3.119 \(\int \frac {\sin ^5(e+f x)}{(a+b \sec ^2(e+f x))^{5/2}} \, dx\)

Optimal. Leaf size=219 \[ \frac {2 (5 a+4 b) \cos ^3(e+f x)}{15 a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {8 b \left (5 a^2+20 a b+16 b^2\right ) \sec (e+f x)}{15 a^5 f \sqrt {a+b \sec ^2(e+f x)}}-\frac {4 b \left (5 a^2+20 a b+16 b^2\right ) \sec (e+f x)}{15 a^4 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\left (5 a^2+20 a b+16 b^2\right ) \cos (e+f x)}{5 a^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\cos ^5(e+f x)}{5 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

[Out]

-1/5*(5*a^2+20*a*b+16*b^2)*cos(f*x+e)/a^3/f/(a+b*sec(f*x+e)^2)^(3/2)+2/15*(5*a+4*b)*cos(f*x+e)^3/a^2/f/(a+b*se
c(f*x+e)^2)^(3/2)-1/5*cos(f*x+e)^5/a/f/(a+b*sec(f*x+e)^2)^(3/2)-4/15*b*(5*a^2+20*a*b+16*b^2)*sec(f*x+e)/a^4/f/
(a+b*sec(f*x+e)^2)^(3/2)-8/15*b*(5*a^2+20*a*b+16*b^2)*sec(f*x+e)/a^5/f/(a+b*sec(f*x+e)^2)^(1/2)

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Rubi [A]  time = 0.22, antiderivative size = 219, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {4134, 462, 453, 271, 192, 191} \[ -\frac {8 b \left (5 a^2+20 a b+16 b^2\right ) \sec (e+f x)}{15 a^5 f \sqrt {a+b \sec ^2(e+f x)}}-\frac {4 b \left (5 a^2+20 a b+16 b^2\right ) \sec (e+f x)}{15 a^4 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac {2 (5 a+4 b) \cos ^3(e+f x)}{15 a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\left (\frac {4 b (5 a+4 b)}{a^2}+5\right ) \cos (e+f x)}{5 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\cos ^5(e+f x)}{5 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^5/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

-((5 + (4*b*(5*a + 4*b))/a^2)*Cos[e + f*x])/(5*a*f*(a + b*Sec[e + f*x]^2)^(3/2)) + (2*(5*a + 4*b)*Cos[e + f*x]
^3)/(15*a^2*f*(a + b*Sec[e + f*x]^2)^(3/2)) - Cos[e + f*x]^5/(5*a*f*(a + b*Sec[e + f*x]^2)^(3/2)) - (4*b*(5*a^
2 + 20*a*b + 16*b^2)*Sec[e + f*x])/(15*a^4*f*(a + b*Sec[e + f*x]^2)^(3/2)) - (8*b*(5*a^2 + 20*a*b + 16*b^2)*Se
c[e + f*x])/(15*a^5*f*Sqrt[a + b*Sec[e + f*x]^2])

Rule 191

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^(p + 1))/a, x] /; FreeQ[{a, b, n, p}, x] &
& EqQ[1/n + p + 1, 0]

Rule 192

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1))/(a*n*(p + 1)), x] + Dist[(n*(p +
1) + 1)/(a*n*(p + 1)), Int[(a + b*x^n)^(p + 1), x], x] /; FreeQ[{a, b, n, p}, x] && ILtQ[Simplify[1/n + p + 1
], 0] && NeQ[p, -1]

Rule 271

Int[(x_)^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x^(m + 1)*(a + b*x^n)^(p + 1))/(a*(m + 1)), x]
- Dist[(b*(m + n*(p + 1) + 1))/(a*(m + 1)), Int[x^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, m, n, p}, x]
&& ILtQ[Simplify[(m + 1)/n + p + 1], 0] && NeQ[m, -1]

