3.813 \(\int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx\)

Optimal. Leaf size=77 \[ -\frac {2 a^5 \cos ^3(c+d x)}{d (a-a \sin (c+d x))^2}-\frac {3 a^3 \cos (c+d x)}{d}+3 a^3 x+\frac {\sec ^3(c+d x) (a \sin (c+d x)+a)^3}{3 d} \]

[Out]

3*a^3*x-3*a^3*cos(d*x+c)/d-2*a^5*cos(d*x+c)^3/d/(a-a*sin(d*x+c))^2+1/3*sec(d*x+c)^3*(a+a*sin(d*x+c))^3/d

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Rubi [A]  time = 0.22, antiderivative size = 77, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.172, Rules used = {2871, 2670, 2680, 2682, 8} \[ -\frac {3 a^3 \cos (c+d x)}{d}-\frac {2 a^5 \cos ^3(c+d x)}{d (a-a \sin (c+d x))^2}+3 a^3 x+\frac {\sec ^3(c+d x) (a \sin (c+d x)+a)^3}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

3*a^3*x - (3*a^3*Cos[c + d*x])/d - (2*a^5*Cos[c + d*x]^3)/(d*(a - a*Sin[c + d*x])^2) + (Sec[c + d*x]^3*(a + a*
Sin[c + d*x])^3)/(3*d)

Rule 8

Int[a_, x_Symbol] :> Simp[a*x, x] /; FreeQ[a, x]

Rule 2670

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

Rule 2680

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(2*g*(
g*Cos[e + f*x])^(p - 1)*(a + b*Sin[e + f*x])^(m + 1))/(b*f*(2*m + p + 1)), x] + Dist[(g^2*(p - 1))/(b^2*(2*m +
p + 1)), Int[(g*Cos[e + f*x])^(p - 2)*(a + b*Sin[e + f*x])^(m + 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && Eq
Q[a^2 - b^2, 0] && LeQ[m, -2] && GtQ[p, 1] && NeQ[2*m + p + 1, 0] &&  !ILtQ[m + p + 1, 0] && IntegersQ[2*m, 2*
p]

Rule 2682

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

Rule 2871

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

Rubi steps

\begin {align*} \int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \tan ^2(c+d x) \, dx &=\frac {\sec ^3(c+d x) (a+a \sin (c+d x))^3}{3 d}-\int \sec ^2(c+d x) (a+a \sin (c+d x))^3 \, dx\\ &=\frac {\sec ^3(c+d x) (a+a \sin (c+d x))^3}{3 d}-a^6 \int \frac {\cos ^4(c+d x)}{(a-a \sin (c+d x))^3} \, dx\\ &=-\frac {2 a^5 \cos ^3(c+d x)}{d (a-a \sin (c+d x))^2}+\frac {\sec ^3(c+d x) (a+a \sin (c+d x))^3}{3 d}+\left (3 a^4\right ) \int \frac {\cos ^2(c+d x)}{a-a \sin (c+d x)} \, dx\\ &=-\frac {3 a^3 \cos (c+d x)}{d}-\frac {2 a^5 \cos ^3(c+d x)}{d (a-a \sin (c+d x))^2}+\frac {\sec ^3(c+d x) (a+a \sin (c+d x))^3}{3 d}+\left (3 a^3\right ) \int 1 \, dx\\ &=3 a^3 x-\frac {3 a^3 \cos (c+d x)}{d}-\frac {2 a^5 \cos ^3(c+d x)}{d (a-a \sin (c+d x))^2}+\frac {\sec ^3(c+d x) (a+a \sin (c+d x))^3}{3 d}\\ \end {align*}

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Mathematica [A]  time = 1.24, size = 133, normalized size = 1.73 \[ \frac {a^3 (\sin (c+d x)+1)^3 \left (-3 \cos (c+d x)+\frac {2 \sin \left (\frac {1}{2} (c+d x)\right ) (13 \sin (c+d x)-11)}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}+\frac {2}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^2}+9 c+9 d x\right )}{3 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^6} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^3*(1 + Sin[c + d*x])^3*(9*c + 9*d*x - 3*Cos[c + d*x] + 2/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^2 + (2*Sin[(
c + d*x)/2]*(-11 + 13*Sin[c + d*x]))/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3))/(3*d*(Cos[(c + d*x)/2] + Sin[(c
+ d*x)/2])^6)

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fricas [B]  time = 0.47, size = 169, normalized size = 2.19 \[ -\frac {3 \, a^{3} \cos \left (d x + c\right )^{3} + 18 \, a^{3} d x + 2 \, a^{3} - {\left (9 \, a^{3} d x + 16 \, a^{3}\right )} \cos \left (d x + c\right )^{2} + {\left (9 \, a^{3} d x - 17 \, a^{3}\right )} \cos \left (d x + c\right ) - {\left (18 \, a^{3} d x - 3 \, a^{3} \cos \left (d x + c\right )^{2} - 2 \, a^{3} + {\left (9 \, a^{3} d x - 19 \, a^{3}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

