3.12 \(\int \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx\)

Optimal. Leaf size=84 \[ \frac {a^3}{8 d (a-a \sin (c+d x))^2}+\frac {a^2}{4 d (a-a \sin (c+d x))}-\frac {a^2}{8 d (a \sin (c+d x)+a)}+\frac {3 a \tanh ^{-1}(\sin (c+d x))}{8 d} \]

[Out]

3/8*a*arctanh(sin(d*x+c))/d+1/8*a^3/d/(a-a*sin(d*x+c))^2+1/4*a^2/d/(a-a*sin(d*x+c))-1/8*a^2/d/(a+a*sin(d*x+c))

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Rubi [A]  time = 0.06, antiderivative size = 84, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 19, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.158, Rules used = {2667, 44, 206} \[ \frac {a^3}{8 d (a-a \sin (c+d x))^2}+\frac {a^2}{4 d (a-a \sin (c+d x))}-\frac {a^2}{8 d (a \sin (c+d x)+a)}+\frac {3 a \tanh ^{-1}(\sin (c+d x))}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*(a + a*Sin[c + d*x]),x]

[Out]

(3*a*ArcTanh[Sin[c + d*x]])/(8*d) + a^3/(8*d*(a - a*Sin[c + d*x])^2) + a^2/(4*d*(a - a*Sin[c + d*x])) - a^2/(8
*d*(a + a*Sin[c + d*x]))

Rule 44

Int[((a_) + (b_.)*(x_))^(m_.)*((c_.) + (d_.)*(x_))^(n_.), x_Symbol] :> Int[ExpandIntegrand[(a + b*x)^m*(c + d*
x)^n, x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && ILtQ[m, 0] && IntegerQ[n] &&  !(IGtQ[n, 0] && L
tQ[m + n + 2, 0])

Rule 206

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])

Rule 2667

Int[cos[(e_.) + (f_.)*(x_)]^(p_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.), x_Symbol] :> Dist[1/(b^p*f), S
ubst[Int[(a + x)^(m + (p - 1)/2)*(a - x)^((p - 1)/2), x], x, b*Sin[e + f*x]], x] /; FreeQ[{a, b, e, f, m}, x]
&& IntegerQ[(p - 1)/2] && EqQ[a^2 - b^2, 0] && (GeQ[p, -1] ||  !IntegerQ[m + 1/2])

Rubi steps

\begin {align*} \int \sec ^5(c+d x) (a+a \sin (c+d x)) \, dx &=\frac {a^5 \operatorname {Subst}\left (\int \frac {1}{(a-x)^3 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^5 \operatorname {Subst}\left (\int \left (\frac {1}{4 a^2 (a-x)^3}+\frac {1}{4 a^3 (a-x)^2}+\frac {1}{8 a^3 (a+x)^2}+\frac {3}{8 a^3 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac {a^3}{8 d (a-a \sin (c+d x))^2}+\frac {a^2}{4 d (a-a \sin (c+d x))}-\frac {a^2}{8 d (a+a \sin (c+d x))}+\frac {\left (3 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{8 d}\\ &=\frac {3 a \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac {a^3}{8 d (a-a \sin (c+d x))^2}+\frac {a^2}{4 d (a-a \sin (c+d x))}-\frac {a^2}{8 d (a+a \sin (c+d x))}\\ \end {align*}

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Mathematica [A]  time = 0.08, size = 68, normalized size = 0.81 \[ \frac {a \sec ^4(c+d x)}{4 d}+\frac {a \tan (c+d x) \sec ^3(c+d x)}{4 d}+\frac {3 a \left (\tanh ^{-1}(\sin (c+d x))+\tan (c+d x) \sec (c+d x)\right )}{8 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5*(a + a*Sin[c + d*x]),x]

[Out]

(a*Sec[c + d*x]^4)/(4*d) + (a*Sec[c + d*x]^3*Tan[c + d*x])/(4*d) + (3*a*(ArcTanh[Sin[c + d*x]] + Sec[c + d*x]*
Tan[c + d*x]))/(8*d)