Rule 453

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(c*(e*x)^(m
+ 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] + Dist[(a*d*(m + 1) - b*c*(m + n*(p + 1) + 1))/(a*e^n*(m + 1)), In
t[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a, b, c, d, e, p}, x] && NeQ[b*c - a*d, 0] && (IntegerQ[n] ||
GtQ[e, 0]) && ((GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -1])) &&  !ILtQ[p, -1]

Rule 462

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^2, x_Symbol] :> Simp[(c^2*(e*x)^(
m + 1)*(a + b*x^n)^(p + 1))/(a*e*(m + 1)), x] - Dist[1/(a*e^n*(m + 1)), Int[(e*x)^(m + n)*(a + b*x^n)^p*Simp[b
*c^2*n*(p + 1) + c*(b*c - 2*a*d)*(m + 1) - a*(m + 1)*d^2*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, p}, x] && Ne
Q[b*c - a*d, 0] && IGtQ[n, 0] && LtQ[m, -1] && GtQ[n, 0]

Rule 4134

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Cos[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^
n)^p)/x^(m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (Gt
Q[m, 0] || EqQ[n, 2] || EqQ[n, 4])

Rubi steps

\begin {align*} \int \frac {\sin ^5(e+f x)}{\left (a+b \sec ^2(e+f x)\right )^{5/2}} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (-1+x^2\right )^2}{x^6 \left (a+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {\cos ^5(e+f x)}{5 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac {\operatorname {Subst}\left (\int \frac {-2 (5 a+4 b)+5 a x^2}{x^4 \left (a+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{5 a f}\\ &=\frac {2 (5 a+4 b) \cos ^3(e+f x)}{15 a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\cos ^5(e+f x)}{5 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac {\left (5 a^2+20 a b+16 b^2\right ) \operatorname {Subst}\left (\int \frac {1}{x^2 \left (a+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{5 a^2 f}\\ &=-\frac {\left (5 a^2+20 a b+16 b^2\right ) \cos (e+f x)}{5 a^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac {2 (5 a+4 b) \cos ^3(e+f x)}{15 a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\cos ^5(e+f x)}{5 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\left (4 b \left (5 a^2+20 a b+16 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\left (a+b x^2\right )^{5/2}} \, dx,x,\sec (e+f x)\right )}{5 a^3 f}\\ &=-\frac {\left (5 a^2+20 a b+16 b^2\right ) \cos (e+f x)}{5 a^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac {2 (5 a+4 b) \cos ^3(e+f x)}{15 a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\cos ^5(e+f x)}{5 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {4 b \left (5 a^2+20 a b+16 b^2\right ) \sec (e+f x)}{15 a^4 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\left (8 b \left (5 a^2+20 a b+16 b^2\right )\right ) \operatorname {Subst}\left (\int \frac {1}{\left (a+b x^2\right )^{3/2}} \, dx,x,\sec (e+f x)\right )}{15 a^4 f}\\ &=-\frac {\left (5 a^2+20 a b+16 b^2\right ) \cos (e+f x)}{5 a^3 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}+\frac {2 (5 a+4 b) \cos ^3(e+f x)}{15 a^2 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {\cos ^5(e+f x)}{5 a f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {4 b \left (5 a^2+20 a b+16 b^2\right ) \sec (e+f x)}{15 a^4 f \left (a+b \sec ^2(e+f x)\right )^{3/2}}-\frac {8 b \left (5 a^2+20 a b+16 b^2\right ) \sec (e+f x)}{15 a^5 f \sqrt {a+b \sec ^2(e+f x)}}\\ \end {align*}

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Mathematica [A]  time = 3.21, size = 182, normalized size = 0.83 \[ -\frac {\sec ^5(e+f x) (a \cos (2 (e+f x))+a+2 b) \left (-16 a^4 \cos (6 (e+f x))+3 a^4 \cos (8 (e+f x))+425 a^4-32 a^3 b \cos (6 (e+f x))+6400 a^3 b+12 a^2 \left (7 a^2+64 a b+64 b^2\right ) \cos (4 (e+f x))+22784 a^2 b^2+48 a \left (11 a^3+150 a^2 b+384 a b^2+256 b^3\right ) \cos (2 (e+f x))+32768 a b^3+16384 b^4\right )}{3840 a^5 f \left (a+b \sec ^2(e+f x)\right )^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^5/(a + b*Sec[e + f*x]^2)^(5/2),x]

