3.155 \(\int \sec ^{10}(c+d x) (a+a \sin (c+d x))^{7/2} \, dx\)

Optimal. Leaf size=233 \[ -\frac{11 a^5 \cos (c+d x)}{64 d (a \sin (c+d x)+a)^{3/2}}+\frac{11 a^2 \sec ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{140 d}+\frac{11 a^3 \sec ^3(c+d x) \sqrt{a \sin (c+d x)+a}}{120 d}+\frac{11 a^4 \sec (c+d x)}{48 d \sqrt{a \sin (c+d x)+a}}-\frac{11 a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{64 \sqrt{2} d}+\frac{\sec ^9(c+d x) (a \sin (c+d x)+a)^{7/2}}{9 d}+\frac{11 a \sec ^7(c+d x) (a \sin (c+d x)+a)^{5/2}}{126 d} \]

[Out]

(-11*a^(7/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(64*Sqrt[2]*d) - (11*a^5*Cos[
c + d*x])/(64*d*(a + a*Sin[c + d*x])^(3/2)) + (11*a^4*Sec[c + d*x])/(48*d*Sqrt[a + a*Sin[c + d*x]]) + (11*a^3*
Sec[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(120*d) + (11*a^2*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(140*d)
+ (11*a*Sec[c + d*x]^7*(a + a*Sin[c + d*x])^(5/2))/(126*d) + (Sec[c + d*x]^9*(a + a*Sin[c + d*x])^(7/2))/(9*d)

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Rubi [A]  time = 0.354378, antiderivative size = 233, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used = {2675, 2687, 2650, 2649, 206} \[ -\frac{11 a^5 \cos (c+d x)}{64 d (a \sin (c+d x)+a)^{3/2}}+\frac{11 a^2 \sec ^5(c+d x) (a \sin (c+d x)+a)^{3/2}}{140 d}+\frac{11 a^3 \sec ^3(c+d x) \sqrt{a \sin (c+d x)+a}}{120 d}+\frac{11 a^4 \sec (c+d x)}{48 d \sqrt{a \sin (c+d x)+a}}-\frac{11 a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a \sin (c+d x)+a}}\right )}{64 \sqrt{2} d}+\frac{\sec ^9(c+d x) (a \sin (c+d x)+a)^{7/2}}{9 d}+\frac{11 a \sec ^7(c+d x) (a \sin (c+d x)+a)^{5/2}}{126 d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^10*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

(-11*a^(7/2)*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(64*Sqrt[2]*d) - (11*a^5*Cos[
c + d*x])/(64*d*(a + a*Sin[c + d*x])^(3/2)) + (11*a^4*Sec[c + d*x])/(48*d*Sqrt[a + a*Sin[c + d*x]]) + (11*a^3*
Sec[c + d*x]^3*Sqrt[a + a*Sin[c + d*x]])/(120*d) + (11*a^2*Sec[c + d*x]^5*(a + a*Sin[c + d*x])^(3/2))/(140*d)
+ (11*a*Sec[c + d*x]^7*(a + a*Sin[c + d*x])^(5/2))/(126*d) + (Sec[c + d*x]^9*(a + a*Sin[c + d*x])^(7/2))/(9*d)

Rule 2675

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^m)/(a*f*g*(p + 1)), x] + Dist[(a*(m + p + 1))/(g^2*(p + 1)), Int[(
g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& GtQ[m, 0] && LeQ[p, -2*m] && IntegersQ[m + 1/2, 2*p]

Rule 2687

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> -Simp[(b*(g*
Cos[e + f*x])^(p + 1))/(a*f*g*(p + 1)*Sqrt[a + b*Sin[e + f*x]]), x] + Dist[(a*(2*p + 1))/(2*g^2*(p + 1)), Int[
(g*Cos[e + f*x])^(p + 2)/(a + b*Sin[e + f*x])^(3/2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2 - b^2, 0]
&& LtQ[p, -1] && IntegerQ[2*p]

