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

Optimal. Leaf size=63 \[ \frac {a^4 \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {5 a^2 \cos (c+d x)}{3 d (1-\sin (c+d x))}+a^2 x \]

[Out]

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

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Rubi [A]  time = 0.21, antiderivative size = 63, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.138, Rules used = {2869, 2758, 2735, 2648} \[ \frac {a^4 \cos (c+d x)}{3 d (a-a \sin (c+d x))^2}-\frac {5 a^2 \cos (c+d x)}{3 d (1-\sin (c+d x))}+a^2 x \]

Antiderivative was successfully verified.

[In]

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

[Out]

a^2*x - (5*a^2*Cos[c + d*x])/(3*d*(1 - Sin[c + d*x])) + (a^4*Cos[c + d*x])/(3*d*(a - a*Sin[c + d*x])^2)

Rule 2648

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(-1), x_Symbol] :> -Simp[Cos[c + d*x]/(d*(b + a*Sin[c + d*x])), x]
/; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]

Rule 2735

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]), x_Symbol] :> Simp[(b*x)/d
, x] - Dist[(b*c - a*d)/d, Int[1/(c + d*Sin[e + f*x]), x], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d
, 0]

Rule 2758

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

Rule 2869

Int[cos[(e_.) + (f_.)*(x_)]^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(
m_), x_Symbol] :> Dist[a^(2*m), Int[(d*Sin[e + f*x])^n/(a - b*Sin[e + f*x])^m, x], x] /; FreeQ[{a, b, d, e, f,
n}, x] && EqQ[a^2 - b^2, 0] && IntegersQ[m, p] && EqQ[2*m + p, 0]

Rubi steps

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

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Mathematica [A]  time = 0.05, size = 79, normalized size = 1.25 \[ \frac {a^2 \tan ^{-1}(\tan (c+d x))}{d}+\frac {2 a^2 \tan ^3(c+d x)}{3 d}-\frac {a^2 \tan (c+d x)}{d}+\frac {2 a^2 \sec ^3(c+d x)}{3 d}-\frac {2 a^2 \sec (c+d x)}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*ArcTan[Tan[c + d*x]])/d - (2*a^2*Sec[c + d*x])/d + (2*a^2*Sec[c + d*x]^3)/(3*d) - (a^2*Tan[c + d*x])/d +
(2*a^2*Tan[c + d*x]^3)/(3*d)

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fricas [B]  time = 0.45, size = 141, normalized size = 2.24 \[ -\frac {6 \, a^{2} d x - {\left (3 \, a^{2} d x + 5 \, a^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + {\left (3 \, a^{2} d x - 4 \, a^{2}\right )} \cos \left (d x + c\right ) - {\left (6 \, a^{2} d x - a^{2} + {\left (3 \, a^{2} d x - 5 \, a^{2}\right )} \cos \left (d x + c\right )\right )} \sin \left (d x + c\right )}{3 \, {\left (d \cos \left (d x + c\right )^{2} - d \cos \left (d x + c\right ) + {\left (d \cos \left (d x + c\right ) + 2 \, d\right )} \sin \left (d x + c\right ) - 2 \, d\right )}} \]

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))^2,x, algorithm="fricas")

[Out]

-1/3*(6*a^2*d*x - (3*a^2*d*x + 5*a^2)*cos(d*x + c)^2 + a^2 + (3*a^2*d*x - 4*a^2)*cos(d*x + c) - (6*a^2*d*x - a
^2 + (3*a^2*d*x - 5*a^2)*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 = 0.21, size = 67, normalized size = 1.06 \[ \frac {3 \, {\left (d x + c\right )} a^{2} + \frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 9 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 4 \, a^{2}\right )}}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{3 \, d} \]

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))^2,x, algorithm="giac")

[Out]

1/3*(3*(d*x + c)*a^2 + 2*(3*a^2*tan(1/2*d*x + 1/2*c)^2 - 9*a^2*tan(1/2*d*x + 1/2*c) + 4*a^2)/(tan(1/2*d*x + 1/
2*c) - 1)^3)/d

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maple [A]  time = 0.40, size = 114, normalized size = 1.81 \[ \frac {a^{2} \left (\frac {\left (\tan ^{3}\left (d x +c \right )\right )}{3}-\tan \left (d x +c \right )+d x +c \right )+2 a^{2} \left (\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )^{3}}-\frac {\sin ^{4}\left (d x +c \right )}{3 \cos \left (d x +c \right )}-\frac {\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )}{3}\right )+\frac {a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}}{d} \]

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))^2,x)

[Out]

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

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maxima [A]  time = 0.42, size = 71, normalized size = 1.13 \[ \frac {a^{2} \tan \left (d x + c\right )^{3} + {\left (\tan \left (d x + c\right )^{3} + 3 \, d x + 3 \, c - 3 \, \tan \left (d x + c\right )\right )} a^{2} - \frac {2 \, {\left (3 \, \cos \left (d x + c\right )^{2} - 1\right )} a^{2}}{\cos \left (d x + c\right )^{3}}}{3 \, d} \]

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))^2,x, algorithm="maxima")

[Out]

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

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mupad [B]  time = 9.29, size = 102, normalized size = 1.62 \[ a^2\,x+\frac {\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (\frac {a^2\,\left (9\,d\,x-18\right )}{3}-3\,a^2\,d\,x\right )-{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (\frac {a^2\,\left (9\,d\,x-6\right )}{3}-3\,a^2\,d\,x\right )-\frac {a^2\,\left (3\,d\,x-8\right )}{3}+a^2\,d\,x}{d\,{\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

a^2*x + (tan(c/2 + (d*x)/2)*((a^2*(9*d*x - 18))/3 - 3*a^2*d*x) - tan(c/2 + (d*x)/2)^2*((a^2*(9*d*x - 6))/3 - 3
*a^2*d*x) - (a^2*(3*d*x - 8))/3 + a^2*d*x)/(d*(tan(c/2 + (d*x)/2) - 1)^3)

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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))**2,x)

[Out]

Timed out

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