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

Optimal. Leaf size=55 \[ \frac{a^2 (A+2 B) \cos (c+d x)}{d}+a^2 x (-(A+2 B))+\frac{(A+B) \sec (c+d x) (a \sin (c+d x)+a)^2}{d} \]

[Out]

-(a^2*(A + 2*B)*x) + (a^2*(A + 2*B)*Cos[c + d*x])/d + ((A + B)*Sec[c + d*x]*(a + a*Sin[c + d*x])^2)/d

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Rubi [A]  time = 0.0933537, antiderivative size = 55, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {2855, 2638} \[ \frac{a^2 (A+2 B) \cos (c+d x)}{d}+a^2 x (-(A+2 B))+\frac{(A+B) \sec (c+d x) (a \sin (c+d x)+a)^2}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-(a^2*(A + 2*B)*x) + (a^2*(A + 2*B)*Cos[c + d*x])/d + ((A + B)*Sec[c + d*x]*(a + a*Sin[c + d*x])^2)/d

Rule 2855

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.)
+ (f_.)*(x_)]), x_Symbol] :> -Simp[((b*c + a*d)*(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(p +
1)), x] + Dist[(b*(a*d*m + b*c*(m + p + 1)))/(a*g^2*(p + 1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x]
)^(m - 1), x], x] /; FreeQ[{a, b, c, d, e, f, g}, x] && EqQ[a^2 - b^2, 0] && GtQ[m, -1] && LtQ[p, -1]

Rule 2638

Int[sin[(c_.) + (d_.)*(x_)], x_Symbol] :> -Simp[Cos[c + d*x]/d, x] /; FreeQ[{c, d}, x]

Rubi steps

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

Mathematica [A]  time = 0.241612, size = 91, normalized size = 1.65 \[ \frac{a^2 \sec (c+d x) \left (4 (A+2 B) \sin ^{-1}\left (\frac{\sqrt{1-\sin (c+d x)}}{\sqrt{2}}\right ) \sqrt{\cos ^2(c+d x)}+4 A \sin (c+d x)+4 A+4 B \sin (c+d x)+B \cos (2 (c+d x))+5 B\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*Sec[c + d*x]*(4*A + 5*B + 4*(A + 2*B)*ArcSin[Sqrt[1 - Sin[c + d*x]]/Sqrt[2]]*Sqrt[Cos[c + d*x]^2] + B*Cos
[2*(c + d*x)] + 4*A*Sin[c + d*x] + 4*B*Sin[c + d*x]))/(2*d)

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Maple [B]  time = 0.079, size = 123, normalized size = 2.2 \begin{align*}{\frac{1}{d} \left ({a}^{2}A \left ( \tan \left ( dx+c \right ) -dx-c \right ) +B{a}^{2} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{\cos \left ( dx+c \right ) }}+ \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) \right ) +2\,{\frac{{a}^{2}A}{\cos \left ( dx+c \right ) }}+2\,B{a}^{2} \left ( \tan \left ( dx+c \right ) -dx-c \right ) +{a}^{2}A\tan \left ( dx+c \right ) +{\frac{B{a}^{2}}{\cos \left ( dx+c \right ) }} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.52915, size = 140, normalized size = 2.55 \begin{align*} -\frac{{\left (d x + c - \tan \left (d x + c\right )\right )} A a^{2} + 2 \,{\left (d x + c - \tan \left (d x + c\right )\right )} B a^{2} - B a^{2}{\left (\frac{1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - A a^{2} \tan \left (d x + c\right ) - \frac{2 \, A a^{2}}{\cos \left (d x + c\right )} - \frac{B a^{2}}{\cos \left (d x + c\right )}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [B]  time = 2.09241, size = 302, normalized size = 5.49 \begin{align*} -\frac{{\left (A + 2 \, B\right )} a^{2} d x - B a^{2} \cos \left (d x + c\right )^{2} - 2 \,{\left (A + B\right )} a^{2} +{\left ({\left (A + 2 \, B\right )} a^{2} d x -{\left (2 \, A + 3 \, B\right )} a^{2}\right )} \cos \left (d x + c\right ) -{\left ({\left (A + 2 \, B\right )} a^{2} d x - B a^{2} \cos \left (d x + c\right ) + 2 \,{\left (A + B\right )} a^{2}\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [B]  time = 1.3381, size = 169, normalized size = 3.07 \begin{align*} -\frac{{\left (A a^{2} + 2 \, B a^{2}\right )}{\left (d x + c\right )} + \frac{2 \,{\left (2 \, A a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 2 \, B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - B a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, A a^{2} + 3 \, B a^{2}\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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