3.11 \(\int \sec ^4(c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=44 \[ \frac {a \tan ^3(c+d x)}{3 d}+\frac {a \tan (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d} \]

[Out]

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

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Rubi [A]  time = 0.04, antiderivative size = 44, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.105, Rules used = {2669, 3767} \[ \frac {a \tan ^3(c+d x)}{3 d}+\frac {a \tan (c+d x)}{d}+\frac {a \sec ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sec[c + d*x]^3)/(3*d) + (a*Tan[c + d*x])/d + (a*Tan[c + d*x]^3)/(3*d)

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]

Rubi steps

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

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Mathematica [A]  time = 0.06, size = 41, normalized size = 0.93 \[ \frac {a \left (\frac {1}{3} \tan ^3(c+d x)+\tan (c+d x)\right )}{d}+\frac {a \sec ^3(c+d x)}{3 d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*Sec[c + d*x]^3)/(3*d) + (a*(Tan[c + d*x] + Tan[c + d*x]^3/3))/d

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fricas [A]  time = 0.64, size = 52, normalized size = 1.18 \[ -\frac {2 \, a \cos \left (d x + c\right )^{2} + 2 \, a \sin \left (d x + c\right ) - a}{3 \, {\left (d \cos \left (d x + c\right ) \sin \left (d x + c\right ) - d \cos \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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giac [A]  time = 0.36, size = 66, normalized size = 1.50 \[ -\frac {\frac {3 \, a}{\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1} + \frac {9 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 12 \, a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 7 \, a}{{\left (\tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1\right )}^{3}}}{6 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*(3*a/(tan(1/2*d*x + 1/2*c) + 1) + (9*a*tan(1/2*d*x + 1/2*c)^2 - 12*a*tan(1/2*d*x + 1/2*c) + 7*a)/(tan(1/2
*d*x + 1/2*c) - 1)^3)/d

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maple [A]  time = 0.16, size = 38, normalized size = 0.86 \[ \frac {\frac {a}{3 \cos \left (d x +c \right )^{3}}-a \left (-\frac {2}{3}-\frac {\left (\sec ^{2}\left (d x +c \right )\right )}{3}\right ) \tan \left (d x +c \right )}{d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maxima [A]  time = 0.38, size = 35, normalized size = 0.80 \[ \frac {{\left (\tan \left (d x + c\right )^{3} + 3 \, \tan \left (d x + c\right )\right )} a + \frac {a}{\cos \left (d x + c\right )^{3}}}{3 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [B]  time = 4.60, size = 63, normalized size = 1.43 \[ \frac {2\,a\,\left (\cos \left (c+d\,x\right )+2\,\sin \left (c+d\,x\right )+\cos \left (2\,c+2\,d\,x\right )-\frac {\sin \left (2\,c+2\,d\,x\right )}{2}\right )}{3\,d\,\left (2\,\cos \left (c+d\,x\right )-\sin \left (2\,c+2\,d\,x\right )\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4*(a+a*sin(d*x+c)),x)

[Out]

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

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