3.195 \(\int \frac {\sec ^4(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx\)

Optimal. Leaf size=233 \[ -\frac {1155 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{4096 \sqrt {2} a^{5/2} d}+\frac {11 \sec ^3(c+d x)}{64 a^2 d \sqrt {a \sin (c+d x)+a}}+\frac {385 \sec (c+d x)}{1024 a^2 d \sqrt {a \sin (c+d x)+a}}-\frac {1155 \cos (c+d x)}{4096 a d (a \sin (c+d x)+a)^{3/2}}-\frac {11 \sec ^3(c+d x)}{96 a d (a \sin (c+d x)+a)^{3/2}}-\frac {\sec ^3(c+d x)}{8 d (a \sin (c+d x)+a)^{5/2}}-\frac {77 \sec (c+d x)}{512 a d (a \sin (c+d x)+a)^{3/2}} \]

[Out]

-1/8*sec(d*x+c)^3/d/(a+a*sin(d*x+c))^(5/2)-1155/4096*cos(d*x+c)/a/d/(a+a*sin(d*x+c))^(3/2)-77/512*sec(d*x+c)/a
/d/(a+a*sin(d*x+c))^(3/2)-11/96*sec(d*x+c)^3/a/d/(a+a*sin(d*x+c))^(3/2)-1155/8192*arctanh(1/2*cos(d*x+c)*a^(1/
2)*2^(1/2)/(a+a*sin(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)+385/1024*sec(d*x+c)/a^2/d/(a+a*sin(d*x+c))^(1/2)+11/64*se
c(d*x+c)^3/a^2/d/(a+a*sin(d*x+c))^(1/2)