[Out]

-1/3840*((a + 2*b + a*Cos[2*(e + f*x)])*(425*a^4 + 6400*a^3*b + 22784*a^2*b^2 + 32768*a*b^3 + 16384*b^4 + 48*a
*(11*a^3 + 150*a^2*b + 384*a*b^2 + 256*b^3)*Cos[2*(e + f*x)] + 12*a^2*(7*a^2 + 64*a*b + 64*b^2)*Cos[4*(e + f*x
)] - 16*a^4*Cos[6*(e + f*x)] - 32*a^3*b*Cos[6*(e + f*x)] + 3*a^4*Cos[8*(e + f*x)])*Sec[e + f*x]^5)/(a^5*f*(a +
b*Sec[e + f*x]^2)^(5/2))

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fricas [A]  time = 0.96, size = 189, normalized size = 0.86 \[ -\frac {{\left (3 \, a^{4} \cos \left (f x + e\right )^{9} - 2 \, {\left (5 \, a^{4} + 4 \, a^{3} b\right )} \cos \left (f x + e\right )^{7} + 3 \, {\left (5 \, a^{4} + 20 \, a^{3} b + 16 \, a^{2} b^{2}\right )} \cos \left (f x + e\right )^{5} + 12 \, {\left (5 \, a^{3} b + 20 \, a^{2} b^{2} + 16 \, a b^{3}\right )} \cos \left (f x + e\right )^{3} + 8 \, {\left (5 \, a^{2} b^{2} + 20 \, a b^{3} + 16 \, b^{4}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, {\left (a^{7} f \cos \left (f x + e\right )^{4} + 2 \, a^{6} b f \cos \left (f x + e\right )^{2} + a^{5} b^{2} f\right )}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="fricas")

[Out]

