3.25 \(\int \sec ^7(c+d x) (a+a \sin (c+d x))^2 \, dx\)

Optimal. Leaf size=109 \[ \frac{a^5}{12 d (a-a \sin (c+d x))^3}+\frac{a^4}{8 d (a-a \sin (c+d x))^2}+\frac{3 a^3}{16 d (a-a \sin (c+d x))}-\frac{a^3}{16 d (a \sin (c+d x)+a)}+\frac{a^2 \tanh ^{-1}(\sin (c+d x))}{4 d} \]

[Out]

(a^2*ArcTanh[Sin[c + d*x]])/(4*d) + a^5/(12*d*(a - a*Sin[c + d*x])^3) + a^4/(8*d*(a - a*Sin[c + d*x])^2) + (3*
a^3)/(16*d*(a - a*Sin[c + d*x])) - a^3/(16*d*(a + a*Sin[c + d*x]))

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Rubi [A]  time = 0.0886078, antiderivative size = 109, 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} \[ \frac{a^5}{12 d (a-a \sin (c+d x))^3}+\frac{a^4}{8 d (a-a \sin (c+d x))^2}+\frac{3 a^3}{16 d (a-a \sin (c+d x))}-\frac{a^3}{16 d (a \sin (c+d x)+a)}+\frac{a^2 \tanh ^{-1}(\sin (c+d x))}{4 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^7*(a + a*Sin[c + d*x])^2,x]

[Out]

(a^2*ArcTanh[Sin[c + d*x]])/(4*d) + a^5/(12*d*(a - a*Sin[c + d*x])^3) + a^4/(8*d*(a - a*Sin[c + d*x])^2) + (3*
a^3)/(16*d*(a - a*Sin[c + d*x])) - a^3/(16*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))^2 \, dx &=\frac{a^7 \operatorname{Subst}\left (\int \frac{1}{(a-x)^4 (a+x)^2} \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^7 \operatorname{Subst}\left (\int \left (\frac{1}{4 a^2 (a-x)^4}+\frac{1}{4 a^3 (a-x)^3}+\frac{3}{16 a^4 (a-x)^2}+\frac{1}{16 a^4 (a+x)^2}+\frac{1}{4 a^4 \left (a^2-x^2\right )}\right ) \, dx,x,a \sin (c+d x)\right )}{d}\\ &=\frac{a^5}{12 d (a-a \sin (c+d x))^3}+\frac{a^4}{8 d (a-a \sin (c+d x))^2}+\frac{3 a^3}{16 d (a-a \sin (c+d x))}-\frac{a^3}{16 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^5}{12 d (a-a \sin (c+d x))^3}+\frac{a^4}{8 d (a-a \sin (c+d x))^2}+\frac{3 a^3}{16 d (a-a \sin (c+d x))}-\frac{a^3}{16 d (a+a \sin (c+d x))}\\ \end{align*}

Mathematica [A]  time = 0.142689, size = 85, normalized size = 0.78 \[ -\frac{a^2 (\sin (c+d x)+1)^2 \sec ^6(c+d x) \left (-3 \sin ^3(c+d x)+6 \sin ^2(c+d x)-\sin (c+d x)+3 (\sin (c+d x)+1) (\sin (c+d x)-1)^3 \tanh ^{-1}(\sin (c+d x))-4\right )}{12 d} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^7*(a + a*Sin[c + d*x])^2,x]

[Out]

-(a^2*Sec[c + d*x]^6*(1 + Sin[c + d*x])^2*(-4 - Sin[c + d*x] + 6*Sin[c + d*x]^2 - 3*Sin[c + d*x]^3 + 3*ArcTanh
[Sin[c + d*x]]*(-1 + Sin[c + d*x])^3*(1 + Sin[c + d*x])))/(12*d)

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Maple [A]  time = 0.108, size = 190, normalized size = 1.7 \begin{align*}{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{6\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{8\,d \left ( \cos \left ( dx+c \right ) \right ) ^{4}}}+{\frac{{a}^{2} \left ( \sin \left ( dx+c \right ) \right ) ^{3}}{16\,d \left ( \cos \left ( dx+c \right ) \right ) ^{2}}}+{\frac{{a}^{2}\sin \left ( dx+c \right ) }{16\,d}}+{\frac{{a}^{2}\ln \left ( \sec \left ( dx+c \right ) +\tan \left ( dx+c \right ) \right ) }{4\,d}}+{\frac{{a}^{2}}{3\,d \left ( \cos \left ( dx+c \right ) \right ) ^{6}}}+{\frac{{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{5}}{6\,d}}+{\frac{5\,{a}^{2}\tan \left ( dx+c \right ) \left ( \sec \left ( dx+c \right ) \right ) ^{3}}{24\,d}}+{\frac{5\,{a}^{2}\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))^2,x)