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Rubi [A]  time = 0.36, antiderivative size = 233, 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 = {2681, 2687, 2650, 2649, 206} \[ \frac {11 \sec ^3(c+d x)}{64 a^2 d \sqrt {a \sin (c+d x)+a}}+\frac {385 \sec (c+d x)}{1024 a^2 d \sqrt {a \sin (c+d x)+a}}-\frac {1155 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a \sin (c+d x)+a}}\right )}{4096 \sqrt {2} a^{5/2} d}-\frac {1155 \cos (c+d x)}{4096 a d (a \sin (c+d x)+a)^{3/2}}-\frac {11 \sec ^3(c+d x)}{96 a d (a \sin (c+d x)+a)^{3/2}}-\frac {\sec ^3(c+d x)}{8 d (a \sin (c+d x)+a)^{5/2}}-\frac {77 \sec (c+d x)}{512 a d (a \sin (c+d x)+a)^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[c + d*x]^4/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-1155*ArcTanh[(Sqrt[a]*Cos[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sin[c + d*x]])])/(4096*Sqrt[2]*a^(5/2)*d) - Sec[c +
d*x]^3/(8*d*(a + a*Sin[c + d*x])^(5/2)) - (1155*Cos[c + d*x])/(4096*a*d*(a + a*Sin[c + d*x])^(3/2)) - (77*Sec[
c + d*x])/(512*a*d*(a + a*Sin[c + d*x])^(3/2)) - (11*Sec[c + d*x]^3)/(96*a*d*(a + a*Sin[c + d*x])^(3/2)) + (38
5*Sec[c + d*x])/(1024*a^2*d*Sqrt[a + a*Sin[c + d*x]]) + (11*Sec[c + d*x]^3)/(64*a^2*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 ^4(c+d x)}{(a+a \sin (c+d x))^{5/2}} \, dx &=-\frac {\sec ^3(c+d x)}{8 d (a+a \sin (c+d x))^{5/2}}+\frac {11 \int \frac {\sec ^4(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx}{16 a}\\ &=-\frac {\sec ^3(c+d x)}{8 d (a+a \sin (c+d x))^{5/2}}-\frac {11 \sec ^3(c+d x)}{96 a d (a+a \sin (c+d x))^{3/2}}+\frac {33 \int \frac {\sec ^4(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{64 a^2}\\ &=-\frac {\sec ^3(c+d x)}{8 d (a+a \sin (c+d x))^{5/2}}-\frac {11 \sec ^3(c+d x)}{96 a d (a+a \sin (c+d x))^{3/2}}+\frac {11 \sec ^3(c+d x)}{64 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {77 \int \frac {\sec ^2(c+d x)}{(a+a \sin (c+d x))^{3/2}} \, dx}{128 a}\\ &=-\frac {\sec ^3(c+d x)}{8 d (a+a \sin (c+d x))^{5/2}}-\frac {77 \sec (c+d x)}{512 a d (a+a \sin (c+d x))^{3/2}}-\frac {11 \sec ^3(c+d x)}{96 a d (a+a \sin (c+d x))^{3/2}}+\frac {11 \sec ^3(c+d x)}{64 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {385 \int \frac {\sec ^2(c+d x)}{\sqrt {a+a \sin (c+d x)}} \, dx}{1024 a^2}\\ &=-\frac {\sec ^3(c+d x)}{8 d (a+a \sin (c+d x))^{5/2}}-\frac {77 \sec (c+d x)}{512 a d (a+a \sin (c+d x))^{3/2}}-\frac {11 \sec ^3(c+d x)}{96 a d (a+a \sin (c+d x))^{3/2}}+\frac {385 \sec (c+d x)}{1024 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {11 \sec ^3(c+d x)}{64 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {1155 \int \frac {1}{(a+a \sin (c+d x))^{3/2}} \, dx}{2048 a}\\ &=-\frac {\sec ^3(c+d x)}{8 d (a+a \sin (c+d x))^{5/2}}-\frac {1155 \cos (c+d x)}{4096 a d (a+a \sin (c+d x))^{3/2}}-\frac {77 \sec (c+d x)}{512 a d (a+a \sin (c+d x))^{3/2}}-\frac {11 \sec ^3(c+d x)}{96 a d (a+a \sin (c+d x))^{3/2}}+\frac {385 \sec (c+d x)}{1024 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {11 \sec ^3(c+d x)}{64 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {1155 \int \frac {1}{\sqrt {a+a \sin (c+d x)}} \, dx}{8192 a^2}\\ &=-\frac {\sec ^3(c+d x)}{8 d (a+a \sin (c+d x))^{5/2}}-\frac {1155 \cos (c+d x)}{4096 a d (a+a \sin (c+d x))^{3/2}}-\frac {77 \sec (c+d x)}{512 a d (a+a \sin (c+d x))^{3/2}}-\frac {11 \sec ^3(c+d x)}{96 a d (a+a \sin (c+d x))^{3/2}}+\frac {385 \sec (c+d x)}{1024 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {11 \sec ^3(c+d x)}{64 a^2 d \sqrt {a+a \sin (c+d x)}}-\frac {1155 \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 )}{4096 a^2 d}\\ &=-\frac {1155 \tanh ^{-1}\left (\frac {\sqrt {a} \cos (c+d x)}{\sqrt {2} \sqrt {a+a \sin (c+d x)}}\right )}{4096 \sqrt {2} a^{5/2} d}-\frac {\sec ^3(c+d x)}{8 d (a+a \sin (c+d x))^{5/2}}-\frac {1155 \cos (c+d x)}{4096 a d (a+a \sin (c+d x))^{3/2}}-\frac {77 \sec (c+d x)}{512 a d (a+a \sin (c+d x))^{3/2}}-\frac {11 \sec ^3(c+d x)}{96 a d (a+a \sin (c+d x))^{3/2}}+\frac {385 \sec (c+d x)}{1024 a^2 d \sqrt {a+a \sin (c+d x)}}+\frac {11 \sec ^3(c+d x)}{64 a^2 d \sqrt {a+a \sin (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 0.55, size = 394, normalized size = 1.69 \[ \frac {\frac {1920 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^5}{\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )}+\frac {256 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^5}{\left (\cos \left (\frac {1}{2} (c+d x)\right )-\sin \left (\frac {1}{2} (c+d x)\right )\right )^3}-1545 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^4+3090 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3-1036 \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2+2072 \sin \left (\frac {1}{2} (c+d x)\right ) \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )+\frac {1472 \sin \left (\frac {1}{2} (c+d x)\right )}{\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )}-\frac {384}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^2}+\frac {768 \sin \left (\frac {1}{2} (c+d x)\right )}{\left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^3}+(3465+3465 i) (-1)^{3/4} \left (\sin \left (\frac {1}{2} (c+d x)\right )+\cos \left (\frac {1}{2} (c+d x)\right )\right )^5 \tanh ^{-1}\left (\left (\frac {1}{2}+\frac {i}{2}\right ) (-1)^{3/4} \left (\tan \left (\frac {1}{4} (c+d x)\right )-1\right )\right )-736}{12288 d (a (\sin (c+d x)+1))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[c + d*x]^4/(a + a*Sin[c + d*x])^(5/2),x]

