3.168 \(\int \frac {\sec ^6(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\)

Optimal. Leaf size=221 \[ -\frac {231 a \cos (c+d x)}{512 d (a \sin (c+d x)+a)^{3/2}}+\frac {\sec ^5(c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}+\frac {11 \sec ^3(c+d x)}{40 d \sqrt {a \sin (c+d x)+a}}-\frac {11 a \sec ^3(c+d x)}{60 d (a \sin (c+d x)+a)^{3/2}}+\frac {77 \sec (c+d x)}{128 d \sqrt {a \sin (c+d x)+a}}-\frac {77 a \sec (c+d x)}{320 d (a \sin (c+d x)+a)^{3/2}}-\frac {231 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{512 \sqrt {2} \sqrt {a} d} \]

[Out]

-231/512*a*cos(d*x+c)/d/(a+a*sin(d*x+c))^(3/2)-77/320*a*sec(d*x+c)/d/(a+a*sin(d*x+c))^(3/2)-11/60*a*sec(d*x+c)
^3/d/(a+a*sin(d*x+c))^(3/2)-231/1024*arctanh(1/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))*2^(1/2)/d/
a^(1/2)+77/128*sec(d*x+c)/d/(a+a*sin(d*x+c))^(1/2)+11/40*sec(d*x+c)^3/d/(a+a*sin(d*x+c))^(1/2)+1/5*sec(d*x+c)^
5/d/(a+a*sin(d*x+c))^(1/2)

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Rubi [A]  time = 0.36, antiderivative size = 221, normalized size of antiderivative = 1.00, 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 = {2687, 2681, 2650, 2649, 206} \[ -\frac {231 a \cos (c+d x)}{512 d (a \sin (c+d x)+a)^{3/2}}+\frac {\sec ^5(c+d x)}{5 d \sqrt {a \sin (c+d x)+a}}+\frac {11 \sec ^3(c+d x)}{40 d \sqrt {a \sin (c+d x)+a}}-\frac {11 a \sec ^3(c+d x)}{60 d (a \sin (c+d x)+a)^{3/2}}+\frac {77 \sec (c+d x)}{128 d \sqrt {a \sin (c+d x)+a}}-\frac {77 a \sec (c+d x)}{320 d (a \sin (c+d x)+a)^{3/2}}-\frac {231 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{512 \sqrt {2} \sqrt {a} d} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^6/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

(-231*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(512*Sqrt[2]*Sqrt[a]*d) - (231*a*Cos
[c + d*x])/(512*d*(a + a*Sin[c + d*x])^(3/2)) - (77*a*Sec[c + d*x])/(320*d*(a + a*Sin[c + d*x])^(3/2)) - (11*a
*Sec[c + d*x]^3)/(60*d*(a + a*Sin[c + d*x])^(3/2)) + (77*Sec[c + d*x])/(128*d*Sqrt[a + a*Sin[c + d*x]]) + (11*
Sec[c + d*x]^3)/(40*d*Sqrt[a + a*Sin[c + d*x]]) + Sec[c + d*x]^5/(5*d*Sqrt[a + a*Sin[c + d*x]])

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 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 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 2681

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*(2*m + p + 1)), x] + Dist[(m + p + 1)/(a*(2*m + p + 1)),
Int[(g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && LtQ[m, -1] && NeQ[2*m + p + 1, 0] && IntegersQ[2*m, 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]

