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

Optimal. Leaf size=50 \[ \frac {2 a^5 \cos ^3(c+d x)}{d (a-a \sin (c+d x))^2}+\frac {3 a^3 \cos (c+d x)}{d}-3 a^3 x \]

[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

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Rubi [A]  time = 0.14, antiderivative size = 50, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.190, Rules used = {2670, 2680, 2682, 8} \[ \frac {3 a^3 \cos (c+d x)}{d}+\frac {2 a^5 \cos ^3(c+d x)}{d (a-a \sin (c+d x))^2}-3 a^3 x \]

Antiderivative was successfully verified.

[In]

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

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]

Rubi steps

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

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Mathematica [C]  time = 0.03, size = 55, normalized size = 1.10 \[ \frac {4 \sqrt {2} a^3 \sqrt {\sin (c+d x)+1} \sec (c+d x) \, _2F_1\left (-\frac {3}{2},-\frac {1}{2};\frac {1}{2};\frac {1}{2} (1-\sin (c+d x))\right )}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [A]  time = 0.68, size = 101, normalized size = 2.02 \[ -\frac {3 \, a^{3} d x - a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3} + {\left (3 \, a^{3} d x - 5 \, a^{3}\right )} \cos \left (d x + c\right ) - {\left (3 \, a^{3} d x - a^{3} \cos \left (d x + c\right ) + 4 \, a^{3}\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right ) - d \sin \left (d x + c\right ) + d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.54, size = 91, normalized size = 1.82 \[ -\frac {3 \, {\left (d x + c\right )} a^{3} + \frac {2 \, {\left (4 \, a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a^{3} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 5 \, a^{3}\right )}}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [A]  time = 0.22, size = 87, normalized size = 1.74 \[ \frac {a^{3} \left (\frac {\sin ^{4}\left (d x +c \right )}{\cos \left (d x +c \right )}+\left (2+\sin ^{2}\left (d x +c \right )\right ) \cos \left (d x +c \right )\right )+3 a^{3} \left (\tan \left (d x +c \right )-d x -c \right )+\frac {3 a^{3}}{\cos \left (d x +c \right )}+a^{3} \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/d*(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)+3*a^3/cos(d*x+c)+a^3*t
an(d*x+c))

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maxima [A]  time = 0.41, size = 68, normalized size = 1.36 \[ -\frac {3 \, {\left (d x + c - \tan \left (d x + c\right )\right )} a^{3} - a^{3} {\left (\frac {1}{\cos \left (d x + c\right )} + \cos \left (d x + c\right )\right )} - a^{3} \tan \left (d x + c\right ) - \frac {3 \, a^{3}}{\cos \left (d x + c\right )}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.78, size = 138, normalized size = 2.76 \[ -3\,a^3\,x-\frac {3\,a^3\,\left (c+d\,x\right )-\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )\,\left (3\,a^3\,\left (c+d\,x\right )-a^3\,\left (3\,c+3\,d\,x-2\right )\right )-a^3\,\left (3\,c+3\,d\,x-10\right )+{\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2\,\left (3\,a^3\,\left (c+d\,x\right )-a^3\,\left (3\,c+3\,d\,x-8\right )\right )}{d\,\left (\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )-1\right )\,\left ({\mathrm {tan}\left (\frac {c}{2}+\frac {d\,x}{2}\right )}^2+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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