3.39 \(\int \sec ^7(c+d x) (a+a \sin (c+d x))^3 \, dx\)

Optimal. Leaf size=87 \[ \frac{a^6}{6 d (a-a \sin (c+d x))^3}+\frac{a^5}{8 d (a-a \sin (c+d x))^2}+\frac{a^4}{8 d (a-a \sin (c+d x))}+\frac{a^3 \tanh ^{-1}(\sin (c+d x))}{8 d} \]

[Out]

(a^3*ArcTanh[Sin[c + d*x]])/(8*d) + a^6/(6*d*(a - a*Sin[c + d*x])^3) + a^5/(8*d*(a - a*Sin[c + d*x])^2) + a^4/
(8*d*(a - a*Sin[c + d*x]))

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Rubi [A]  time = 0.0724086, antiderivative size = 87, normalized size of antiderivative = 1., 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} \[ \frac{a^6}{6 d (a-a \sin (c+d x))^3}+\frac{a^5}{8 d (a-a \sin (c+d x))^2}+\frac{a^4}{8 d (a-a \sin (c+d x))}+\frac{a^3 \tanh ^{-1}(\sin (c+d x))}{8 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^7*(a + a*Sin[c + d*x])^3,x]

[Out]

(a^3*ArcTanh[Sin[c + d*x]])/(8*d) + a^6/(6*d*(a - a*Sin[c + d*x])^3) + a^5/(8*d*(a - a*Sin[c + d*x])^2) + a^4/
(8*d*(a - a*Sin[c + d*x]))

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])

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])

Rubi steps

\begin{align*} \int \sec ^7(c+d x) (a+a \sin (c+d x))^3 \, dx &=\frac{a^7 \operatorname{Subst}\left (\int \frac{1}{(a-x)^4 (a+x)} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^7 \operatorname{Subst}\left (\int \left (\frac{1}{2 a (a-x)^4}+\frac{1}{4 a^2 (a-x)^3}+\frac{1}{8 a^3 (a-x)^2}+\frac{1}{8 a^3 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^6}{6 d (a-a \sin (c+d x))^3}+\frac{a^5}{8 d (a-a \sin (c+d x))^2}+\frac{a^4}{8 d (a-a \sin (c+d x))}+\frac{a^4 \operatorname{Subst}\left (\int \frac{1}{a^2-x^2} \, dx,x,a \sin (c+d x)\right )}{8 d}\\ &=\frac{a^3 \tanh ^{-1}(\sin (c+d x))}{8 d}+\frac{a^6}{6 d (a-a \sin (c+d x))^3}+\frac{a^5}{8 d (a-a \sin (c+d x))^2}+\frac{a^4}{8 d (a-a \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.100502, size = 67, normalized size = 0.77 \[ -\frac{a^3 (\sin (c+d x)+1)^3 \sec ^6(c+d x) \left (-3 \sin ^2(c+d x)+9 \sin (c+d x)+3 (\sin (c+d x)-1)^3 \tanh ^{-1}(\sin (c+d x))-10\right )}{24 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^7*(a + a*Sin[c + d*x])^3,x]

[Out]

-(a^3*Sec[c + d*x]^6*(1 + Sin[c + d*x])^3*(-10 + 3*ArcTanh[Sin[c + d*x]]*(-1 + Sin[c + d*x])^3 + 9*Sin[c + d*x
] - 3*Sin[c + d*x]^2))/(24*d)

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Maple [B]  time = 0.08, size = 238, normalized size = 2.7 \begin{align*}{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{6\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{4}}{12\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{3\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{3\,{a}^{3} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{16\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{3\,{a}^{3}\sin \left ( dx+c \right ) }{16\,d}}+{\frac{{a}^{3}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{8\,d}}+{\frac{{a}^{3}}{2\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{5}}{6\,d}}+{\frac{5\,{a}^{3}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{24\,d}}+{\frac{5\,{a}^{3}\sec \left ( dx+c \right ) \tan \left ( dx+c \right ) }{16\,d}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^7*(a+a*sin(d*x+c))^3,x)

[Out]