[Out]

(-736 + (768*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^3 - 384/(Cos[(c + d*x)/2] + Sin[(c + d*x)
/2])^2 + (1472*Sin[(c + d*x)/2])/(Cos[(c + d*x)/2] + Sin[(c + d*x)/2]) + 2072*Sin[(c + d*x)/2]*(Cos[(c + d*x)/
2] + Sin[(c + d*x)/2]) - 1036*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^2 + 3090*Sin[(c + d*x)/2]*(Cos[(c + d*x)/2
] + Sin[(c + d*x)/2])^3 - 1545*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^4 + (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])^5 + (256*(Cos[(c + d*x)/2]
+ Sin[(c + d*x)/2])^5)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2])^3 + (1920*(Cos[(c + d*x)/2] + Sin[(c + d*x)/2])^
5)/(Cos[(c + d*x)/2] - Sin[(c + d*x)/2]))/(12288*d*(a*(1 + Sin[c + d*x]))^(5/2))

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fricas [A]  time = 0.75, size = 308, normalized size = 1.32 \[ \frac {3465 \, \sqrt {2} {\left (3 \, \cos \left (d x + c\right )^{5} - 4 \, \cos \left (d x + c\right )^{3} + {\left (\cos \left (d x + c\right )^{5} - 4 \, \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )} \sqrt {a} \log \left (-\frac {a \cos \left (d x + c\right )^{2} - 2 \, \sqrt {2} \sqrt {a \sin \left (d x + c\right ) + a} \sqrt {a} {\left (\cos \left (d x + c\right ) - \sin \left (d x + c\right ) + 1\right )} + 3 \, a \cos \left (d x + c\right ) - {\left (a \cos \left (d x + c\right ) - 2 \, a\right )} \sin \left (d x + c\right ) + 2 \, a}{\cos \left (d x + c\right )^{2} - {\left (\cos \left (d x + c\right ) + 2\right )} \sin \left (d x + c\right ) - \cos \left (d x + c\right ) - 2}\right ) + 4 \, {\left (8085 \, \cos \left (d x + c\right )^{4} - 5280 \, \cos \left (d x + c\right )^{2} + 11 \, {\left (315 \, \cos \left (d x + c\right )^{4} - 672 \, \cos \left (d x + c\right )^{2} - 256\right )} \sin \left (d x + c\right ) - 1280\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{49152 \, {\left (3 \, a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3} + {\left (a^{3} d \cos \left (d x + c\right )^{5} - 4 \, a^{3} d \cos \left (d x + c\right )^{3}\right )} \sin \left (d x + c\right )\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+a*sin(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

1/49152*(3465*sqrt(2)*(3*cos(d*x + c)^5 - 4*cos(d*x + c)^3 + (cos(d*x + c)^5 - 4*cos(d*x + c)^3)*sin(d*x + c))
*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*(8085*cos(d*x + c)^4 - 5280*cos(d*x + c)^2 + 11*(315*cos(d*x + c)^4 - 672*cos(d*
x + c)^2 - 256)*sin(d*x + c) - 1280)*sqrt(a*sin(d*x + c) + a))/(3*a^3*d*cos(d*x + c)^5 - 4*a^3*d*cos(d*x + c)^
3 + (a^3*d*cos(d*x + c)^5 - 4*a^3*d*cos(d*x + c)^3)*sin(d*x + c))

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giac [B]  time = 9.34, size = 1074, normalized size = 4.61 \[ \text {result too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^4/(a+a*sin(d*x+c))^(5/2),x, algorithm="giac")

[Out]

