3.767 \(\int \sec (c+d x) (a+a \sin (c+d x))^3 \tan (c+d x) \, dx\)

Optimal. Leaf size=67 \[ \frac {6 a^3 \cos (c+d x)}{d}+\frac {3 a^3 \sin (c+d x) \cos (c+d x)}{2 d}-\frac {9 a^3 x}{2}+\frac {\sec (c+d x) (a \sin (c+d x)+a)^3}{d} \]

[Out]

-9/2*a^3*x+6*a^3*cos(d*x+c)/d+3/2*a^3*cos(d*x+c)*sin(d*x+c)/d+sec(d*x+c)*(a+a*sin(d*x+c))^3/d

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Rubi [A]  time = 0.06, antiderivative size = 67, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.080, Rules used = {2855, 2644} \[ \frac {6 a^3 \cos (c+d x)}{d}+\frac {3 a^3 \sin (c+d x) \cos (c+d x)}{2 d}-\frac {9 a^3 x}{2}+\frac {\sec (c+d x) (a \sin (c+d x)+a)^3}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-9*a^3*x)/2 + (6*a^3*Cos[c + d*x])/d + (3*a^3*Cos[c + d*x]*Sin[c + d*x])/(2*d) + (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 (c+d x) (a+a \sin (c+d x))^3 \tan (c+d x) \, dx &=\frac {\sec (c+d x) (a+a \sin (c+d x))^3}{d}-(3 a) \int (a+a \sin (c+d x))^2 \, dx\\ &=-\frac {9 a^3 x}{2}+\frac {6 a^3 \cos (c+d x)}{d}+\frac {3 a^3 \cos (c+d x) \sin (c+d x)}{2 d}+\frac {\sec (c+d x) (a+a \sin (c+d x))^3}{d}\\ \end {align*}

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Mathematica [B]  time = 0.50, size = 145, normalized size = 2.16 \[ -\frac {a^3 (\sin (c+d x)+1)^3 \left (\cos \left (\frac {1}{2} (c+d x)\right ) (18 (c+d x)-\sin (2 (c+d x))-12 \cos (c+d x))+\sin \left (\frac {1}{2} (c+d x)\right ) (-2 (9 c+9 d x+16)+\sin (2 (c+d x))+12 \cos (c+d x))\right )}{4 d \left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

-1/4*(a^3*(1 + Sin[c + d*x])^3*(Cos[(c + d*x)/2]*(18*(c + d*x) - 12*Cos[c + d*x] - Sin[2*(c + d*x)]) + Sin[(c
+ d*x)/2]*(-2*(16 + 9*c + 9*d*x) + 12*Cos[c + d*x] + Sin[2*(c + d*x)])))/(d*(Cos[(c + d*x)/2] - Sin[(c + d*x)/
2])*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^6)

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fricas [A]  time = 0.44, size = 125, normalized size = 1.87 \[ \frac {a^{3} \cos \left (d x + c\right )^{3} - 9 \, a^{3} d x + 6 \, a^{3} \cos \left (d x + c\right )^{2} + 8 \, a^{3} - {\left (9 \, a^{3} d x - 13 \, a^{3}\right )} \cos \left (d x + c\right ) + {\left (9 \, a^{3} d x + a^{3} \cos \left (d x + c\right )^{2} - 5 \, a^{3} \cos \left (d x + c\right ) + 8 \, a^{3}\right )} \sin \left (d x + c\right )}{2 \, {\left (d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/2*(a^3*cos(d*x + c)^3 - 9*a^3*d*x + 6*a^3*cos(d*x + c)^2 + 8*a^3 - (9*a^3*d*x - 13*a^3)*cos(d*x + c) + (9*a^
3*d*x + a^3*cos(d*x + c)^2 - 5*a^3*cos(d*x + c) + 8*a^3)*sin(d*x + c))/(d*cos(d*x + c) - d*sin(d*x + c) + d)

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giac [A]  time = 0.18, size = 102, normalized size = 1.52 \[ -\frac {9 \, {\left (d x + c\right )} a^{3} + \frac {16 \, a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1} + \frac {2 \, {\left (a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 6 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 6 \, a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1\right )}^{2}}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

-1/2*(9*(d*x + c)*a^3 + 16*a^3/(tan(1/2*d*x + 1/2*c) - 1) + 2*(a^3*tan(1/2*d*x + 1/2*c)^3 - 6*a^3*tan(1/2*d*x
+ 1/2*c)^2 - a^3*tan(1/2*d*x + 1/2*c) - 6*a^3)/(tan(1/2*d*x + 1/2*c)^2 + 1)^2)/d

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maple [B]  time = 0.42, size = 130, normalized size = 1.94 \[ \frac {a^{3} \left (\frac {\sin ^{5}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (\sin ^{3}\left (d x +c \right )+\frac {3 \sin \left (d x +c \right )}{2}\right ) \cos \left (d x +c \right )-\frac {3 d x}{2}-\frac {3 c}{2}\right )+3 a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+3 a^{3} \left (\tan \left (d x +c \right )-d x -c \right )+\frac {a^{3}}{\cos \left (d x +c \right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^2*sin(d*x+c)*(a+a*sin(d*x+c))^3,x)

[Out]

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

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maxima [A]  time = 0.42, size = 97, normalized size = 1.45 \[ -\frac {{\left (3 \, d x + 3 \, c - \frac {\tan \left (d x + c\right )}{\tan \left (d x + c\right )^{2} + 1} - 2 \, \tan \left (d x + c\right )\right )} a^{3} + 6 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{3} - 6 \, a^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - \frac {2 \, a^{3}}{\cos \left (d x + c\right )}}{2 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^2*sin(d*x+c)*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 11.39, size = 183, normalized size = 2.73 \[ -\frac {9\,a^3\,x}{2}-\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {a^3\,\left (9\,d\,x-10\right )}{2}-\frac {9\,a^3\,d\,x}{2}\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a^3\,\left (9\,d\,x-18\right )}{2}-\frac {9\,a^3\,d\,x}{2}\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {a^3\,\left (18\,d\,x-14\right )}{2}-9\,a^3\,d\,x\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^3\,\left (18\,d\,x-42\right )}{2}-9\,a^3\,d\,x\right )-\frac {a^3\,\left (9\,d\,x-28\right )}{2}+\frac {9\,a^3\,d\,x}{2}}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,{\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )}^2} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

- (9*a^3*x)/2 - (tan(c/2 + (d*x)/2)*((a^3*(9*d*x - 10))/2 - (9*a^3*d*x)/2) - tan(c/2 + (d*x)/2)^4*((a^3*(9*d*x
- 18))/2 - (9*a^3*d*x)/2) + tan(c/2 + (d*x)/2)^3*((a^3*(18*d*x - 14))/2 - 9*a^3*d*x) - tan(c/2 + (d*x)/2)^2*(
(a^3*(18*d*x - 42))/2 - 9*a^3*d*x) - (a^3*(9*d*x - 28))/2 + (9*a^3*d*x)/2)/(d*(tan(c/2 + (d*x)/2) - 1)*(tan(c/
2 + (d*x)/2)^2 + 1)^2)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**2*sin(d*x+c)*(a+a*sin(d*x+c))**3,x)

[Out]

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

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