-1/15*(3*a^4*cos(f*x + e)^9 - 2*(5*a^4 + 4*a^3*b)*cos(f*x + e)^7 + 3*(5*a^4 + 20*a^3*b + 16*a^2*b^2)*cos(f*x +
e)^5 + 12*(5*a^3*b + 20*a^2*b^2 + 16*a*b^3)*cos(f*x + e)^3 + 8*(5*a^2*b^2 + 20*a*b^3 + 16*b^4)*cos(f*x + e))*
sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/(a^7*f*cos(f*x + e)^4 + 2*a^6*b*f*cos(f*x + e)^2 + a^5*b^2*f)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check si
gn: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)Unable to check sign: (4*pi/x/2)>(-4*pi/x/2)2/f*(2*(tan((f*x+exp(1))/2)^2*(tan((f*x+exp(1))/2)^2*(-1/2304*
tan((f*x+exp(1))/2)^2*(-2112*a^13*b^6*sign(tan((f*x+exp(1))/2)^2-1)-5376*a^14*b^5*sign(tan((f*x+exp(1))/2)^2-1
)-4416*a^15*b^4*sign(tan((f*x+exp(1))/2)^2-1)-1152*a^16*b^3*sign(tan((f*x+exp(1))/2)^2-1))/a^18/b^2-1/2304*(-6
336*a^13*b^6*sign(tan((f*x+exp(1))/2)^2-1)-6912*a^14*b^5*sign(tan((f*x+exp(1))/2)^2-1)+576*a^15*b^4*sign(tan((
f*x+exp(1))/2)^2-1)+1152*a^16*b^3*sign(tan((f*x+exp(1))/2)^2-1))/a^18/b^2)-1/2304*(-6336*a^13*b^6*sign(tan((f*
x+exp(1))/2)^2-1)-6912*a^14*b^5*sign(tan((f*x+exp(1))/2)^2-1)+576*a^15*b^4*sign(tan((f*x+exp(1))/2)^2-1)+1152*
a^16*b^3*sign(tan((f*x+exp(1))/2)^2-1))/a^18/b^2)-1/2304*(-2112*a^13*b^6*sign(tan((f*x+exp(1))/2)^2-1)-5376*a^
14*b^5*sign(tan((f*x+exp(1))/2)^2-1)-4416*a^15*b^4*sign(tan((f*x+exp(1))/2)^2-1)-1152*a^16*b^3*sign(tan((f*x+e
xp(1))/2)^2-1))/a^18/b^2)/sqrt(a*tan((f*x+exp(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+2*b*t
an((f*x+exp(1))/2)^2+a+b)/(a*tan((f*x+exp(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+2*b*tan((
f*x+exp(1))/2)^2+a+b)+2/15*((45*b^2+30*a*b)*(-tan((f*x+exp(1))/2)^2*sqrt(a+b)+sqrt(a*tan((f*x+exp(1))/2)^4+b*t
an((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+2*b*tan((f*x+exp(1))/2)^2+a+b))^9+sqrt(a+b)*(-315*b^2-330*a*b)*
(-tan((f*x+exp(1))/2)^2*sqrt(a+b)+sqrt(a*tan((f*x+exp(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)
^2+2*b*tan((f*x+exp(1))/2)^2+a+b))^8+(-tan((f*x+exp(1))/2)^2*sqrt(a+b)+sqrt(a*tan((f*x+exp(1))/2)^4+b*tan((f*x
+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+2*b*tan((f*x+exp(1))/2)^2+a+b))^7*(320*a^3+900*b^3+1260*a*b^2+760*a^2*
b)+sqrt(a+b)*(-tan((f*x+exp(1))/2)^2*sqrt(a+b)+sqrt(a*tan((f*x+exp(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f
*x+exp(1))/2)^2+2*b*tan((f*x+exp(1))/2)^2+a+b))^6*(640*a^3-1260*b^3+1020*a*b^2+2360*a^2*b)+(-tan((f*x+exp(1))/
2)^2*sqrt(a+b)+sqrt(a*tan((f*x+exp(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+2*b*tan((f*x+exp
(1))/2)^2+a+b))^5*(-832*a^4+630*b^4-5280*a*b^3-9330*a^2*b^2-5180*a^3*b)+(-tan((f*x+exp(1))/2)^2*sqrt(a+b)+sqrt
(a*tan((f*x+exp(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+2*b*tan((f*x+exp(1))/2)^2+a+b))*(28
80*a^6-315*b^6+2610*a*b^5-3610*a^2*b^4-13480*a^3*b^3+14125*a^4*b^2+20110*a^5*b)+sqrt(a+b)*(-tan((f*x+exp(1))/2
)^2*sqrt(a+b)+sqrt(a*tan((f*x+exp(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+2*b*tan((f*x+exp(
1))/2)^2+a+b))^4*(-2560*a^4+630*b^4+6480*a*b^3-4050*a^2*b^2-12620*a^3*b)+(-tan((f*x+exp(1))/2)^2*sqrt(a+b)+sqr
t(a*tan((f*x+exp(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+2*b*tan((f*x+exp(1))/2)^2+a+b))^3*
(-320*a^5-1260*b^5-1500*a*b^4+16740*a^2*b^3+18220*a^3*b^2+2200*a^4*b)+sqrt(a+b)*(-tan((f*x+exp(1))/2)^2*sqrt(a
+b)+sqrt(a*tan((f*x+exp(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+2*b*tan((f*x+exp(1))/2)^2+a
+b))^2*(3200*a^5+900*b^5-3660*a*b^4-7500*a^2*b^3+20380*a^3*b^2+22680*a^4*b)+sqrt(a+b)*(768*a^6+45*b^6-630*a*b^
5+3190*a^2*b^4-4920*a^3*b^3-907*a^4*b^2+4806*a^5*b))/a^4/(2*sqrt(a+b)*(-tan((f*x+exp(1))/2)^2*sqrt(a+b)+sqrt(a
*tan((f*x+exp(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+2*b*tan((f*x+exp(1))/2)^2+a+b))-(-tan
((f*x+exp(1))/2)^2*sqrt(a+b)+sqrt(a*tan((f*x+exp(1))/2)^4+b*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+2*
b*tan((f*x+exp(1))/2)^2+a+b))^2+3*a-b)^5/sign(tan((f*x+exp(1))/2)^2-1))