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fricas [A]  time = 0.80, size = 136, normalized size = 1.62 \[ -\frac {6 \, a \cos \left (d x + c\right )^{2} - 3 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \, {\left (a \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - a \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) + 6 \, a \sin \left (d x + c\right ) - 2 \, a}{16 \, {\left (d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - d \cos \left (d x + c\right )^{2}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="fricas")

[Out]

-1/16*(6*a*cos(d*x + c)^2 - 3*(a*cos(d*x + c)^2*sin(d*x + c) - a*cos(d*x + c)^2)*log(sin(d*x + c) + 1) + 3*(a*
cos(d*x + c)^2*sin(d*x + c) - a*cos(d*x + c)^2)*log(-sin(d*x + c) + 1) + 6*a*sin(d*x + c) - 2*a)/(d*cos(d*x +
c)^2*sin(d*x + c) - d*cos(d*x + c)^2)

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giac [A]  time = 0.51, size = 92, normalized size = 1.10 \[ \frac {6 \, a \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 6 \, a \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac {2 \, {\left (3 \, a \sin \left (d x + c\right ) + 5 \, a\right )}}{\sin \left (d x + c\right ) + 1} + \frac {9 \, a \sin \left (d x + c\right )^{2} - 26 \, a \sin \left (d x + c\right ) + 21 \, a}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{32 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="giac")

[Out]

1/32*(6*a*log(abs(sin(d*x + c) + 1)) - 6*a*log(abs(sin(d*x + c) - 1)) - 2*(3*a*sin(d*x + c) + 5*a)/(sin(d*x +
c) + 1) + (9*a*sin(d*x + c)^2 - 26*a*sin(d*x + c) + 21*a)/(sin(d*x + c) - 1)^2)/d

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maple [A]  time = 0.18, size = 74, normalized size = 0.88 \[ \frac {a}{4 d \cos \left (d x +c \right )^{4}}+\frac {a \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 a \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d}+\frac {3 a \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*(a+a*sin(d*x+c)),x)

[Out]

1/4/d*a/cos(d*x+c)^4+1/4/d*a*tan(d*x+c)*sec(d*x+c)^3+3/8*a*sec(d*x+c)*tan(d*x+c)/d+3/8/d*a*ln(sec(d*x+c)+tan(d
*x+c))

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maxima [A]  time = 0.53, size = 86, normalized size = 1.02 \[ \frac {3 \, a \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (3 \, a \sin \left (d x + c\right )^{2} - 3 \, a \sin \left (d x + c\right ) - 2 \, a\right )}}{\sin \left (d x + c\right )^{3} - \sin \left (d x + c\right )^{2} - \sin \left (d x + c\right ) + 1}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c)),x, algorithm="maxima")

[Out]

1/16*(3*a*log(sin(d*x + c) + 1) - 3*a*log(sin(d*x + c) - 1) - 2*(3*a*sin(d*x + c)^2 - 3*a*sin(d*x + c) - 2*a)/
(sin(d*x + c)^3 - sin(d*x + c)^2 - sin(d*x + c) + 1))/d

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mupad [B]  time = 4.50, size = 71, normalized size = 0.85 \[ \frac {3\,a\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{8\,d}-\frac {-\frac {3\,a\,{\sin \left (c+d\,x\right )}^2}{8}+\frac {3\,a\,\sin \left (c+d\,x\right )}{8}+\frac {a}{4}}{d\,\left (-{\sin \left (c+d\,x\right )}^3+{\sin \left (c+d\,x\right )}^2+\sin \left (c+d\,x\right )-1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))/cos(c + d*x)^5,x)

[Out]

(3*a*atanh(sin(c + d*x)))/(8*d) - (a/4 + (3*a*sin(c + d*x))/8 - (3*a*sin(c + d*x)^2)/8)/(d*(sin(c + d*x) + sin
(c + d*x)^2 - sin(c + d*x)^3 - 1))

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a \left (\int \sin {\left (c + d x \right )} \sec ^{5}{\left (c + d x \right )}\, dx + \int \sec ^{5}{\left (c + d x \right )}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5*(a+a*sin(d*x+c)),x)

[Out]

a*(Integral(sin(c + d*x)*sec(c + d*x)**5, x) + Integral(sec(c + d*x)**5, x))

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