1/6/d*a^3*sin(d*x+c)^4/cos(d*x+c)^6+1/12/d*a^3*sin(d*x+c)^4/cos(d*x+c)^4+1/2/d*a^3*sin(d*x+c)^3/cos(d*x+c)^6+3
/8/d*a^3*sin(d*x+c)^3/cos(d*x+c)^4+3/16/d*a^3*sin(d*x+c)^3/cos(d*x+c)^2+3/16*a^3*sin(d*x+c)/d+1/8/d*a^3*ln(sec
(d*x+c)+tan(d*x+c))+1/2/d*a^3/cos(d*x+c)^6+1/6/d*a^3*tan(d*x+c)*sec(d*x+c)^5+5/24/d*a^3*tan(d*x+c)*sec(d*x+c)^
3+5/16/d*a^3*sec(d*x+c)*tan(d*x+c)

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Maxima [A]  time = 0.957635, size = 130, normalized size = 1.49 \begin{align*} \frac{3 \, a^{3} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a^{3} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left (3 \, a^{3} \sin \left (d x + c\right )^{2} - 9 \, a^{3} \sin \left (d x + c\right ) + 10 \, a^{3}\right )}}{\sin \left (d x + c\right )^{3} - 3 \, \sin \left (d x + c\right )^{2} + 3 \, \sin \left (d x + c\right ) - 1}}{48 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^3,x, algorithm="maxima")

[Out]

1/48*(3*a^3*log(sin(d*x + c) + 1) - 3*a^3*log(sin(d*x + c) - 1) - 2*(3*a^3*sin(d*x + c)^2 - 9*a^3*sin(d*x + c)
+ 10*a^3)/(sin(d*x + c)^3 - 3*sin(d*x + c)^2 + 3*sin(d*x + c) - 1))/d

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Fricas [B]  time = 1.70695, size = 446, normalized size = 5.13 \begin{align*} \frac{6 \, a^{3} \cos \left (d x + c\right )^{2} + 18 \, a^{3} \sin \left (d x + c\right ) - 26 \, a^{3} + 3 \,{\left (3 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3} -{\left (a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \,{\left (3 \, a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3} -{\left (a^{3} \cos \left (d x + c\right )^{2} - 4 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \log \left (-\sin \left (d x + c\right ) + 1\right )}{48 \,{\left (3 \, d \cos \left (d x + c\right )^{2} -{\left (d \cos \left (d x + c\right )^{2} - 4 \, d\right )} \sin \left (d x + c\right ) - 4 \, d\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^3,x, algorithm="fricas")

[Out]

1/48*(6*a^3*cos(d*x + c)^2 + 18*a^3*sin(d*x + c) - 26*a^3 + 3*(3*a^3*cos(d*x + c)^2 - 4*a^3 - (a^3*cos(d*x + c
)^2 - 4*a^3)*sin(d*x + c))*log(sin(d*x + c) + 1) - 3*(3*a^3*cos(d*x + c)^2 - 4*a^3 - (a^3*cos(d*x + c)^2 - 4*a
^3)*sin(d*x + c))*log(-sin(d*x + c) + 1))/(3*d*cos(d*x + c)^2 - (d*cos(d*x + c)^2 - 4*d)*sin(d*x + c) - 4*d)

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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)**7*(a+a*sin(d*x+c))**3,x)

[Out]

Timed out

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Giac [A]  time = 1.24394, size = 122, normalized size = 1.4 \begin{align*} \frac{6 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 6 \, a^{3} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) + \frac{11 \, a^{3} \sin \left (d x + c\right )^{3} - 45 \, a^{3} \sin \left (d x + c\right )^{2} + 69 \, a^{3} \sin \left (d x + c\right ) - 51 \, a^{3}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{3}}}{96 \, d} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^7*(a+a*sin(d*x+c))^3,x, algorithm="giac")

[Out]

1/96*(6*a^3*log(abs(sin(d*x + c) + 1)) - 6*a^3*log(abs(sin(d*x + c) - 1)) + (11*a^3*sin(d*x + c)^3 - 45*a^3*si
n(d*x + c)^2 + 69*a^3*sin(d*x + c) - 51*a^3)/(sin(d*x + c) - 1)^3)/d