[Out]

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

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Maxima [A]  time = 0.97087, size = 146, normalized size = 1.34 \begin{align*} \frac{3 \, a^{2} \log \left (\sin \left (d x + c\right ) + 1\right ) - 3 \, a^{2} \log \left (\sin \left (d x + c\right ) - 1\right ) - \frac{2 \,{\left (3 \, a^{2} \sin \left (d x + c\right )^{3} - 6 \, a^{2} \sin \left (d x + c\right )^{2} + a^{2} \sin \left (d x + c\right ) + 4 \, a^{2}\right )}}{\sin \left (d x + c\right )^{4} - 2 \, \sin \left (d x + c\right )^{3} + 2 \, \sin \left (d x + c\right ) - 1}}{24 \, 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))^2,x, algorithm="maxima")

[Out]

1/24*(3*a^2*log(sin(d*x + c) + 1) - 3*a^2*log(sin(d*x + c) - 1) - 2*(3*a^2*sin(d*x + c)^3 - 6*a^2*sin(d*x + c)
^2 + a^2*sin(d*x + c) + 4*a^2)/(sin(d*x + c)^4 - 2*sin(d*x + c)^3 + 2*sin(d*x + c) - 1))/d

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Fricas [A]  time = 1.64871, size = 505, normalized size = 4.63 \begin{align*} -\frac{12 \, a^{2} \cos \left (d x + c\right )^{2} - 4 \, a^{2} - 3 \,{\left (a^{2} \cos \left (d x + c\right )^{4} + 2 \, a^{2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, a^{2} \cos \left (d x + c\right )^{2}\right )} \log \left (\sin \left (d x + c\right ) + 1\right ) + 3 \,{\left (a^{2} \cos \left (d x + c\right )^{4} + 2 \, a^{2} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, a^{2} \cos \left (d x + c\right )^{2}\right )} \log \left (-\sin \left (d x + c\right ) + 1\right ) - 2 \,{\left (3 \, a^{2} \cos \left (d x + c\right )^{2} - 4 \, a^{2}\right )} \sin \left (d x + c\right )}{24 \,{\left (d \cos \left (d x + c\right )^{4} + 2 \, d \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, d \cos \left (d x + c\right )^{2}\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))^2,x, algorithm="fricas")

[Out]

-1/24*(12*a^2*cos(d*x + c)^2 - 4*a^2 - 3*(a^2*cos(d*x + c)^4 + 2*a^2*cos(d*x + c)^2*sin(d*x + c) - 2*a^2*cos(d
*x + c)^2)*log(sin(d*x + c) + 1) + 3*(a^2*cos(d*x + c)^4 + 2*a^2*cos(d*x + c)^2*sin(d*x + c) - 2*a^2*cos(d*x +
c)^2)*log(-sin(d*x + c) + 1) - 2*(3*a^2*cos(d*x + c)^2 - 4*a^2)*sin(d*x + c))/(d*cos(d*x + c)^4 + 2*d*cos(d*x
+ c)^2*sin(d*x + c) - 2*d*cos(d*x + c)^2)

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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))**2,x)

[Out]

Timed out

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Giac [A]  time = 1.20027, size = 161, normalized size = 1.48 \begin{align*} \frac{6 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) + 1 \right |}\right ) - 6 \, a^{2} \log \left ({\left | \sin \left (d x + c\right ) - 1 \right |}\right ) - \frac{3 \,{\left (2 \, a^{2} \sin \left (d x + c\right ) + 3 \, a^{2}\right )}}{\sin \left (d x + c\right ) + 1} + \frac{11 \, a^{2} \sin \left (d x + c\right )^{3} - 42 \, a^{2} \sin \left (d x + c\right )^{2} + 57 \, a^{2} \sin \left (d x + c\right ) - 30 \, a^{2}}{{\left (\sin \left (d x + c\right ) - 1\right )}^{3}}}{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))^2,x, algorithm="giac")

[Out]

1/48*(6*a^2*log(abs(sin(d*x + c) + 1)) - 6*a^2*log(abs(sin(d*x + c) - 1)) - 3*(2*a^2*sin(d*x + c) + 3*a^2)/(si
n(d*x + c) + 1) + (11*a^2*sin(d*x + c)^3 - 42*a^2*sin(d*x + c)^2 + 57*a^2*sin(d*x + c) - 30*a^2)/(sin(d*x + c)
- 1)^3)/d