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]

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

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Rubi [A]  time = 0.10, antiderivative size = 91, normalized size of antiderivative = 1.00, 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 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]

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]

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*}

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Mathematica [C]  time = 0.26, size = 82, normalized size = 0.90 \[ \frac {\sec (c+d x) \left (4 \sqrt {2} a^3 (2 A+3 B) \sqrt {\sin (c+d x)+1} \, _2F_1\left (-\frac {3}{2},-\frac {1}{2};\frac {1}{2};\frac {1}{2} (1-\sin (c+d x))\right )-B (a \sin (c+d x)+a)^3\right )}{2 d} \]

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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fricas [A]  time = 0.68, size = 173, normalized size = 1.90 \[ \frac {B a^{3} \cos \left (d x + c\right )^{3} - 3 \, {\left (2 \, A + 3 \, B\right )} a^{3} d x + 2 \, {\left (A + 3 \, B\right )} a^{3} \cos \left (d x + c\right )^{2} + 8 \, {\left (A + B\right )} a^{3} - {\left (3 \, {\left (2 \, A + 3 \, B\right )} a^{3} d x - {\left (10 \, A + 13 \, B\right )} a^{3}\right )} \cos \left (d x + c\right ) + {\left (3 \, {\left (2 \, A + 3 \, B\right )} a^{3} d x + B a^{3} \cos \left (d x + c\right )^{2} - {\left (2 \, A + 5 \, B\right )} a^{3} \cos \left (d x + c\right ) + 8 \, {\left (A + B\right )} a^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d\right )}} \]

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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giac [A]  time = 0.20, size = 147, normalized size = 1.62 \[ -\frac {3 \, {\left (2 \, A a^{3} + 3 \, B a^{3}\right )} {\left (d x + c\right )} + \frac {16 \, {\left (A a^{3} + B a^{3}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1} + \frac {2 \, {\left (B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 2 \, A a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 6 \, B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - B a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 2 \, A a^{3} - 6 \, B a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

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

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maple [B]  time = 0.68, size = 219, normalized size = 2.41 \[ \frac {a^{3} A \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+B \,a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+3 a^{3} A \left (\tan \left (d x +c \right )-d x -c \right )+3 B \,a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+\frac {3 a^{3} A}{\cos \left (d x +c \right )}+3 B \,a^{3} \left (\tan \left (d x +c \right )-d x -c \right )+a^{3} A \tan \left (d x +c \right )+\frac {B \,a^{3}}{\cos \left (d x +c \right )}}{d} \]

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 = 0.41, size = 167, normalized size = 1.84 \[ -\frac {6 \, {\left (d x + c - \tan \left (d x + c\right )\right )} A a^{3} + {\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} B a^{3} + 6 \, {\left (d x + c - \tan \left (d x + c\right )\right )} B a^{3} - 2 \, A a^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - 6 \, B a^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - 2 \, A a^{3} \tan \left (d x + c\right ) - \frac {6 \, A a^{3}}{\cos \left (d x + c\right )} - \frac {2 \, B a^{3}}{\cos \left (d x + c\right )}}{2 \, d} \]

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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mupad [B]  time = 11.51, size = 234, normalized size = 2.57 \[ -\frac {10\,A\,a^3-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A\,a^3+5\,B\,a^3\right )+14\,B\,a^3-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (2\,A\,a^3+7\,B\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (8\,A\,a^3+9\,B\,a^3\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (18\,A\,a^3+21\,B\,a^3\right )}{d\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^5-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4+2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3-2\,{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}-\frac {3\,a^3\,\mathrm {atan}\left (\frac {3\,a^3\,\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (2\,A+3\,B\right )}{6\,A\,a^3+9\,B\,a^3}\right )\,\left (2\,A+3\,B\right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (10*A*a^3 - tan(c/2 + (d*x)/2)*(2*A*a^3 + 5*B*a^3) + 14*B*a^3 - tan(c/2 + (d*x)/2)^3*(2*A*a^3 + 7*B*a^3) + t
an(c/2 + (d*x)/2)^4*(8*A*a^3 + 9*B*a^3) + tan(c/2 + (d*x)/2)^2*(18*A*a^3 + 21*B*a^3))/(d*(tan(c/2 + (d*x)/2) -
2*tan(c/2 + (d*x)/2)^2 + 2*tan(c/2 + (d*x)/2)^3 - tan(c/2 + (d*x)/2)^4 + tan(c/2 + (d*x)/2)^5 - 1)) - (3*a^3*
atan((3*a^3*tan(c/2 + (d*x)/2)*(2*A + 3*B))/(6*A*a^3 + 9*B*a^3))*(2*A + 3*B))/d

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{3} \left (\int A \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 A \sin {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 A \sin ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int A \sin ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int B \sin {\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 B \sin ^{2}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int 3 B \sin ^{3}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx + \int B \sin ^{4}{\left (c + d x \right )} \sec ^{2}{\left (c + d x \right )}\, dx\right ) \]

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]

a**3*(Integral(A*sec(c + d*x)**2, x) + Integral(3*A*sin(c + d*x)*sec(c + d*x)**2, x) + Integral(3*A*sin(c + d*
x)**2*sec(c + d*x)**2, x) + Integral(A*sin(c + d*x)**3*sec(c + d*x)**2, x) + Integral(B*sin(c + d*x)*sec(c + d
*x)**2, x) + Integral(3*B*sin(c + d*x)**2*sec(c + d*x)**2, x) + Integral(3*B*sin(c + d*x)**3*sec(c + d*x)**2,
x) + Integral(B*sin(c + d*x)**4*sec(c + d*x)**2, x))

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