3.23 \(\int \sec ^5(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=64 \[ \frac {a^4}{4 d (a-a \sin (c+d x))^2}+\frac {a^3}{4 d (a-a \sin (c+d x))}+\frac {a^2 \tanh ^{-1}(\sin (c+d x))}{4 d} \]

[Out]

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

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Rubi [A]  time = 0.07, antiderivative size = 64, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 3, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.143, Rules used = {2667, 44, 206} \[ \frac {a^4}{4 d (a-a \sin (c+d x))^2}+\frac {a^3}{4 d (a-a \sin (c+d x))}+\frac {a^2 \tanh ^{-1}(\sin (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^2,x]

[Out]

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

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Mathematica [A]  time = 0.07, size = 56, normalized size = 0.88 \[ \frac {a^2 (\sin (c+d x)+1)^2 \sec ^4(c+d x) \left (-\sin (c+d x)+(\sin (c+d x)-1)^2 \tanh ^{-1}(\sin (c+d x))+2\right )}{4 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^5*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*Sec[c + d*x]^4*(2 + ArcTanh[Sin[c + d*x]]*(-1 + Sin[c + d*x])^2 - Sin[c + d*x])*(1 + Sin[c + d*x])^2)/(4*
d)

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fricas [B]  time = 0.67, size = 125, normalized size = 1.95 \[ \frac {2 \, a^{2} \sin \left (d x + c\right ) - 4 \, a^{2} + {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - {\left (a^{2} \cos \left (d x + c\right )^{2} + 2 \, a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{8 \, {\left (d \cos \left (d x + c\right )^{2} + 2 \, d \sin \left (d x + c\right ) - 2 \, d\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="fricas")

[Out]

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

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giac [A]  time = 0.62, size = 77, normalized size = 1.20 \[ \frac {2 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 2 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac {3 \, a^{2} \sin \left (d x + c\right )^{2} - 10 \, a^{2} \sin \left (d x + c\right ) + 11 \, a^{2}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{2}}}{16 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="giac")

[Out]

1/16*(2*a^2*log(abs(sin(d*x + c) + 1)) - 2*a^2*log(abs(sin(d*x + c) - 1)) + (3*a^2*sin(d*x + c)^2 - 10*a^2*sin
(d*x + c) + 11*a^2)/(sin(d*x + c) - 1)^2)/d

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maple [B]  time = 0.23, size = 144, normalized size = 2.25 \[ \frac {a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{4 d \cos \left (d x +c \right )^{4}}+\frac {a^{2} \left (\sin ^{3}\left (d x +c \right )\right )}{8 d \cos \left (d x +c \right )^{2}}+\frac {a^{2} \sin \left (d x +c \right )}{8 d}+\frac {a^{2} \ln \left (\sec \left (d x +c \right )+\tan \left (d x +c \right )\right )}{4 d}+\frac {a^{2}}{2 d \cos \left (d x +c \right )^{4}}+\frac {a^{2} \tan \left (d x +c \right ) \left (\sec ^{3}\left (d x +c \right )\right )}{4 d}+\frac {3 a^{2} \sec \left (d x +c \right ) \tan \left (d x +c \right )}{8 d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^5*(a+a*sin(d*x+c))^2,x)

[Out]

1/4/d*a^2*sin(d*x+c)^3/cos(d*x+c)^4+1/8/d*a^2*sin(d*x+c)^3/cos(d*x+c)^2+1/8*a^2*sin(d*x+c)/d+1/4/d*a^2*ln(sec(
d*x+c)+tan(d*x+c))+1/2/d*a^2/cos(d*x+c)^4+1/4/d*a^2*tan(d*x+c)*sec(d*x+c)^3+3/8/d*a^2*sec(d*x+c)*tan(d*x+c)

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maxima [A]  time = 0.35, size = 71, normalized size = 1.11 \[ \frac {a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac {2 \, {\left (a^{2} \sin \left (d x + c\right ) - 2 \, a^{2}\right )}}{\sin \left (d x + c\right )^{2} - 2 \, \sin \left (d x + c\right ) + 1}}{8 \, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^5*(a+a*sin(d*x+c))^2,x, algorithm="maxima")

[Out]

1/8*(a^2*log(sin(d*x + c) + 1) - a^2*log(sin(d*x + c) - 1) - 2*(a^2*sin(d*x + c) - 2*a^2)/(sin(d*x + c)^2 - 2*
sin(d*x + c) + 1))/d

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mupad [B]  time = 4.47, size = 58, normalized size = 0.91 \[ \frac {a^2\,\mathrm {atanh}\left (\sin \left (c+d\,x\right )\right )}{4\,d}-\frac {\frac {a^2\,\sin \left (c+d\,x\right )}{4}-\frac {a^2}{2}}{d\,\left ({\sin \left (c+d\,x\right )}^2-2\,\sin \left (c+d\,x\right )+1\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^2/cos(c + d*x)^5,x)

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**5*(a+a*sin(d*x+c))**2,x)

[Out]

Timed out

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