Rubi steps

\begin {align*} \int \frac {\sec ^6(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx &=\frac {\sec ^5(c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}+\frac {1}{10} (11 a) \int \frac {\sec ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac {11 a \sec ^3(c+d x)}{60 d (a+a \sin (c+d x))^{3/2}}+\frac {\sec ^5(c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}+\frac {33}{40} \int \frac {\sec ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=-\frac {11 a \sec ^3(c+d x)}{60 d (a+a \sin (c+d x))^{3/2}}+\frac {11 \sec ^3(c+d x)}{40 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^5(c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}+\frac {1}{80} (77 a) \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac {77 a \sec (c+d x)}{320 d (a+a \sin (c+d x))^{3/2}}-\frac {11 a \sec ^3(c+d x)}{60 d (a+a \sin (c+d x))^{3/2}}+\frac {11 \sec ^3(c+d x)}{40 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^5(c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}+\frac {77}{128} \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=-\frac {77 a \sec (c+d x)}{320 d (a+a \sin (c+d x))^{3/2}}-\frac {11 a \sec ^3(c+d x)}{60 d (a+a \sin (c+d x))^{3/2}}+\frac {77 \sec (c+d x)}{128 d \sqrt {a+a \sin (c+d x)}}+\frac {11 \sec ^3(c+d x)}{40 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^5(c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}+\frac {1}{256} (231 a) \int \frac {1}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac {231 a \cos (c+d x)}{512 d (a+a \sin (c+d x))^{3/2}}-\frac {77 a \sec (c+d x)}{320 d (a+a \sin (c+d x))^{3/2}}-\frac {11 a \sec ^3(c+d x)}{60 d (a+a \sin (c+d x))^{3/2}}+\frac {77 \sec (c+d x)}{128 d \sqrt {a+a \sin (c+d x)}}+\frac {11 \sec ^3(c+d x)}{40 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^5(c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}+\frac {231 \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx}{1024}\\ &=-\frac {231 a \cos (c+d x)}{512 d (a+a \sin (c+d x))^{3/2}}-\frac {77 a \sec (c+d x)}{320 d (a+a \sin (c+d x))^{3/2}}-\frac {11 a \sec ^3(c+d x)}{60 d (a+a \sin (c+d x))^{3/2}}+\frac {77 \sec (c+d x)}{128 d \sqrt {a+a \sin (c+d x)}}+\frac {11 \sec ^3(c+d x)}{40 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^5(c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}-\frac {231 \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 )}{512 d}\\ &=-\frac {231 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{512 \sqrt {2} \sqrt {a} d}-\frac {231 a \cos (c+d x)}{512 d (a+a \sin (c+d x))^{3/2}}-\frac {77 a \sec (c+d x)}{320 d (a+a \sin (c+d x))^{3/2}}-\frac {11 a \sec ^3(c+d x)}{60 d (a+a \sin (c+d x))^{3/2}}+\frac {77 \sec (c+d x)}{128 d \sqrt {a+a \sin (c+d x)}}+\frac {11 \sec ^3(c+d x)}{40 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^5(c+d x)}{5 d \sqrt {a+a \sin (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 0.72, size = 140, normalized size = 0.63 \[ \frac {\frac {1}{16} \sec ^5(c+d x) (36850 \sin (c+d x)+17787 \sin (3 (c+d x))+3465 \sin (5 (c+d x))+11352 \cos (2 (c+d x))+2310 \cos (4 (c+d x))+11090)+(3465+3465 i) (-1)^{3/4} \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right ) \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac {1}{4} (c+d x)\right )-1\right )\right )}{7680 d \sqrt {a (\sin (c+d x)+1)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^6/Sqrt[a + a*Sin[c + d*x]],x]

[Out]

((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]) + (Sec[c + d*x]^5*(11090 + 11352*Cos[2*(c + d*x)] + 2310*Cos[4*(c + d*x)] + 36850*Sin[c + d*x] + 1
7787*Sin[3*(c + d*x)] + 3465*Sin[5*(c + d*x)]))/16)/(7680*d*Sqrt[a*(1 + Sin[c + d*x])])

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fricas [A]  time = 0.56, size = 250, normalized size = 1.13 \[ \frac {3465 \, \sqrt {2} {\left (\cos \left (d x + c\right )^{5} \sin \left (d x + c\right ) + \cos \left (d x + c\right )^{5}\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 4 \, {\left (1155 \, \cos \left (d x + c\right )^{4} + 264 \, \cos \left (d x + c\right )^{2} + 11 \, {\left (315 \, \cos \left (d x + c\right )^{4} + 168 \, \cos \left (d x + c\right )^{2} + 128\right )} \sin \left (d x + c\right ) + 128\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{30720 \, {\left (a d \cos \left (d x + c\right )^{5} \sin \left (d x + c\right ) + a d \cos \left (d x + c\right )^{5}\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a+a*sin(d*x+c))^(1/2),x, algorithm="fricas")

