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

Optimal. Leaf size=79 \[ \frac{2 b \left (a^2+b^2\right ) \cos (c+d x)}{d}+\frac{a b^2 \sin (c+d x) \cos (c+d x)}{d}-3 a b^2 x+\frac{\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{d} \]

[Out]

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

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Rubi [A]  time = 0.0704697, antiderivative size = 79, normalized size of antiderivative = 1., number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.095, Rules used = {2691, 2734} \[ \frac{2 b \left (a^2+b^2\right ) \cos (c+d x)}{d}+\frac{a b^2 \sin (c+d x) \cos (c+d x)}{d}-3 a b^2 x+\frac{\sec (c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))^2}{d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 2691

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

Rule 2734

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

Rubi steps

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

Mathematica [A]  time = 0.30729, size = 68, normalized size = 0.86 \[ \frac{2 a \left (a^2+3 b^2\right ) \tan (c+d x)+\sec (c+d x) \left (6 a^2 b+b^3 \cos (2 (c+d x))+3 b^3\right )-6 a b^2 (c+d x)}{2 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-6*a*b^2*(c + d*x) + (6*a^2*b + 3*b^3 + b^3*Cos[2*(c + d*x)])*Sec[c + d*x] + 2*a*(a^2 + 3*b^2)*Tan[c + d*x])/
(2*d)

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Maple [A]  time = 0.046, size = 89, normalized size = 1.1 \begin{align*}{\frac{1}{d} \left ({a}^{3}\tan \left ( dx+c \right ) +3\,{\frac{{a}^{2}b}{\cos \left ( dx+c \right ) }}+3\,a{b}^{2} \left ( \tan \left ( dx+c \right ) -dx-c \right ) +{b}^{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 ) \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 1.44526, size = 95, normalized size = 1.2 \begin{align*} -\frac{3 \,{\left (d x + c - \tan \left (d x + c\right )\right )} a b^{2} - b^{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^{2} b}{\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+b*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

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

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Fricas [A]  time = 2.50838, size = 154, normalized size = 1.95 \begin{align*} -\frac{3 \, a b^{2} d x \cos \left (d x + c\right ) - b^{3} \cos \left (d x + c\right )^{2} - 3 \, a^{2} b - b^{3} -{\left (a^{3} + 3 \, a b^{2}\right )} \sin \left (d x + c\right )}{d \cos \left (d x + c\right )} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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+b*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.10934, size = 166, normalized size = 2.1 \begin{align*} -\frac{3 \,{\left (d x + c\right )} a b^{2} + \frac{2 \,{\left (a^{3} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, a^{2} b \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 ) + 3 \, a b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 3 \, a^{2} b + 2 \, b^{3}\right )}}{\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 1}}{d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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