3.958 \(\int \sec ^3(c+d x) (a+a \sin (c+d x)) (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=47 \[ \frac {a^2 (A+B)}{2 d (a-a \sin (c+d x))}+\frac {a (A-B) \tanh ^{-1}(\sin (c+d x))}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.09, antiderivative size = 47, 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} \[ \frac {a^2 (A+B)}{2 d (a-a \sin (c+d x))}+\frac {a (A-B) \tanh ^{-1}(\sin (c+d x))}{2 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^3*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]

[Out]

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

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Mathematica [C]  time = 0.63, size = 260, normalized size = 5.53 \[ \frac {a \left (2 i (A-B) (\sin (c+d x)-1) \tan ^{-1}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )+(A-B) \sin (c+d x) \left (2 \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-\log \left (\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2\right )-i d x\right )-2 A \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )+A \log \left (\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2\right )+i A d x+2 A+2 B \log \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )-B \log \left (\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2\right )-i B d x+2 B\right )}{4 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^3*(a + a*Sin[c + d*x])*(A + B*Sin[c + d*x]),x]

[Out]

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

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fricas [B]  time = 0.71, size = 90, normalized size = 1.91 \[ -\frac {2 \, {\left (A + B\right )} a - {\left ({\left (A - B\right )} a \sin \left (d x + c\right ) - {\left (A - B\right )} a\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (A - B\right )} a \sin \left (d x + c\right ) - {\left (A - B\right )} a\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{4 \, {\left (d \sin \left (d x + c\right ) - d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.23, size = 84, normalized size = 1.79 \[ \frac {{\left (A a - B a\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - {\left (A a - B a\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac {A a \sin \left (d x + c\right ) - B a \sin \left (d x + c\right ) - 3 \, A a - B a}{\sin \left (d x + c\right ) - 1}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

1/4*((A*a - B*a)*log(abs(sin(d*x + c) + 1)) - (A*a - B*a)*log(abs(sin(d*x + c) - 1)) + (A*a*sin(d*x + c) - B*a
*sin(d*x + c) - 3*A*a - B*a)/(sin(d*x + c) - 1))/d

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maple [B]  time = 0.57, size = 129, normalized size = 2.74 \[ \frac {a A}{2 d \cos \left (d x +c \right )^{2}}+\frac {a B \left (\sin ^{3}\left (d x +c \right )\right )}{2 d \cos \left (d x +c \right )^{2}}+\frac {a B \sin \left (d x +c \right )}{2 d}-\frac {a B \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {a A \sec \left (d x +c \right ) \tan \left (d x +c \right )}{2 d}+\frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{2 d}+\frac {a B}{2 d \cos \left (d x +c \right )^{2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

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

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maxima [A]  time = 0.34, size = 55, normalized size = 1.17 \[ \frac {{\left (A - B\right )} a \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (A - B\right )} a \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (A + B\right )} a}{\sin \left (d x + c\right ) - 1}}{4 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 9.12, size = 43, normalized size = 0.91 \[ \frac {a\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )\,\left (A-B\right )}{2\,d}-\frac {\frac {A\,a}{2}+\frac {B\,a}{2}}{d\,\left (\sin \left (c+d\,x\right )-1\right )} \]

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)^3,x)

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int A \sec ^{3}{\left (c + d x \right )}\, dx + \int A \sin {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int B \sin {\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx + \int B \sin ^{2}{\left (c + d x \right )} \sec ^{3}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3*(a+a*sin(d*x+c))*(A+B*sin(d*x+c)),x)

[Out]

a*(Integral(A*sec(c + d*x)**3, x) + Integral(A*sin(c + d*x)*sec(c + d*x)**3, x) + Integral(B*sin(c + d*x)*sec(
c + d*x)**3, x) + Integral(B*sin(c + d*x)**2*sec(c + d*x)**3, x))

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