Rule 2650

Int[((a_) + (b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Cos[c + d*x]*(a + b*Sin[c + d*x])^n)/(a*
d*(2*n + 1)), x] + Dist[(n + 1)/(a*(2*n + 1)), Int[(a + b*Sin[c + d*x])^(n + 1), x], x] /; FreeQ[{a, b, c, d},
x] && EqQ[a^2 - b^2, 0] && LtQ[n, -1] && IntegerQ[2*n]

Rule 2649

Int[1/Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[-2/d, Subst[Int[1/(2*a - x^2), x], x, (b*C
os[c + d*x])/Sqrt[a + b*Sin[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^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 ^{10}(c+d x) (a+a \sin (c+d x))^{7/2} \, dx &=\frac{\sec ^9(c+d x) (a+a \sin (c+d x))^{7/2}}{9 d}+\frac{1}{18} (11 a) \int \sec ^8(c+d x) (a+a \sin (c+d x))^{5/2} \, dx\\ &=\frac{11 a \sec ^7(c+d x) (a+a \sin (c+d x))^{5/2}}{126 d}+\frac{\sec ^9(c+d x) (a+a \sin (c+d x))^{7/2}}{9 d}+\frac{1}{28} \left (11 a^2\right ) \int \sec ^6(c+d x) (a+a \sin (c+d x))^{3/2} \, dx\\ &=\frac{11 a^2 \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{140 d}+\frac{11 a \sec ^7(c+d x) (a+a \sin (c+d x))^{5/2}}{126 d}+\frac{\sec ^9(c+d x) (a+a \sin (c+d x))^{7/2}}{9 d}+\frac{1}{40} \left (11 a^3\right ) \int \sec ^4(c+d x) \sqrt{a+a \sin (c+d x)} \, dx\\ &=\frac{11 a^3 \sec ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{120 d}+\frac{11 a^2 \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{140 d}+\frac{11 a \sec ^7(c+d x) (a+a \sin (c+d x))^{5/2}}{126 d}+\frac{\sec ^9(c+d x) (a+a \sin (c+d x))^{7/2}}{9 d}+\frac{1}{48} \left (11 a^4\right ) \int \frac{\sec ^2(c+d x)}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=\frac{11 a^4 \sec (c+d x)}{48 d \sqrt{a+a \sin (c+d x)}}+\frac{11 a^3 \sec ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{120 d}+\frac{11 a^2 \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{140 d}+\frac{11 a \sec ^7(c+d x) (a+a \sin (c+d x))^{5/2}}{126 d}+\frac{\sec ^9(c+d x) (a+a \sin (c+d x))^{7/2}}{9 d}+\frac{1}{32} \left (11 a^5\right ) \int \frac{1}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac{11 a^5 \cos (c+d x)}{64 d (a+a \sin (c+d x))^{3/2}}+\frac{11 a^4 \sec (c+d x)}{48 d \sqrt{a+a \sin (c+d x)}}+\frac{11 a^3 \sec ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{120 d}+\frac{11 a^2 \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{140 d}+\frac{11 a \sec ^7(c+d x) (a+a \sin (c+d x))^{5/2}}{126 d}+\frac{\sec ^9(c+d x) (a+a \sin (c+d x))^{7/2}}{9 d}+\frac{1}{128} \left (11 a^4\right ) \int \frac{1}{\sqrt{a+a \sin (c+d x)}} \, dx\\ &=-\frac{11 a^5 \cos (c+d x)}{64 d (a+a \sin (c+d x))^{3/2}}+\frac{11 a^4 \sec (c+d x)}{48 d \sqrt{a+a \sin (c+d x)}}+\frac{11 a^3 \sec ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{120 d}+\frac{11 a^2 \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{140 d}+\frac{11 a \sec ^7(c+d x) (a+a \sin (c+d x))^{5/2}}{126 d}+\frac{\sec ^9(c+d x) (a+a \sin (c+d x))^{7/2}}{9 d}-\frac{\left (11 a^4\right ) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,\frac{a \cos (c+d x)}{\sqrt{a+a \sin (c+d x)}}\right )}{64 d}\\ &=-\frac{11 a^{7/2} \tanh ^{-1}\left (\frac{\sqrt{a} \cos (c+d x)}{\sqrt{2} \sqrt{a+a \sin (c+d x)}}\right )}{64 \sqrt{2} d}-\frac{11 a^5 \cos (c+d x)}{64 d (a+a \sin (c+d x))^{3/2}}+\frac{11 a^4 \sec (c+d x)}{48 d \sqrt{a+a \sin (c+d x)}}+\frac{11 a^3 \sec ^3(c+d x) \sqrt{a+a \sin (c+d x)}}{120 d}+\frac{11 a^2 \sec ^5(c+d x) (a+a \sin (c+d x))^{3/2}}{140 d}+\frac{11 a \sec ^7(c+d x) (a+a \sin (c+d x))^{5/2}}{126 d}+\frac{\sec ^9(c+d x) (a+a \sin (c+d x))^{7/2}}{9 d}\\ \end{align*}

