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

Optimal. Leaf size=100 \[ \frac{a^3 (A+B)}{8 d (a-a \sin (c+d x))^2}-\frac{a^2 (A-B)}{8 d (a \sin (c+d x)+a)}+\frac{a^2 A}{4 d (a-a \sin (c+d x))}+\frac{a (3 A-B) \tanh ^{-1}(\sin (c+d x))}{8 d} \]

[Out]

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

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Rubi [A]  time = 0.122077, antiderivative size = 100, normalized size of antiderivative = 1., 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} \[ \frac{a^3 (A+B)}{8 d (a-a \sin (c+d x))^2}-\frac{a^2 (A-B)}{8 d (a \sin (c+d x)+a)}+\frac{a^2 A}{4 d (a-a \sin (c+d x))}+\frac{a (3 A-B) \tanh ^{-1}(\sin (c+d x))}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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]

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])

Rubi steps

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

Mathematica [C]  time = 1.52678, size = 357, normalized size = 3.57 \[ \frac{a (\sin (c+d x)+1) \left (i x (3 A-B) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2+\frac{2 (A+B) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}{d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^4}-\frac{2 (3 A-B) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 \log \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )}{d}+\frac{(3 A-B) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 \log \left (\left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2\right )}{d}-\frac{2 i (3 A-B) \tan ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}{d}+\frac{2 (B-A)}{d}+\frac{4 A \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}{d \left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^2}\right )}{16 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^4} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*((2*(-A + B))/d + I*(3*A - B)*x*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 - ((2*I)*(3*A - B)*ArcTan[Tan[(c +
d*x)/2]]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/d - (2*(3*A - B)*Log[Cos[(c + d*x)/2] - Sin[(c + d*x)/2]]*(C
os[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/d + ((3*A - B)*Log[(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2]*(Cos[(c + d
*x)/2] + Sin[(c + d*x)/2])^2)/d + (2*(A + B)*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(d*(Cos[(c + d*x)/2] - S
in[(c + d*x)/2])^4) + (4*A*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2
))*(1 + Sin[c + d*x]))/(16*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4)

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Maple [A]  time = 0.096, size = 173, normalized size = 1.7 \begin{align*}{\frac{aA}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{aB \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{aB \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{aB\sin \left ( dx+c \right ) }{8\,d}}-{\frac{aB\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{aA\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{4\,d}}+{\frac{3\,aA\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{8\,d}}+{\frac{3\,aA\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{aB}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.02367, size = 155, normalized size = 1.55 \begin{align*} \frac{{\left (3 \, A - B\right )} a \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (3 \, A - B\right )} a \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left ({\left (3 \, A - B\right )} a \sin \left (d x + c\right )^{2} -{\left (3 \, A - B\right )} a \sin \left (d x + c\right ) - 2 \,{\left (A + B\right )} a\right )}}{\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )^{2} - \sin \left (d x + c\right ) + 1}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((3*A - B)*a*log(sin(d*x + c) + 1) - (3*A - B)*a*log(sin(d*x + c) - 1) - 2*((3*A - B)*a*sin(d*x + c)^2 -
(3*A - B)*a*sin(d*x + c) - 2*(A + B)*a)/(sin(d*x + c)^3 - sin(d*x + c)^2 - sin(d*x + c) + 1))/d

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Fricas [A]  time = 1.81289, size = 439, normalized size = 4.39 \begin{align*} -\frac{2 \,{\left (3 \, A - B\right )} a \cos \left (d x + c\right )^{2} + 2 \,{\left (3 \, A - B\right )} a \sin \left (d x + c\right ) - 2 \,{\left (A - 3 \, B\right )} a -{\left ({\left (3 \, A - B\right )} a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) -{\left (3 \, A - B\right )} a \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) +{\left ({\left (3 \, A - B\right )} a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) -{\left (3 \, A - B\right )} a \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{16 \,{\left (d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/16*(2*(3*A - B)*a*cos(d*x + c)^2 + 2*(3*A - B)*a*sin(d*x + c) - 2*(A - 3*B)*a - ((3*A - B)*a*cos(d*x + c)^2
*sin(d*x + c) - (3*A - B)*a*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + ((3*A - B)*a*cos(d*x + c)^2*sin(d*x + c) -
(3*A - B)*a*cos(d*x + c)^2)*log(-sin(d*x + c) + 1))/(d*cos(d*x + c)^2*sin(d*x + c) - d*cos(d*x + c)^2)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [A]  time = 1.34597, size = 205, normalized size = 2.05 \begin{align*} \frac{2 \,{\left (3 \, A a - B a\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 2 \,{\left (3 \, A a - B a\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{2 \,{\left (3 \, A a \sin \left (d x + c\right ) - B a \sin \left (d x + c\right ) + 5 \, A a - 3 \, B a\right )}}{\sin \left (d x + c\right ) + 1} + \frac{9 \, A a \sin \left (d x + c\right )^{2} - 3 \, B a \sin \left (d x + c\right )^{2} - 26 \, A a \sin \left (d x + c\right ) + 6 \, B a \sin \left (d x + c\right ) + 21 \, A a + B a}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{32 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/32*(2*(3*A*a - B*a)*log(abs(sin(d*x + c) + 1)) - 2*(3*A*a - B*a)*log(abs(sin(d*x + c) - 1)) - 2*(3*A*a*sin(d
*x + c) - B*a*sin(d*x + c) + 5*A*a - 3*B*a)/(sin(d*x + c) + 1) + (9*A*a*sin(d*x + c)^2 - 3*B*a*sin(d*x + c)^2
- 26*A*a*sin(d*x + c) + 6*B*a*sin(d*x + c) + 21*A*a + B*a)/(sin(d*x + c) - 1)^2)/d