-1/3*(3*a^3*cos(d*x + c)^3 + 18*a^3*d*x + 2*a^3 - (9*a^3*d*x + 16*a^3)*cos(d*x + c)^2 + (9*a^3*d*x - 17*a^3)*c
os(d*x + c) - (18*a^3*d*x - 3*a^3*cos(d*x + c)^2 - 2*a^3 + (9*a^3*d*x - 19*a^3)*cos(d*x + c))*sin(d*x + c))/(d
*cos(d*x + c)^2 - d*cos(d*x + c) + (d*cos(d*x + c) + 2*d)*sin(d*x + c) - 2*d)

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giac [A]  time = 0.23, size = 87, normalized size = 1.13 \[ \frac {9 \, {\left (d x + c\right )} a^{3} - \frac {6 \, a^{3}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + 1} + \frac {2 \, {\left (9 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 24 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 11 \, a^{3}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/3*(9*(d*x + c)*a^3 - 6*a^3/(tan(1/2*d*x + 1/2*c)^2 + 1) + 2*(9*a^3*tan(1/2*d*x + 1/2*c)^2 - 24*a^3*tan(1/2*d
*x + 1/2*c) + 11*a^3)/(tan(1/2*d*x + 1/2*c) - 1)^3)/d

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maple [B]  time = 0.50, size = 184, normalized size = 2.39 \[ \frac {a^{3} \left (\frac {\sin ^{6}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{6}\left (d x +c \right )}{\cos \left (d x +c \right )}-\left (\frac {8}{3}+\sin ^{4}\left (d x +c \right )+\frac {4 \left (\sin ^{2}\left (d x +c \right )\right )}{3}\right ) \cos \left (d x +c \right )\right )+3 a^{3} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+d x +c \right )+3 a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )+\frac {a^{3} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(a^3*(1/3*sin(d*x+c)^6/cos(d*x+c)^3-sin(d*x+c)^6/cos(d*x+c)-(8/3+sin(d*x+c)^4+4/3*sin(d*x+c)^2)*cos(d*x+c)
)+3*a^3*(1/3*tan(d*x+c)^3-tan(d*x+c)+d*x+c)+3*a^3*(1/3*sin(d*x+c)^4/cos(d*x+c)^3-1/3*sin(d*x+c)^4/cos(d*x+c)-1
/3*(2+sin(d*x+c)^2)*cos(d*x+c))+1/3*a^3*sin(d*x+c)^3/cos(d*x+c)^3)

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maxima [A]  time = 0.42, size = 107, normalized size = 1.39 \[ \frac {a^{3} \tan \left (d x + c\right )^{3} + 3 \, {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{3} - a^{3} {\left (\frac {6 \, \cos \left (d x + c\right )^{2} - 1}{\cos \left (d x + c\right )^{3}} + 3 \, \cos \left (d x + c\right )\right )} - \frac {3 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{3}}{\cos \left (d x + c\right )^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4*sin(d*x+c)^2*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 12.30, size = 182, normalized size = 2.36 \[ 3\,a^3\,x+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {a^3\,\left (27\,d\,x-66\right )}{3}-9\,a^3\,d\,x\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^4\,\left (\frac {a^3\,\left (27\,d\,x-18\right )}{3}-9\,a^3\,d\,x\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^3\,\left (\frac {a^3\,\left (36\,d\,x-54\right )}{3}-12\,a^3\,d\,x\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^3\,\left (36\,d\,x-58\right )}{3}-12\,a^3\,d\,x\right )-\frac {a^3\,\left (9\,d\,x-28\right )}{3}+3\,a^3\,d\,x}{d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((sin(c + d*x)^2*(a + a*sin(c + d*x))^3)/cos(c + d*x)^4,x)

[Out]

3*a^3*x + (tan(c/2 + (d*x)/2)*((a^3*(27*d*x - 66))/3 - 9*a^3*d*x) - tan(c/2 + (d*x)/2)^4*((a^3*(27*d*x - 18))/
3 - 9*a^3*d*x) + tan(c/2 + (d*x)/2)^3*((a^3*(36*d*x - 54))/3 - 12*a^3*d*x) - tan(c/2 + (d*x)/2)^2*((a^3*(36*d*
x - 58))/3 - 12*a^3*d*x) - (a^3*(9*d*x - 28))/3 + 3*a^3*d*x)/(d*(tan(c/2 + (d*x)/2) - 1)^3*(tan(c/2 + (d*x)/2)
^2 + 1))

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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