3.1540 \(\int \sec (c+d x) (a+b \sin (c+d x))^2 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=94 \[ -\frac {b (2 a B+A b) \sin (c+d x)}{d}+\frac {(a-b)^2 (A-B) \log (\sin (c+d x)+1)}{2 d}-\frac {(a+b)^2 (A+B) \log (1-\sin (c+d x))}{2 d}-\frac {b^2 B \sin ^2(c+d x)}{2 d} \]

[Out]

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

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Rubi [A]  time = 0.17, antiderivative size = 94, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2837, 801, 633, 31} \[ -\frac {b (2 a B+A b) \sin (c+d x)}{d}+\frac {(a-b)^2 (A-B) \log (\sin (c+d x)+1)}{2 d}-\frac {(a+b)^2 (A+B) \log (1-\sin (c+d x))}{2 d}-\frac {b^2 B \sin ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-((a + b)^2*(A + B)*Log[1 - Sin[c + d*x]])/(2*d) + ((a - b)^2*(A - B)*Log[1 + Sin[c + d*x]])/(2*d) - (b*(A*b +
2*a*B)*Sin[c + d*x])/d - (b^2*B*Sin[c + d*x]^2)/(2*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 801

Int[(((d_.) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_)))/((a_) + (c_.)*(x_)^2), x_Symbol] :> Int[ExpandIntegrand[(
(d + e*x)^m*(f + g*x))/(a + c*x^2), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[c*d^2 + a*e^2, 0] && Integer
Q[m]

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

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Mathematica [A]  time = 0.21, size = 81, normalized size = 0.86 \[ -\frac {2 b (2 a B+A b) \sin (c+d x)-\left ((a-b)^2 (A-B) \log (\sin (c+d x)+1)\right )+(a+b)^2 (A+B) \log (1-\sin (c+d x))+b^2 B \sin ^2(c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.48, size = 111, normalized size = 1.18 \[ \frac {B b^{2} \cos \left (d x + c\right )^{2} + {\left ({\left (A - B\right )} a^{2} - 2 \, {\left (A - B\right )} a b + {\left (A - B\right )} b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left ({\left (A + B\right )} a^{2} + 2 \, {\left (A + B\right )} a b + {\left (A + B\right )} b^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \, {\left (2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.20, size = 129, normalized size = 1.37 \[ -\frac {B b^{2} \sin \left (d x + c\right )^{2} + 4 \, B a b \sin \left (d x + c\right ) + 2 \, A b^{2} \sin \left (d x + c\right ) - {\left (A a^{2} - B a^{2} - 2 \, A a b + 2 \, B a b + A b^{2} - B b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) + {\left (A a^{2} + B a^{2} + 2 \, A a b + 2 \, B a b + A b^{2} + B b^{2}\right )} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.38, size = 161, normalized size = 1.71 \[ \frac {a^{2} A \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-\frac {B \,a^{2} \ln \left (\cos \left (d x +c \right )\right )}{d}-\frac {2 A a b \ln \left (\cos \left (d x +c \right )\right )}{d}+\frac {2 B a b \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-\frac {2 B a b \sin \left (d x +c \right )}{d}+\frac {A \,b^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{d}-\frac {A \,b^{2} \sin \left (d x +c \right )}{d}-\frac {b^{2} B \left (\sin ^{2}\left (d x +c \right )\right )}{2 d}-\frac {B \,b^{2} \ln \left (\cos \left (d x +c \right )\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.34, size = 109, normalized size = 1.16 \[ -\frac {B b^{2} \sin \left (d x + c\right )^{2} - {\left ({\left (A - B\right )} a^{2} - 2 \, {\left (A - B\right )} a b + {\left (A - B\right )} b^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + {\left ({\left (A + B\right )} a^{2} + 2 \, {\left (A + B\right )} a b + {\left (A + B\right )} b^{2}\right )} \log \left (\sin \left (d x + c\right ) - 1\right ) + 2 \, {\left (2 \, B a b + A b^{2}\right )} \sin \left (d x + c\right )}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 12.37, size = 80, normalized size = 0.85 \[ -\frac {\sin \left (c+d\,x\right )\,\left (A\,b^2+2\,B\,a\,b\right )+\frac {\ln \left (\sin \left (c+d\,x\right )-1\right )\,{\left (a+b\right )}^2\,\left (A+B\right )}{2}+\frac {B\,b^2\,{\sin \left (c+d\,x\right )}^2}{2}-\frac {\ln \left (\sin \left (c+d\,x\right )+1\right )\,\left (A-B\right )\,{\left (a-b\right )}^2}{2}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(((A + B*sin(c + d*x))*(a + b*sin(c + d*x))^2)/cos(c + d*x),x)

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)*(a+b*sin(d*x+c))**2*(A+B*sin(d*x+c)),x)

[Out]

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

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