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

Optimal. Leaf size=197 \[ -\frac {63 a^2 \cos (c+d x)}{128 d (a \sin (c+d x)+a)^{3/2}}-\frac {21 a^2 \sec (c+d x)}{80 d (a \sin (c+d x)+a)^{3/2}}+\frac {\sec ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{5 d}+\frac {3 a \sec ^3(c+d x)}{10 d \sqrt {a \sin (c+d x)+a}}+\frac {21 a \sec (c+d x)}{32 d \sqrt {a \sin (c+d x)+a}}-\frac {63 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{128 \sqrt {2} d} \]

[Out]

-63/128*a^2*cos(d*x+c)/d/(a+a*sin(d*x+c))^(3/2)-21/80*a^2*sec(d*x+c)/d/(a+a*sin(d*x+c))^(3/2)-63/256*arctanh(1
/2*cos(d*x+c)*a^(1/2)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))*a^(1/2)/d*2^(1/2)+21/32*a*sec(d*x+c)/d/(a+a*sin(d*x+c))^
(1/2)+3/10*a*sec(d*x+c)^3/d/(a+a*sin(d*x+c))^(1/2)+1/5*sec(d*x+c)^5*(a+a*sin(d*x+c))^(1/2)/d

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Rubi [A]  time = 0.29, antiderivative size = 197, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {2675, 2687, 2681, 2650, 2649, 206} \[ -\frac {63 a^2 \cos (c+d x)}{128 d (a \sin (c+d x)+a)^{3/2}}-\frac {21 a^2 \sec (c+d x)}{80 d (a \sin (c+d x)+a)^{3/2}}+\frac {\sec ^5(c+d x) \sqrt {a \sin (c+d x)+a}}{5 d}+\frac {3 a \sec ^3(c+d x)}{10 d \sqrt {a \sin (c+d x)+a}}+\frac {21 a \sec (c+d x)}{32 d \sqrt {a \sin (c+d x)+a}}-\frac {63 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{128 \sqrt {2} d} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-63*Sqrt[a]*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(128*Sqrt[2]*d) - (63*a^2*Cos
[c + d*x])/(128*d*(a + a*Sin[c + d*x])^(3/2)) - (21*a^2*Sec[c + d*x])/(80*d*(a + a*Sin[c + d*x])^(3/2)) + (21*
a*Sec[c + d*x])/(32*d*Sqrt[a + a*Sin[c + d*x]]) + (3*a*Sec[c + d*x]^3)/(10*d*Sqrt[a + a*Sin[c + d*x]]) + (Sec[
c + d*x]^5*Sqrt[a + a*Sin[c + d*x]])/(5*d)

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 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 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 \sec ^6(c+d x) \sqrt {a+a \sin (c+d x)} \, dx &=\frac {\sec ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}+\frac {1}{10} (9 a) \int \frac {\sec ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=\frac {3 a \sec ^3(c+d x)}{10 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}+\frac {1}{20} \left (21 a^2\right ) \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac {21 a^2 \sec (c+d x)}{80 d (a+a \sin (c+d x))^{3/2}}+\frac {3 a \sec ^3(c+d x)}{10 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}+\frac {1}{32} (21 a) \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=-\frac {21 a^2 \sec (c+d x)}{80 d (a+a \sin (c+d x))^{3/2}}+\frac {21 a \sec (c+d x)}{32 d \sqrt {a+a \sin (c+d x)}}+\frac {3 a \sec ^3(c+d x)}{10 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}+\frac {1}{64} \left (63 a^2\right ) \int \frac {1}{(a+a \sin (c+d x))^{3/2}} \, dx\\ &=-\frac {63 a^2 \cos (c+d x)}{128 d (a+a \sin (c+d x))^{3/2}}-\frac {21 a^2 \sec (c+d x)}{80 d (a+a \sin (c+d x))^{3/2}}+\frac {21 a \sec (c+d x)}{32 d \sqrt {a+a \sin (c+d x)}}+\frac {3 a \sec ^3(c+d x)}{10 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}+\frac {1}{256} (63 a) \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx\\ &=-\frac {63 a^2 \cos (c+d x)}{128 d (a+a \sin (c+d x))^{3/2}}-\frac {21 a^2 \sec (c+d x)}{80 d (a+a \sin (c+d x))^{3/2}}+\frac {21 a \sec (c+d x)}{32 d \sqrt {a+a \sin (c+d x)}}+\frac {3 a \sec ^3(c+d x)}{10 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}-\frac {(63 a) \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 )}{128 d}\\ &=-\frac {63 \sqrt {a} \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{128 \sqrt {2} d}-\frac {63 a^2 \cos (c+d x)}{128 d (a+a \sin (c+d x))^{3/2}}-\frac {21 a^2 \sec (c+d x)}{80 d (a+a \sin (c+d x))^{3/2}}+\frac {21 a \sec (c+d x)}{32 d \sqrt {a+a \sin (c+d x)}}+\frac {3 a \sec ^3(c+d x)}{10 d \sqrt {a+a \sin (c+d x)}}+\frac {\sec ^5(c+d x) \sqrt {a+a \sin (c+d x)}}{5 d}\\ \end {align*}