[Out]

1/30720*(3465*sqrt(2)*(cos(d*x + c)^5*sin(d*x + c) + cos(d*x + c)^5)*sqrt(a)*log(-(a*cos(d*x + c)^2 - 2*sqrt(2
)*sqrt(a*sin(d*x + c) + a)*sqrt(a)*(cos(d*x + c) - sin(d*x + c) + 1) + 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*(1155*cos(d*
x + c)^4 + 264*cos(d*x + c)^2 + 11*(315*cos(d*x + c)^4 + 168*cos(d*x + c)^2 + 128)*sin(d*x + c) + 128)*sqrt(a*
sin(d*x + c) + a))/(a*d*cos(d*x + c)^5*sin(d*x + c) + a*d*cos(d*x + c)^5)

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: TypeError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a+a*sin(d*x+c))^(1/2),x, algorithm="giac")

[Out]

Exception raised: TypeError >> An error occurred running a Giac command:INPUT:sage2:=int(sage0,x):;OUTPUT:Unab
le to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Unable to check sign: (2*p
i/x/2)>(-2*pi/x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)Warning, integration of abs or sign assumes cons
tant sign by intervals (correct if the argument is real):Check [abs(cos((d*t_nostep+c)/2-pi/4))]Discontinuitie
s at zeroes of cos((d*t_nostep+c)/2-pi/4) were not checkedWarning, integration of abs or sign assumes constant
sign by intervals (correct if the argument is real):Check [abs(t_nostep+1)]Evaluation time: 0.47Not invertibl
e Error: Bad Argument Value

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maple [A]  time = 0.28, size = 308, normalized size = 1.39 \[ -\frac {-6930 a^{\frac {11}{2}} \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )+\left (-3696 a^{\frac {11}{2}}-3465 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3} \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+\left (-2816 a^{\frac {11}{2}}+13860 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3} \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}}\right ) \sin \left (d x +c \right )-2310 a^{\frac {11}{2}} \left (\cos ^{4}\left (d x +c \right )\right )+\left (-528 a^{\frac {11}{2}}-10395 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3} \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}}\right ) \left (\cos ^{2}\left (d x +c \right )\right )-256 a^{\frac {11}{2}}+13860 \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{3} \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}}}{15360 a^{\frac {11}{2}} \left (\sin \left (d x +c \right )-1\right )^{2} \left (1+\sin \left (d x +c \right )\right )^{2} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^6/(a+a*sin(d*x+c))^(1/2),x)

[Out]

-1/15360*(-6930*a^(11/2)*sin(d*x+c)*cos(d*x+c)^4+(-3696*a^(11/2)-3465*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/
2)*2^(1/2)/a^(1/2))*a^3*(a-a*sin(d*x+c))^(5/2))*cos(d*x+c)^2*sin(d*x+c)+(-2816*a^(11/2)+13860*2^(1/2)*arctanh(
1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^3*(a-a*sin(d*x+c))^(5/2))*sin(d*x+c)-2310*a^(11/2)*cos(d*x+c)^4+
(-528*a^(11/2)-10395*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^3*(a-a*sin(d*x+c))^(5/2))*c
os(d*x+c)^2-256*a^(11/2)+13860*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^3*(a-a*sin(d*x+c)
)^(5/2))/a^(11/2)/(sin(d*x+c)-1)^2/(1+sin(d*x+c))^2/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (d x + c\right )^{6}}{\sqrt {a \sin \left (d x + c\right ) + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^6/(a+a*sin(d*x+c))^(1/2),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^6/sqrt(a*sin(d*x + c) + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^6\,\sqrt {a+a\,\sin \left (c+d\,x\right )}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^6*(a + a*sin(c + d*x))^(1/2)),x)

[Out]

int(1/(cos(c + d*x)^6*(a + a*sin(c + d*x))^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{6}{\left (c + d x \right )}}{\sqrt {a \left (\sin {\left (c + d x \right )} + 1\right )}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**6/(a+a*sin(d*x+c))**(1/2),x)

[Out]

Integral(sec(c + d*x)**6/sqrt(a*(sin(c + d*x) + 1)), x)

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