3.975 \(\int \sec ^5(c+d x) (a+a \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=77 \[ \frac {a^4 (A+B)}{4 d (a-a \sin (c+d x))^2}+\frac {a^3 (A-B)}{4 d (a-a \sin (c+d x))}+\frac {a^2 (A-B) \tanh ^{-1}(\sin (c+d x))}{4 d} \]

[Out]

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

________________________________________________________________________________________

Rubi [A]  time = 0.12, antiderivative size = 77, 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} \[ \frac {a^4 (A+B)}{4 d (a-a \sin (c+d x))^2}+\frac {a^3 (A-B)}{4 d (a-a \sin (c+d x))}+\frac {a^2 (A-B) \tanh ^{-1}(\sin (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

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

________________________________________________________________________________________

Mathematica [A]  time = 0.14, size = 75, normalized size = 0.97 \[ \frac {a^5 \left (\frac {(A-B) \tanh ^{-1}(\sin (c+d x))}{4 a^3}+\frac {A-B}{4 a^2 (a-a \sin (c+d x))}+\frac {A+B}{4 a (a-a \sin (c+d x))^2}\right )}{d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^2*(A + B*Sin[c + d*x]),x]

[Out]

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

________________________________________________________________________________________

fricas [B]  time = 0.69, size = 161, normalized size = 2.09 \[ \frac {2 \, {\left (A - B\right )} a^{2} \sin \left (d x + c\right ) - 4 \, A a^{2} + {\left ({\left (A - B\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (A - B\right )} a^{2} \sin \left (d x + c\right ) - 2 \, {\left (A - B\right )} a^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (A - B\right )} a^{2} \cos \left (d x + c\right )^{2} + 2 \, {\left (A - B\right )} a^{2} \sin \left (d x + c\right ) - 2 \, {\left (A - B\right )} a^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{8 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

giac [A]  time = 0.24, size = 130, normalized size = 1.69 \[ \frac {2 \, {\left (A a^{2} - B a^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 2 \, {\left (A a^{2} - B a^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac {3 \, A a^{2} \sin \left (d x + c\right )^{2} - 3 \, B a^{2} \sin \left (d x + c\right )^{2} - 10 \, A a^{2} \sin \left (d x + c\right ) + 10 \, B a^{2} \sin \left (d x + c\right ) + 11 \, A a^{2} - 3 \, B a^{2}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/16*(2*(A*a^2 - B*a^2)*log(abs(sin(d*x + c) + 1)) - 2*(A*a^2 - B*a^2)*log(abs(sin(d*x + c) - 1)) + (3*A*a^2*s
in(d*x + c)^2 - 3*B*a^2*sin(d*x + c)^2 - 10*A*a^2*sin(d*x + c) + 10*B*a^2*sin(d*x + c) + 11*A*a^2 - 3*B*a^2)/(
sin(d*x + c) - 1)^2)/d

________________________________________________________________________________________

maple [B]  time = 0.57, size = 281, normalized size = 3.65 \[ \frac {a^{2} A \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {a^{2} A \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}+\frac {a^{2} A \sin \left (d x +c \right )}{8 d}+\frac {a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {B \,a^{2} \left (\sin ^{4}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {a^{2} A}{2 d \cos \left (d x +c \right )^{4}}+\frac {B \,a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{4}}+\frac {B \,a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{2}}+\frac {B \,a^{2} \sin \left (d x +c \right )}{4 d}-\frac {B \,a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {a^{2} A \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 a^{2} A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {B \,a^{2}}{4 d \cos \left (d x +c \right )^{4}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x)

[Out]

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

________________________________________________________________________________________

maxima [A]  time = 0.49, size = 87, normalized size = 1.13 \[ \frac {{\left (A - B\right )} a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A - B\right )} a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left ({\left (A - B\right )} a^{2} \sin \left (d x + c\right ) - 2 \, A a^{2}\right )}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^2*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

________________________________________________________________________________________

mupad [B]  time = 0.11, size = 73, normalized size = 0.95 \[ \frac {\frac {A\,a^2}{2}-\sin \left (c+d\,x\right )\,\left (\frac {A\,a^2}{4}-\frac {B\,a^2}{4}\right )}{d\,\left ({\sin \left (c+d\,x\right )}^2-2\,\sin \left (c+d\,x\right )+1\right )}+\frac {a^2\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (A-B\right )}{4\,d} \]

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)^5,x)

[Out]

((A*a^2)/2 - sin(c + d*x)*((A*a^2)/4 - (B*a^2)/4))/(d*(sin(c + d*x)^2 - 2*sin(c + d*x) + 1)) + (a^2*atanh(sin(
c + d*x))*(A - B))/(4*d)

________________________________________________________________________________________

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)**5*(a+a*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)

[Out]

Timed out

________________________________________________________________________________________