1/12288*(3465*sqrt(2)*arctan(-1/2*sqrt(2)*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a) +
sqrt(a))/sqrt(-a))/(sqrt(-a)*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)) + 256*(21*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqr
t(a*tan(1/2*d*x + 1/2*c)^2 + a))^5 - 51*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*
sqrt(a) - 34*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^3*a + 102*(sqrt(a)*tan(1/2*d*
x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a^(3/2) + 81*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2
*d*x + 1/2*c)^2 + a))*a^2 + 17*a^(5/2))/(((sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^
2 - 2*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))*sqrt(a) - a)^3*a^2*sgn(tan(1/2*d*x +
1/2*c) + 1)) + 2*(18423*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^15 + 165753*(sqrt
(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^14*sqrt(a) + 644313*(sqrt(a)*tan(1/2*d*x + 1/2*
c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^13*a + 1072899*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1
/2*c)^2 + a))^12*a^(3/2) + 94635*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^11*a^2 -
1907635*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^10*a^(5/2) - 875803*(sqrt(a)*tan(1
/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^9*a^3 + 2261311*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*t
an(1/2*d*x + 1/2*c)^2 + a))^8*a^(7/2) + 723029*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 +
a))^7*a^4 - 2030229*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^6*a^(9/2) + 509147*(s
qrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^5*a^5 + 688777*(sqrt(a)*tan(1/2*d*x + 1/2*c)
- sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^4*a^(11/2) - 599223*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x
+ 1/2*c)^2 + a))^3*a^6 + 219151*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2*a^(13/2)
- 40793*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))*a^7 + 3701*a^(15/2))/(((sqrt(a)*t
an(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2*d*x + 1/2*c)^2 + a))^2 + 2*(sqrt(a)*tan(1/2*d*x + 1/2*c) - sqrt(a*tan(1/2
*d*x + 1/2*c)^2 + a))*sqrt(a) - a)^8*a^2*sgn(tan(1/2*d*x + 1/2*c) + 1)))/d

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maple [A]  time = 0.29, size = 355, normalized size = 1.52 \[ \frac {6930 a^{\frac {11}{2}} \sin \left (d x +c \right ) \left (\cos ^{4}\left (d x +c \right )\right )-924 \left (16 a^{\frac {11}{2}}+15 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4}\right ) \left (\cos ^{2}\left (d x +c \right )\right ) \sin \left (d x +c \right )+\left (-5632 a^{\frac {11}{2}}+27720 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4}\right ) \sin \left (d x +c \right )+\left (16170 a^{\frac {11}{2}}+3465 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4}\right ) \left (\cos ^{4}\left (d x +c \right )\right )-1320 \left (8 a^{\frac {11}{2}}+21 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4}\right ) \left (\cos ^{2}\left (d x +c \right )\right )-2560 a^{\frac {11}{2}}+27720 \left (a -a \sin \left (d x +c \right )\right )^{\frac {3}{2}} \sqrt {2}\, \arctanh \left (\frac {\sqrt {a -a \sin \left (d x +c \right )}\, \sqrt {2}}{2 \sqrt {a}}\right ) a^{4}}{24576 a^{\frac {15}{2}} \left (\sin \left (d x +c \right )-1\right ) \left (1+\sin \left (d x +c \right )\right )^{3} \cos \left (d x +c \right ) \sqrt {a +a \sin \left (d x +c \right )}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^4/(a+a*sin(d*x+c))^(5/2),x)

[Out]

1/24576/a^(15/2)*(6930*a^(11/2)*sin(d*x+c)*cos(d*x+c)^4-924*(16*a^(11/2)+15*(a-a*sin(d*x+c))^(3/2)*2^(1/2)*arc
tanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^4)*cos(d*x+c)^2*sin(d*x+c)+(-5632*a^(11/2)+27720*(a-a*sin(d
*x+c))^(3/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^4)*sin(d*x+c)+(16170*a^(11/2)+3465*
(a-a*sin(d*x+c))^(3/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^4)*cos(d*x+c)^4-1320*(8*a
^(11/2)+21*(a-a*sin(d*x+c))^(3/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^4)*cos(d*x+c)^
2-2560*a^(11/2)+27720*(a-a*sin(d*x+c))^(3/2)*2^(1/2)*arctanh(1/2*(a-a*sin(d*x+c))^(1/2)*2^(1/2)/a^(1/2))*a^4)/
(sin(d*x+c)-1)/(1+sin(d*x+c))^3/cos(d*x+c)/(a+a*sin(d*x+c))^(1/2)/d

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maxima [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)^4/(a+a*sin(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

Timed out

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {1}{{\cos \left (c+d\,x\right )}^4\,{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^4*(a + a*sin(c + d*x))^(5/2)),x)

[Out]

int(1/(cos(c + d*x)^4*(a + a*sin(c + d*x))^(5/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{4}{\left (c + d x \right )}}{\left (a \left (\sin {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**4/(a+a*sin(d*x+c))**(5/2),x)

[Out]

Integral(sec(c + d*x)**4/(a*(sin(c + d*x) + 1))**(5/2), x)

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