3.998 \(\int \sec ^2(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx\)

Optimal. Leaf size=91 \[ \frac{2 a^3 (2 A+3 B) \cos (c+d x)}{d}+\frac{a^3 (2 A+3 B) \sin (c+d x) \cos (c+d x)}{2 d}-\frac{3}{2} a^3 x (2 A+3 B)+\frac{(A+B) \sec (c+d x) (a \sin (c+d x)+a)^3}{d} \]

[Out]

(-3*a^3*(2*A + 3*B)*x)/2 + (2*a^3*(2*A + 3*B)*Cos[c + d*x])/d + (a^3*(2*A + 3*B)*Cos[c + d*x]*Sin[c + d*x])/(2
*d) + ((A + B)*Sec[c + d*x]*(a + a*Sin[c + d*x])^3)/d

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Rubi [A]  time = 0.104856, antiderivative size = 91, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 31, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.065, Rules used = {2855, 2644} \[ \frac{2 a^3 (2 A+3 B) \cos (c+d x)}{d}+\frac{a^3 (2 A+3 B) \sin (c+d x) \cos (c+d x)}{2 d}-\frac{3}{2} a^3 x (2 A+3 B)+\frac{(A+B) \sec (c+d x) (a \sin (c+d x)+a)^3}{d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^2*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

(-3*a^3*(2*A + 3*B)*x)/2 + (2*a^3*(2*A + 3*B)*Cos[c + d*x])/d + (a^3*(2*A + 3*B)*Cos[c + d*x]*Sin[c + d*x])/(2
*d) + ((A + B)*Sec[c + d*x]*(a + a*Sin[c + d*x])^3)/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 2644

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^2, x_Symbol] :> Simp[((2*a^2 + b^2)*x)/2, x] + (-Simp[(2*a*b*Cos[c
+ d*x])/d, x] - Simp[(b^2*Cos[c + d*x]*Sin[c + d*x])/(2*d), x]) /; FreeQ[{a, b, c, d}, x]

Rubi steps

\begin{align*} \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 (A+B \sin (c+d x)) \, dx &=\frac{(A+B) \sec (c+d x) (a+a \sin (c+d x))^3}{d}-(a (2 A+3 B)) \int (a+a \sin (c+d x))^2 \, dx\\ &=-\frac{3}{2} a^3 (2 A+3 B) x+\frac{2 a^3 (2 A+3 B) \cos (c+d x)}{d}+\frac{a^3 (2 A+3 B) \cos (c+d x) \sin (c+d x)}{2 d}+\frac{(A+B) \sec (c+d x) (a+a \sin (c+d x))^3}{d}\\ \end{align*}

Mathematica [C]  time = 0.243196, size = 82, normalized size = 0.9 \[ \frac{\sec (c+d x) \left (4 \sqrt{2} a^3 (2 A+3 B) \sqrt{\sin (c+d x)+1} \, _2F_1\left (-\frac{3}{2},-\frac{1}{2};\frac{1}{2};\frac{1}{2} (1-\sin (c+d x))\right )-B (a \sin (c+d x)+a)^3\right )}{2 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^2*(a + a*Sin[c + d*x])^3*(A + B*Sin[c + d*x]),x]

[Out]

(Sec[c + d*x]*(4*Sqrt[2]*a^3*(2*A + 3*B)*Hypergeometric2F1[-3/2, -1/2, 1/2, (1 - Sin[c + d*x])/2]*Sqrt[1 + Sin
[c + d*x]] - B*(a + a*Sin[c + d*x])^3))/(2*d)

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Maple [B]  time = 0.087, size = 219, normalized size = 2.4 \begin{align*}{\frac{1}{d} \left ({a}^{3}A \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 ) +B{a}^{3} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{5}}{\cos \left ( dx+c \right ) }}+ \left ( \left ( \sin \left ( dx+c \right ) \right ) ^{3}+{\frac{3\,\sin \left ( dx+c \right ) }{2}} \right ) \cos \left ( dx+c \right ) -{\frac{3\,dx}{2}}-{\frac{3\,c}{2}} \right ) +3\,{a}^{3}A \left ( \tan \left ( dx+c \right ) -dx-c \right ) +3\,B{a}^{3} \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 ) +3\,{\frac{{a}^{3}A}{\cos \left ( dx+c \right ) }}+3\,B{a}^{3} \left ( \tan \left ( dx+c \right ) -dx-c \right ) +{a}^{3}A\tan \left ( dx+c \right ) +{\frac{B{a}^{3}}{\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))^3*(A+B*sin(d*x+c)),x)

[Out]

