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

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

[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)

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Rubi [A]  time = 0.136254, antiderivative size = 50, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, 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 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 8

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

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*}

Mathematica [C]  time = 0.0333694, size = 55, normalized size = 1.1 \[ \frac{4 \sqrt{2} a^3 \sqrt{\sin (c+d x)+1} \sec (c+d x) \, _2F_1\left (-\frac{3}{2},-\frac{1}{2};\frac{1}{2};\frac{1}{2} (1-\sin (c+d x))\right )}{d} \]

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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Maple [A]  time = 0.05, size = 87, normalized size = 1.7 \begin{align*}{\frac{1}{d} \left ({a}^{3} \left ({\frac{ \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{\cos \left ( dx+c \right ) }}+ \left ( 2+ \left ( \sin \left ( dx+c \right ) \right ) ^{2} \right ) \cos \left ( dx+c \right ) \right ) +3\,{a}^{3} \left ( \tan \left ( dx+c \right ) -dx-c \right ) +3\,{\frac{{a}^{3}}{\cos \left ( dx+c \right ) }}+{a}^{3}\tan \left ( dx+c \right ) \right ) } \end{align*}

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 = 1.44948, size = 92, normalized size = 1.84 \begin{align*} -\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} \end{align*}

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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Fricas [A]  time = 1.62659, size = 230, normalized size = 4.6 \begin{align*} -\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} \end{align*}

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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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

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]

Timed out

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Giac [A]  time = 1.16698, size = 123, normalized size = 2.46 \begin{align*} -\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} \end{align*}

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