5*A - B)*a*cos(d*x + c)^4*sin(d*x
+ c) - (5*A - B)*a*cos(d*x + c)^4)*log(-sin(d*x + c) + 1) + 2*(3*(5*A - B)*a*cos(d*x + c)^2 + 2*(5*A - B)*a)*s
in(d*x + c))/(d*cos(d*x + c)^4*sin(d*x + c) - d*cos(d*x + c)^4)

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giac [A]  time = 0.23, size = 201, normalized size = 1.28 \[ \frac {6 \, {\left (5 \, A a - B a\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 6 \, {\left (5 \, A a - B a\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {3 \, {\left (15 \, A a \sin \left (d x + c\right )^{2} - 3 \, B a \sin \left (d x + c\right )^{2} + 38 \, A a \sin \left (d x + c\right ) - 10 \, B a \sin \left (d x + c\right ) + 25 \, A a - 9 \, B a\right )}}{{\left (\sin \left (d x + c\right ) + 1\right )}^{2}} + \frac {55 \, A a \sin \left (d x + c\right )^{3} - 11 \, B a \sin \left (d x + c\right )^{3} - 201 \, A a \sin \left (d x + c\right )^{2} + 33 \, B a \sin \left (d x + c\right )^{2} + 255 \, A a \sin \left (d x + c\right ) - 27 \, B a \sin \left (d x + c\right ) - 117 \, A a - 3 \, B a}{{\left (\sin \left (d x + c\right ) - 1\right )}^{3}}}{192 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/192*(6*(5*A*a - B*a)*log(abs(sin(d*x + c) + 1)) - 6*(5*A*a - B*a)*log(abs(sin(d*x + c) - 1)) - 3*(15*A*a*sin
(d*x + c)^2 - 3*B*a*sin(d*x + c)^2 + 38*A*a*sin(d*x + c) - 10*B*a*sin(d*x + c) + 25*A*a - 9*B*a)/(sin(d*x + c)
+ 1)^2 + (55*A*a*sin(d*x + c)^3 - 11*B*a*sin(d*x + c)^3 - 201*A*a*sin(d*x + c)^2 + 33*B*a*sin(d*x + c)^2 + 25
5*A*a*sin(d*x + c) - 27*B*a*sin(d*x + c) - 117*A*a - 3*B*a)/(sin(d*x + c) - 1)^3)/d

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maple [A]  time = 0.60, size = 217, normalized size = 1.38 \[ \frac {a A}{6 d \cos \left (d x +c \right )^{6}}+\frac {a B \left (\sin ^{3}\left (d x +c \right )\right )}{6 d \cos \left (d x +c \right )^{6}}+\frac {a B \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{4}}+\frac {a B \left (\sin ^{3}\left (d x +c \right )\right )}{16 d \cos \left (d x +c \right )^{2}}+\frac {a B \sin \left (d x +c \right )}{16 d}-\frac {a B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}+\frac {a A \tan \left (d x +c \right ) \left (\sec ^{5}\left (d x +c \right )\right )}{6 d}+\frac {5 a A \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{24 d}+\frac {5 a A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{16 d}+\frac {5 a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{16 d}+\frac {a B}{6 d \cos \left (d x +c \right )^{6}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.35, size = 171, normalized size = 1.09 \[ \frac {3 \, {\left (5 \, A - B\right )} a \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, {\left (5 \, A - B\right )} a \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, {\left (5 \, A - B\right )} a \sin \left (d x + c\right )^{4} - 3 \, {\left (5 \, A - B\right )} a \sin \left (d x + c\right )^{3} - 5 \, {\left (5 \, A - B\right )} a \sin \left (d x + c\right )^{2} + 5 \, {\left (5 \, A - B\right )} a \sin \left (d x + c\right ) + 8 \, {\left (A + B\right )} a\right )}}{\sin \left (d x + c\right )^{5} - \sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{3} + 2 \, \sin \left (d x + c\right )^{2} + \sin \left (d x + c\right ) - 1}}{96 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/96*(3*(5*A - B)*a*log(sin(d*x + c) + 1) - 3*(5*A - B)*a*log(sin(d*x + c) - 1) - 2*(3*(5*A - B)*a*sin(d*x + c
)^4 - 3*(5*A - B)*a*sin(d*x + c)^3 - 5*(5*A - B)*a*sin(d*x + c)^2 + 5*(5*A - B)*a*sin(d*x + c) + 8*(A + B)*a)/
(sin(d*x + c)^5 - sin(d*x + c)^4 - 2*sin(d*x + c)^3 + 2*sin(d*x + c)^2 + sin(d*x + c) - 1))/d

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mupad [B]  time = 9.26, size = 155, normalized size = 0.99 \[ \frac {a\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (5\,A-B\right )}{16\,d}-\frac {\left (\frac {5\,A\,a}{16}-\frac {B\,a}{16}\right )\,{\sin \left (c+d\,x\right )}^4+\left (\frac {B\,a}{16}-\frac {5\,A\,a}{16}\right )\,{\sin \left (c+d\,x\right )}^3+\left (\frac {5\,B\,a}{48}-\frac {25\,A\,a}{48}\right )\,{\sin \left (c+d\,x\right )}^2+\left (\frac {25\,A\,a}{48}-\frac {5\,B\,a}{48}\right )\,\sin \left (c+d\,x\right )+\frac {A\,a}{6}+\frac {B\,a}{6}}{d\,\left ({\sin \left (c+d\,x\right )}^5-{\sin \left (c+d\,x\right )}^4-2\,{\sin \left (c+d\,x\right )}^3+2\,{\sin \left (c+d\,x\right )}^2+\sin \left (c+d\,x\right )-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

(a*atanh(sin(c + d*x))*(5*A - B))/(16*d) - ((A*a)/6 + (B*a)/6 + sin(c + d*x)*((25*A*a)/48 - (5*B*a)/48) - sin(
c + d*x)^3*((5*A*a)/16 - (B*a)/16) + sin(c + d*x)^4*((5*A*a)/16 - (B*a)/16) - sin(c + d*x)^2*((25*A*a)/48 - (5
*B*a)/48))/(d*(sin(c + d*x) + 2*sin(c + d*x)^2 - 2*sin(c + d*x)^3 - sin(c + d*x)^4 + sin(c + d*x)^5 - 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))*(A+B*sin(d*x+c)),x)

[Out]

Timed out

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