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]

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

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Rubi [A]  time = 0.10, antiderivative size = 75, 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 = {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 8

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

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 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 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]

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*}

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Mathematica [A]  time = 0.33, size = 105, normalized size = 1.40 \[ \frac {\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right ) \left (3 \left (2 a^2+b^2\right ) \sin (c+d x)+\left (2 a^2-b^2\right ) \sin (3 (c+d x))+8 a b\right )}{12 d (\sin (c+d x)-1)^2 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3} \]

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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fricas [A]  time = 0.43, size = 52, normalized size = 0.69 \[ \frac {2 \, a b + {\left ({\left (2 \, a^{2} - b^{2}\right )} \cos \left (d x + c\right )^{2} + a^{2} + b^{2}\right )} \sin \left (d x + c\right )}{3 \, d \cos \left (d x + c\right )^{3}} \]

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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giac [A]  time = 0.51, size = 102, normalized size = 1.36 \[ -\frac {2 \, {\left (3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5} + 6 \, a b \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} - 2 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 4 \, b^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} + 3 \, a^{2} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 2 \, a b\right )}}{3 \, {\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{3} d} \]

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)

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maple [A]  time = 0.28, size = 62, normalized size = 0.83 \[ \frac {-a^{2} \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )+\frac {2 a b}{3 \cos \left (d x +c \right )^{3}}+\frac {b^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{3 \cos \left (d x +c \right )^{3}}}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4*(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 = 0.33, size = 51, normalized size = 0.68 \[ \frac {b^{2} \tan \left (d x + c\right )^{3} + {\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a^{2} + \frac {2 \, a b}{\cos \left (d x + c\right )^{3}}}{3 \, d} \]

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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mupad [B]  time = 5.26, size = 71, normalized size = 0.95 \[ \frac {\frac {2\,a\,b}{3}+\frac {a^2\,\sin \left (c+d\,x\right )}{3}+\frac {b^2\,\sin \left (c+d\,x\right )}{3}+{\cos \left (c+d\,x\right )}^2\,\left (\frac {2\,a^2\,\sin \left (c+d\,x\right )}{3}-\frac {b^2\,\sin \left (c+d\,x\right )}{3}\right )}{d\,{\cos \left (c+d\,x\right )}^3} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (a + b \sin {\left (c + d x \right )}\right )^{2} \sec ^{4}{\left (c + d x \right )}\, dx \]

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]

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

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