3.397 \(\int \sec ^4(c+d x) (a+b \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=75 \[ \frac{\left (2 a^2-b^2\right ) \tan (c+d x)}{3 d}+\frac{a b \sec (c+d x)}{3 d}+\frac{\sec ^3(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.0962427, antiderivative size = 75, 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 = {2691, 2669, 3767, 8} \[ \frac{\left (2 a^2-b^2\right ) \tan (c+d x)}{3 d}+\frac{a b \sec (c+d x)}{3 d}+\frac{\sec ^3(c+d x) (a \sin (c+d x)+b) (a+b \sin (c+d x))}{3 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4*(a + b*Sin[c + d*x])^2,x]

[Out]

(a*b*Sec[c + d*x])/(3*d) + (Sec[c + d*x]^3*(b + a*Sin[c + d*x])*(a + b*Sin[c + d*x]))/(3*d) + ((2*a^2 - b^2)*T
an[c + d*x])/(3*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 2669

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

Rule 3767

Int[csc[(c_.) + (d_.)*(x_)]^(n_), x_Symbol] :> -Dist[d^(-1), Subst[Int[ExpandIntegrand[(1 + x^2)^(n/2 - 1), x]
, x], x, Cot[c + d*x]], x] /; FreeQ[{c, d}, x] && IGtQ[n/2, 0]

Rule 8

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

Rubi steps

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

Mathematica [A]  time = 0.322364, size = 105, normalized size = 1.4 \[ \frac{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right ) \left (3 \left (2 a^2+b^2\right ) \sin (c+d x)+\left (2 a^2-b^2\right ) \sin (3 (c+d x))+8 a b\right )}{12 d (\sin (c+d x)-1)^2 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^3} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4*(a + b*Sin[c + d*x])^2,x]

[Out]

((Cos[(c + d*x)/2] - Sin[(c + d*x)/2])*(8*a*b + 3*(2*a^2 + b^2)*Sin[c + d*x] + (2*a^2 - b^2)*Sin[3*(c + d*x)])
)/(12*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3*(-1 + Sin[c + d*x])^2)

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Maple [A]  time = 0.051, size = 62, normalized size = 0.8 \begin{align*}{\frac{1}{d} \left ( -{a}^{2} \left ( -{\frac{2}{3}}-{\frac{ \left ( \sec \left ( dx+c \right ) \right ) ^{2}}{3}} \right ) \tan \left ( dx+c \right ) +{\frac{2\,ab}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}}+{\frac{{b}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{3\, \left ( \cos \left ( dx+c \right ) \right ) ^{3}}} \right ) } \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [A]  time = 0.972691, size = 69, normalized size = 0.92 \begin{align*} \frac{b^{2} \tan \left (d x + c\right )^{3} +{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} + \frac{2 \, a b}{\cos \left (d x + c\right )^{3}}}{3 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [A]  time = 2.12719, size = 122, normalized size = 1.63 \begin{align*} \frac{2 \, a b +{\left ({\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + b^{2}\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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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)**4*(a+b*sin(d*x+c))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.11118, size = 138, normalized size = 1.84 \begin{align*} -\frac{2 \,{\left (3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{5} + 6 \, a b \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{4} - 2 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 4 \, b^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{3} + 3 \, a^{2} \tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right ) + 2 \, a b\right )}}{3 \,{\left (\tan \left (\frac{1}{2} \, d x + \frac{1}{2} \, c\right )^{2} - 1\right )}^{3} d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(3*a^2*tan(1/2*d*x + 1/2*c)^5 + 6*a*b*tan(1/2*d*x + 1/2*c)^4 - 2*a^2*tan(1/2*d*x + 1/2*c)^3 + 4*b^2*tan(1
/2*d*x + 1/2*c)^3 + 3*a^2*tan(1/2*d*x + 1/2*c) + 2*a*b)/((tan(1/2*d*x + 1/2*c)^2 - 1)^3*d)