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]

-((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)

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Rubi [A]  time = 0.173667, antiderivative size = 94, normalized size of antiderivative = 1., 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 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 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 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 31

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

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*}

Mathematica [A]  time = 0.197923, size = 81, normalized size = 0.86 \[ -\frac{2 b (2 a B+A b) \sin (c+d x)+(a-b)^2 (-(A-B)) \log (\sin (c+d x)+1)+(a+b)^2 (A+B) \log (1-\sin (c+d x))+b^2 B \sin ^2(c+d x)}{2 d} \]

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]

-((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)*Sin[c
+ d*x] + b^2*B*Sin[c + d*x]^2)/(2*d)

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Maple [A]  time = 0.079, size = 161, normalized size = 1.7 \begin{align*}{\frac{{a}^{2}A\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-{\frac{B{a}^{2}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-2\,{\frac{Aab\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}}-2\,{\frac{Bab\sin \left ( dx+c \right ) }{d}}+2\,{\frac{Bab\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-{\frac{A{b}^{2}\sin \left ( dx+c \right ) }{d}}+{\frac{A{b}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{d}}-{\frac{B{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{2}}{2\,d}}-{\frac{B{b}^{2}\ln \left ( \cos \left ( dx+c \right ) \right ) }{d}} \end{align*}

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*sin(d*x+c)+2/d
*B*a*b*ln(sec(d*x+c)+tan(d*x+c))-1/d*A*b^2*sin(d*x+c)+1/d*A*b^2*ln(sec(d*x+c)+tan(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 = 0.987584, size = 147, normalized size = 1.56 \begin{align*} -\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} \end{align*}

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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Fricas [A]  time = 1.54315, size = 273, normalized size = 2.9 \begin{align*} \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} \end{align*}

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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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \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 \end{align*}

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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Giac [A]  time = 1.37102, size = 174, normalized size = 1.85 \begin{align*} -\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} \end{align*}

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