Mathematica [C]  time = 5.66159, size = 388, normalized size = 1.67 \[ \frac{(a (\sin (c+d x)+1))^{7/2} \left (630 \sin \left (\frac{1}{2} (c+d x)\right )+\frac{3150 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}{\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )}+\frac{1680 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^3}+\frac{1512 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^5}+\frac{1440 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^7}+\frac{1120 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2}{\left (\cos \left (\frac{1}{2} (c+d x)\right )-\sin \left (\frac{1}{2} (c+d x)\right )\right )^9}-315 \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )+(3465+3465 i) (-1)^{3/4} \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^2 \tanh ^{-1}\left (\left (\frac{1}{2}+\frac{i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac{1}{4} (c+d x)\right )-1\right )\right )\right )}{20160 d \left (\sin \left (\frac{1}{2} (c+d x)\right )+\cos \left (\frac{1}{2} (c+d x)\right )\right )^9} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^10*(a + a*Sin[c + d*x])^(7/2),x]

[Out]

((630*Sin[(c + d*x)/2] - 315*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + (3465 + 3465*I)*(-1)^(3/4)*ArcTanh[(1/2 +
I/2)*(-1)^(3/4)*(-1 + Tan[(c + d*x)/4])]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + (1120*(Cos[(c + d*x)/2] +
Sin[(c + d*x)/2])^2)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^9 + (1440*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/
(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^7 + (1512*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(Cos[(c + d*x)/2] - S
in[(c + d*x)/2])^5 + (1680*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3 +
(3150*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]))*(a*(1 + Sin[c + d*x]))^(
7/2))/(20160*d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^9)

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Maple [A]  time = 0.147, size = 205, normalized size = 0.9 \begin{align*} -{\frac{1}{40320\, \left ( \sin \left ( dx+c \right ) -1 \right ) ^{4}\cos \left ( dx+c \right ) d} \left ( -6930\,{a}^{11/2}\sin \left ( dx+c \right ) \left ( \cos \left ( dx+c \right ) \right ) ^{4}+42504\,{a}^{11/2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +385\, \left ( 9\, \left ( a-a\sin \left ( dx+c \right ) \right ) ^{9/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) a-32\,{a}^{11/2} \right ) \sin \left ( dx+c \right ) +25410\,{a}^{11/2} \left ( \cos \left ( dx+c \right ) \right ) ^{4}-50424\,{a}^{11/2} \left ( \cos \left ( dx+c \right ) \right ) ^{2}+3465\, \left ( a-a\sin \left ( dx+c \right ) \right ) ^{9/2}\sqrt{2}{\it Artanh} \left ( 1/2\,{\frac{\sqrt{a-a\sin \left ( dx+c \right ) }\sqrt{2}}{\sqrt{a}}} \right ) a+7840\,{a}^{11/2} \right ){a}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{a+a\sin \left ( dx+c \right ) }}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^10*(a+a*sin(d*x+c))^(7/2),x)

