3.1534 \(\int \sec ^5(c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=88 \[ \frac{(3 a A-b B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\sec ^4(c+d x) ((a A+b B) \sin (c+d x)+a B+A b)}{4 d}+\frac{(3 a A-b B) \tan (c+d x) \sec (c+d x)}{8 d} \]

[Out]

((3*a*A - b*B)*ArcTanh[Sin[c + d*x]])/(8*d) + (Sec[c + d*x]^4*(A*b + a*B + (a*A + b*B)*Sin[c + d*x]))/(4*d) +
((3*a*A - b*B)*Sec[c + d*x]*Tan[c + d*x])/(8*d)

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Rubi [A]  time = 0.0897351, antiderivative size = 88, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.138, Rules used = {2837, 778, 199, 206} \[ \frac{(3 a A-b B) \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{\sec ^4(c+d x) ((a A+b B) \sin (c+d x)+a B+A b)}{4 d}+\frac{(3 a A-b B) \tan (c+d x) \sec (c+d x)}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a*A - b*B)*ArcTanh[Sin[c + d*x]])/(8*d) + (Sec[c + d*x]^4*(A*b + a*B + (a*A + b*B)*Sin[c + d*x]))/(4*d) +
((3*a*A - b*B)*Sec[c + d*x]*Tan[c + d*x])/(8*d)

Rule 2837

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

Rule 778

Int[((d_.) + (e_.)*(x_))*((f_.) + (g_.)*(x_))*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((a*(e*f + d*g) -
(c*d*f - a*e*g)*x)*(a + c*x^2)^(p + 1))/(2*a*c*(p + 1)), x] - Dist[(a*e*g - c*d*f*(2*p + 3))/(2*a*c*(p + 1)),
Int[(a + c*x^2)^(p + 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && LtQ[p, -1]

Rule 199

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && LtQ[p, -1] && (In
tegerQ[2*p] || (n == 2 && IntegerQ[4*p]) || (n == 2 && IntegerQ[3*p]) || Denominator[p + 1/n] < Denominator[p]
)

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

Mathematica [A]  time = 0.55725, size = 82, normalized size = 0.93 \[ \frac{\sec ^4(c+d x) \left ((b B-3 a A) \sin ^3(c+d x)+(5 a A+b B) \sin (c+d x)+(3 a A-b B) \cos ^4(c+d x) \tanh ^{-1}(\sin (c+d x))+2 (a B+A b)\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B]  time = 0.08, size = 173, normalized size = 2. \begin{align*}{\frac{A\tan \left ( dx+c \right ) a \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}}}+{\frac{Ab}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{B \left ( \sin \left ( dx+c \right ) \right ) ^{3}b}{4\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{B \left ( \sin \left ( dx+c \right ) \right ) ^{3}b}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{Bb\sin \left ( dx+c \right ) }{8\,d}}-{\frac{Bb\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.985599, size = 151, normalized size = 1.72 \begin{align*} \frac{{\left (3 \, A a - B b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (3 \, A a - B b\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left ({\left (3 \, A a - B b\right )} \sin \left (d x + c\right )^{3} - 2 \, B a - 2 \, A b -{\left (5 \, A a + B b\right )} \sin \left (d x + c\right )\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{2} + 1}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 1.49354, size = 286, normalized size = 3.25 \begin{align*} \frac{{\left (3 \, A a - B b\right )} \cos \left (d x + c\right )^{4} \log \left (\sin \left (d x + c\right ) + 1\right ) -{\left (3 \, A a - B b\right )} \cos \left (d x + c\right )^{4} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 4 \, B a + 4 \, A b + 2 \,{\left ({\left (3 \, A a - B b\right )} \cos \left (d x + c\right )^{2} + 2 \, A a + 2 \, B b\right )} \sin \left (d x + c\right )}{16 \, d \cos \left (d x + c\right )^{4}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((3*A*a - B*b)*cos(d*x + c)^4*log(sin(d*x + c) + 1) - (3*A*a - B*b)*cos(d*x + c)^4*log(-sin(d*x + c) + 1)
+ 4*B*a + 4*A*b + 2*((3*A*a - B*b)*cos(d*x + c)^2 + 2*A*a + 2*B*b)*sin(d*x + c))/(d*cos(d*x + c)^4)

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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+b*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.27742, size = 154, normalized size = 1.75 \begin{align*} \frac{{\left (3 \, A a - B b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) -{\left (3 \, A a - B b\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 )^{3} - B b \sin \left (d x + c\right )^{3} - 5 \, A a \sin \left (d x + c\right ) - B b \sin \left (d x + c\right ) - 2 \, B a - 2 \, A b\right )}}{{\left (\sin \left (d x + c\right )^{2} - 1\right )}^{2}}}{16 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*((3*A*a - B*b)*log(abs(sin(d*x + c) + 1)) - (3*A*a - B*b)*log(abs(sin(d*x + c) - 1)) - 2*(3*A*a*sin(d*x +
c)^3 - B*b*sin(d*x + c)^3 - 5*A*a*sin(d*x + c) - B*b*sin(d*x + c) - 2*B*a - 2*A*b)/(sin(d*x + c)^2 - 1)^2)/d