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Mathematica [C]  time = 0.65, size = 191, normalized size = 0.97 \[ \frac {\sqrt {a (\sin (c+d x)+1)} \left (\frac {1572 \sin (c+d x)+420 \sin (3 (c+d x))+1092 \cos (2 (c+d x))+315 \cos (4 (c+d x))+649}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^5}-(2520+2520 i) (-1)^{3/4} \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4 \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \sec \left (\frac {d x}{4}\right ) \left (\cos \left (\frac {1}{4} (2 c+d x)\right )-\sin \left (\frac {1}{4} (2 c+d x)\right )\right )\right )\right )}{5120 d \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^5} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a*(1 + Sin[c + d*x])]*((-2520 - 2520*I)*(-1)^(3/4)*ArcTanh[(1/2 + I/2)*(-1)^(3/4)*Sec[(d*x)/4]*(Cos[(2*c
+ d*x)/4] - Sin[(2*c + d*x)/4])]*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 + (649 + 1092*Cos[2*(c + d*x)] + 315
*Cos[4*(c + d*x)] + 1572*Sin[c + d*x] + 420*Sin[3*(c + d*x)])/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^5))/(5120*
d*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^5)

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fricas [A]  time = 0.62, size = 210, normalized size = 1.07 \[ \frac {315 \, \sqrt {2} \sqrt {a} \cos \left (d x + c\right )^{5} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {a \sin \left (d x + c\right ) + a} {\left (\sqrt {2} \cos \left (d x + c\right ) - \sqrt {2} \sin \left (d x + c\right ) + \sqrt {2}\right )} \sqrt {a} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 4 \, {\left (315 \, \cos \left (d x + c\right )^{4} - 42 \, \cos \left (d x + c\right )^{2} + 6 \, {\left (35 \, \cos \left (d x + c\right )^{2} + 24\right )} \sin \left (d x + c\right ) - 16\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{2560 \, d \cos \left (d x + c\right )^{5}} \]

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]

1/2560*(315*sqrt(2)*sqrt(a)*cos(d*x + c)^5*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*(315*cos(d*x + c)^4 - 42*cos(d*x
+ c)^2 + 6*(35*cos(d*x + c)^2 + 24)*sin(d*x + c) - 16)*sqrt(a*sin(d*x + c) + a))/(d*cos(d*x + c)^5)

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giac [A]  time = 2.53, size = 239, normalized size = 1.21 \[ \frac {\sqrt {2} {\left (315 \, \log \left ({\left | \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + 1 \right |}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - 315 \, \log \left ({\left | \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right ) - 1 \right |}\right ) \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) - \frac {10 \, {\left (15 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{3} - 17 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )}}{{\left (\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - 1\right )}^{2}} - \frac {16 \, {\left (30 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{4} + 5 \, \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right ) \sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \mathrm {sgn}\left (\cos \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )\right )\right )}}{\sin \left (-\frac {1}{4} \, \pi + \frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{5}}\right )} \sqrt {a}}{2560 \, d} \]

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]

1/2560*sqrt(2)*(315*log(abs(sin(-1/4*pi + 1/2*d*x + 1/2*c) + 1))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) - 315*log
(abs(sin(-1/4*pi + 1/2*d*x + 1/2*c) - 1))*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c)) - 10*(15*sgn(cos(-1/4*pi + 1/2*d
*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^3 - 17*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x +
1/2*c))/(sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 - 1)^2 - 16*(30*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1
/2*d*x + 1/2*c)^4 + 5*sgn(cos(-1/4*pi + 1/2*d*x + 1/2*c))*sin(-1/4*pi + 1/2*d*x + 1/2*c)^2 + sgn(cos(-1/4*pi +
1/2*d*x + 1/2*c)))/sin(-1/4*pi + 1/2*d*x + 1/2*c)^5)*sqrt(a)/d

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maple [A]  time = 0.35, size = 244, normalized size = 1.24 \[ -\frac {-420 a^{\frac {9}{2}} \sin \left (d x +c \right ) \left (\cos ^{2}\left (d x +c \right )\right )+\left (630 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}-288 a^{\frac {9}{2}}\right ) \sin \left (d x +c \right )-630 a^{\frac {9}{2}} \left (\cos ^{4}\left (d x +c \right )\right )+\left (-315 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}+84 a^{\frac {9}{2}}\right ) \left (\cos ^{2}\left (d x +c \right )\right )+630 \left (a -a \sin \left (d x +c \right )\right )^{\frac {5}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{2}+32 a^{\frac {9}{2}}}{1280 a^{\frac {7}{2}} \left (\sin \left (d x +c \right )-1\right )^{2} \left (1+\sin \left (d x +c \right )\right ) \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

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]

-1/1280/a^(7/2)*(-420*a^(9/2)*sin(d*x+c)*cos(d*x+c)^2+(630*(a-a*sin(d*x+c))^(5/2)*2^(1/2)*arctanh(1/2*(a-a*sin
(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2-288*a^(9/2))*sin(d*x+c)-630*a^(9/2)*cos(d*x+c)^4+(-315*(a-a*sin(d*x+c))^(5
/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2+84*a^(9/2))*cos(d*x+c)^2+630*(a-a*sin(d*x+
c))^(5/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^2+32*a^(9/2))/(sin(d*x+c)-1)^2/(1+sin(
d*x+c))/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 \[ \int \sqrt {a \sin \left (d x + c\right ) + a} \sec \left (d x + c\right )^{6}\,{d x} \]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {\sqrt {a+a\,\sin \left (c+d\,x\right )}}{{\cos \left (c+d\,x\right )}^6} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

Timed out

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