[Out]

-1/40320/a^(3/2)*(-6930*a^(11/2)*sin(d*x+c)*cos(d*x+c)^4+42504*a^(11/2)*cos(d*x+c)^2*sin(d*x+c)+385*(9*(a-a*si
n(d*x+c))^(9/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a-32*a^(11/2))*sin(d*x+c)+25410*a^
(11/2)*cos(d*x+c)^4-50424*a^(11/2)*cos(d*x+c)^2+3465*(a-a*sin(d*x+c))^(9/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c
))^(1/2)*2^(1/2)/a^(1/2))*a+7840*a^(11/2))/(sin(d*x+c)-1)^4/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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Maxima [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)^10*(a+a*sin(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

Timed out

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Fricas [A]  time = 2.40159, size = 917, normalized size = 3.94 \begin{align*} \frac{3465 \,{\left (3 \, \sqrt{2} a^{3} \cos \left (d x + c\right )^{5} - 4 \, \sqrt{2} a^{3} \cos \left (d x + c\right )^{3} -{\left (\sqrt{2} a^{3} \cos \left (d x + c\right )^{5} - 4 \, \sqrt{2} a^{3} \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt{a} \log \left (-\frac{a \cos \left (d x + c\right )^{2} - 2 \, \sqrt{a \sin \left (d x + c\right ) + a}{\left (\sqrt{2} \cos \left (d x + c\right ) - \sqrt{2} \sin \left (d x + c\right ) + \sqrt{2}\right )} \sqrt{a} + 3 \, a \cos \left (d x + c\right ) -{\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} -{\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 4 \,{\left (12705 \, a^{3} \cos \left (d x + c\right )^{4} - 25212 \, a^{3} \cos \left (d x + c\right )^{2} + 3920 \, a^{3} - 77 \,{\left (45 \, a^{3} \cos \left (d x + c\right )^{4} - 276 \, a^{3} \cos \left (d x + c\right )^{2} + 80 \, a^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{80640 \,{\left (3 \, d \cos \left (d x + c\right )^{5} - 4 \, d \cos \left (d x + c\right )^{3} -{\left (d \cos \left (d x + c\right )^{5} - 4 \, d \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^10*(a+a*sin(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

1/80640*(3465*(3*sqrt(2)*a^3*cos(d*x + c)^5 - 4*sqrt(2)*a^3*cos(d*x + c)^3 - (sqrt(2)*a^3*cos(d*x + c)^5 - 4*s
qrt(2)*a^3*cos(d*x + c)^3)*sin(d*x + c))*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(a*sin(d*x + c) + a)*(sqrt(2)*
cos(d*x + c) - sqrt(2)*sin(d*x + c) + sqrt(2))*sqrt(a) + 3*a*cos(d*x + c) - (a*cos(d*x + c) - 2*a)*sin(d*x + c
) + 2*a)/(cos(d*x + c)^2 - (cos(d*x + c) + 2)*sin(d*x + c) - cos(d*x + c) - 2)) + 4*(12705*a^3*cos(d*x + c)^4
- 25212*a^3*cos(d*x + c)^2 + 3920*a^3 - 77*(45*a^3*cos(d*x + c)^4 - 276*a^3*cos(d*x + c)^2 + 80*a^3)*sin(d*x +
c))*sqrt(a*sin(d*x + c) + a))/(3*d*cos(d*x + c)^5 - 4*d*cos(d*x + c)^3 - (d*cos(d*x + c)^5 - 4*d*cos(d*x + c)
^3)*sin(d*x + c))

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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)**10*(a+a*sin(d*x+c))**(7/2),x)

[Out]

Timed out

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Giac [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)^10*(a+a*sin(d*x+c))^(7/2),x, algorithm="giac")

[Out]

Timed out