3.1532 \(\int \sec (c+d x) (a+b \sin (c+d x)) (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=64 \[ -\frac {(a+b) (A+B) \log (1-\sin (c+d x))}{2 d}+\frac {(a-b) (A-B) \log (\sin (c+d x)+1)}{2 d}-\frac {b B \sin (c+d x)}{d} \]

[Out]

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

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Rubi [A]  time = 0.11, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.148, Rules used = {2837, 774, 633, 31} \[ -\frac {(a+b) (A+B) \log (1-\sin (c+d x))}{2 d}+\frac {(a-b) (A-B) \log (\sin (c+d x)+1)}{2 d}-\frac {b B \sin (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b)*(A + B)*Log[1 - Sin[c + d*x]])/(2*d) + ((a - b)*(A - B)*Log[1 + Sin[c + d*x]])/(2*d) - (b*B*Sin[c +
d*x])/d

Rule 31

Int[((a_) + (b_.)*(x_))^(-1), x_Symbol] :> Simp[Log[RemoveContent[a + b*x, x]]/b, x] /; FreeQ[{a, b}, x]

Rule 633

Int[((d_) + (e_.)*(x_))/((a_) + (c_.)*(x_)^2), x_Symbol] :> With[{q = Rt[-(a*c), 2]}, Dist[e/2 + (c*d)/(2*q),
Int[1/(-q + c*x), x], x] + Dist[e/2 - (c*d)/(2*q), Int[1/(q + c*x), x], x]] /; FreeQ[{a, c, d, e}, x] && NiceS
qrtQ[-(a*c)]

Rule 774

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

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]

Rubi steps

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

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Mathematica [A]  time = 0.03, size = 68, normalized size = 1.06 \[ \frac {a A \tanh ^{-1}(\sin (c+d x))}{d}-\frac {a B \log (\cos (c+d x))}{d}-\frac {A b \log (\cos (c+d x))}{d}-\frac {b B \sin (c+d x)}{d}+\frac {b B \tanh ^{-1}(\sin (c+d x))}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*A*ArcTanh[Sin[c + d*x]])/d + (b*B*ArcTanh[Sin[c + d*x]])/d - (A*b*Log[Cos[c + d*x]])/d - (a*B*Log[Cos[c + d
*x]])/d - (b*B*Sin[c + d*x])/d

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fricas [A]  time = 0.47, size = 66, normalized size = 1.03 \[ -\frac {2 \, B b \sin \left (d x + c\right ) - {\left ({\left (A - B\right )} a - {\left (A - B\right )} b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (A + B\right )} a + {\left (A + B\right )} b\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.17, size = 67, normalized size = 1.05 \[ -\frac {2 \, B b \sin \left (d x + c\right ) - {\left (A a - B a - A b + B b\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + {\left (A a + B a + A b + B b\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/2*(2*B*b*sin(d*x + c) - (A*a - B*a - A*b + B*b)*log(abs(sin(d*x + c) + 1)) + (A*a + B*a + A*b + B*b)*log(ab
s(sin(d*x + c) - 1)))/d

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maple [A]  time = 0.36, size = 83, normalized size = 1.30 \[ \frac {a A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-\frac {A b \ln \left (\cos \left (d x +c \right )\right )}{d}-\frac {b B \sin \left (d x +c \right )}{d}+\frac {B b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-\frac {a B \ln \left (\cos \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.31, size = 64, normalized size = 1.00 \[ -\frac {2 \, B b \sin \left (d x + c\right ) - {\left ({\left (A - B\right )} a - {\left (A - B\right )} b\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (A + B\right )} a + {\left (A + B\right )} b\right )} \log \left (\sin \left (d x + c\right ) - 1\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 0.12, size = 53, normalized size = 0.83 \[ -\frac {B\,b\,\sin \left (c+d\,x\right )-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (A-B\right )\,\left (a-b\right )}{2}+\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,\left (a+b\right )\,\left (A+B\right )}{2}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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