1/d*(a^3*A*(sin(d*x+c)^4/cos(d*x+c)+(2+sin(d*x+c)^2)*cos(d*x+c))+B*a^3*(sin(d*x+c)^5/cos(d*x+c)+(sin(d*x+c)^3+
3/2*sin(d*x+c))*cos(d*x+c)-3/2*d*x-3/2*c)+3*a^3*A*(tan(d*x+c)-d*x-c)+3*B*a^3*(sin(d*x+c)^4/cos(d*x+c)+(2+sin(d
*x+c)^2)*cos(d*x+c))+3*a^3*A/cos(d*x+c)+3*B*a^3*(tan(d*x+c)-d*x-c)+a^3*A*tan(d*x+c)+B*a^3/cos(d*x+c))

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Maxima [A]  time = 1.61656, size = 225, normalized size = 2.47 \begin{align*} -\frac{6 \,{\left (d x + c - \tan \left (d x + c\right )\right )} A a^{3} +{\left (3 \, d x + 3 \, c - \frac{\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} B a^{3} + 6 \,{\left (d x + c - \tan \left (d x + c\right )\right )} B a^{3} - 2 \, A a^{3}{\left (\frac{1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - 6 \, B a^{3}{\left (\frac{1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - 2 \, A a^{3} \tan \left (d x + c\right ) - \frac{6 \, A a^{3}}{\cos \left (d x + c\right )} - \frac{2 \, B a^{3}}{\cos \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)^2*(a+a*sin(d*x+c))^3*(A+B*sin(d*x+c)),x, algorithm="maxima")

[Out]

-1/2*(6*(d*x + c - tan(d*x + c))*A*a^3 + (3*d*x + 3*c - tan(d*x + c)/(tan(d*x + c)^2 + 1) - 2*tan(d*x + c))*B*
a^3 + 6*(d*x + c - tan(d*x + c))*B*a^3 - 2*A*a^3*(1/cos(d*x + c) + cos(d*x + c)) - 6*B*a^3*(1/cos(d*x + c) + c
os(d*x + c)) - 2*A*a^3*tan(d*x + c) - 6*A*a^3/cos(d*x + c) - 2*B*a^3/cos(d*x + c))/d

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Fricas [A]  time = 1.68607, size = 414, normalized size = 4.55 \begin{align*} \frac{B a^{3} \cos \left (d x + c\right )^{3} - 3 \,{\left (2 \, A + 3 \, B\right )} a^{3} d x + 2 \,{\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 8 \,{\left (A + B\right )} a^{3} -{\left (3 \,{\left (2 \, A + 3 \, B\right )} a^{3} d x -{\left (10 \, A + 13 \, B\right )} a^{3}\right )} \cos \left (d x + c\right ) +{\left (3 \,{\left (2 \, A + 3 \, B\right )} a^{3} d x + B a^{3} \cos \left (d x + c\right )^{2} -{\left (2 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right ) + 8 \,{\left (A + B\right )} a^{3}\right )} \sin \left (d x + c\right )}{2 \,{\left (d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d\right )}} \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))^3*(A+B*sin(d*x+c)),x, algorithm="fricas")

[Out]

1/2*(B*a^3*cos(d*x + c)^3 - 3*(2*A + 3*B)*a^3*d*x + 2*(A + 3*B)*a^3*cos(d*x + c)^2 + 8*(A + B)*a^3 - (3*(2*A +
3*B)*a^3*d*x - (10*A + 13*B)*a^3)*cos(d*x + c) + (3*(2*A + 3*B)*a^3*d*x + B*a^3*cos(d*x + c)^2 - (2*A + 5*B)*
a^3*cos(d*x + c) + 8*(A + B)*a^3)*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))**3*(A+B*sin(d*x+c)),x)

[Out]

Timed out

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Giac [A]  time = 1.30437, size = 198, normalized size = 2.18 \begin{align*} -\frac{3 \,{\left (2 \, A a^{3} + 3 \, B a^{3}\right )}{\left (d x + c\right )} + \frac{16 \,{\left (A a^{3} + B a^{3}\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 1} + \frac{2 \,{\left (B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} - 2 \, A a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 6 \, B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - B a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) - 2 \, A a^{3} - 6 \, B a^{3}\right )}}{{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, 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))^3*(A+B*sin(d*x+c)),x, algorithm="giac")

[Out]

-1/2*(3*(2*A*a^3 + 3*B*a^3)*(d*x + c) + 16*(A*a^3 + B*a^3)/(tan(1/2*d*x + 1/2*c) - 1) + 2*(B*a^3*tan(1/2*d*x +
1/2*c)^3 - 2*A*a^3*tan(1/2*d*x + 1/2*c)^2 - 6*B*a^3*tan(1/2*d*x + 1/2*c)^2 - B*a^3*tan(1/2*d*x + 1/2*c) - 2*A
*a^3 - 6*B*a^3)/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d