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maple [A]  time = 3.41, size = 229, normalized size = 1.05 \[ \frac {\left (a +b \right )^{7} \left (b +a \left (\cos ^{2}\left (f x +e \right )\right )\right ) \left (3 \left (\cos ^{8}\left (f x +e \right )\right ) a^{4}-10 \left (\cos ^{6}\left (f x +e \right )\right ) a^{4}-8 \left (\cos ^{6}\left (f x +e \right )\right ) a^{3} b +15 \left (\cos ^{4}\left (f x +e \right )\right ) a^{4}+60 \left (\cos ^{4}\left (f x +e \right )\right ) a^{3} b +48 \left (\cos ^{4}\left (f x +e \right )\right ) a^{2} b^{2}+60 \left (\cos ^{2}\left (f x +e \right )\right ) a^{3} b +240 \left (\cos ^{2}\left (f x +e \right )\right ) a^{2} b^{2}+192 \left (\cos ^{2}\left (f x +e \right )\right ) a \,b^{3}+40 a^{2} b^{2}+160 a \,b^{3}+128 b^{4}\right ) \sqrt {4}\, a^{2}}{30 f \left (\frac {b +a \left (\cos ^{2}\left (f x +e \right )\right )}{\cos \left (f x +e \right )^{2}}\right )^{\frac {5}{2}} \cos \left (f x +e \right )^{5} \left (\sqrt {-a b}+a \right )^{7} \left (\sqrt {-a b}-a \right )^{7}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^5/(a+b*sec(f*x+e)^2)^(5/2),x)

[Out]

1/30/f*(a+b)^7*(b+a*cos(f*x+e)^2)*(3*cos(f*x+e)^8*a^4-10*cos(f*x+e)^6*a^4-8*cos(f*x+e)^6*a^3*b+15*cos(f*x+e)^4
*a^4+60*cos(f*x+e)^4*a^3*b+48*cos(f*x+e)^4*a^2*b^2+60*cos(f*x+e)^2*a^3*b+240*cos(f*x+e)^2*a^2*b^2+192*cos(f*x+
e)^2*a*b^3+40*a^2*b^2+160*a*b^3+128*b^4)*4^(1/2)/((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(5/2)/cos(f*x+e)^5*a^2/((-a
*b)^(1/2)+a)^7/((-a*b)^(1/2)-a)^7

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maxima [A]  time = 0.42, size = 334, normalized size = 1.53 \[ -\frac {\frac {15 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a^{3}} - \frac {10 \, {\left ({\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3} - 9 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} b \cos \left (f x + e\right )\right )}}{a^{4}} + \frac {3 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{5} - 20 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} b \cos \left (f x + e\right )^{3} + 90 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} b^{2} \cos \left (f x + e\right )}{a^{5}} + \frac {5 \, {\left (6 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} b \cos \left (f x + e\right )^{2} - b^{2}\right )}}{{\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} a^{3} \cos \left (f x + e\right )^{3}} + \frac {10 \, {\left (9 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} b^{2} \cos \left (f x + e\right )^{2} - b^{3}\right )}}{{\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} a^{4} \cos \left (f x + e\right )^{3}} + \frac {5 \, {\left (12 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} b^{3} \cos \left (f x + e\right )^{2} - b^{4}\right )}}{{\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} a^{5} \cos \left (f x + e\right )^{3}}}{15 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5/(a+b*sec(f*x+e)^2)^(5/2),x, algorithm="maxima")

[Out]

-1/15*(15*sqrt(a + b/cos(f*x + e)^2)*cos(f*x + e)/a^3 - 10*((a + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3 - 9*sq
rt(a + b/cos(f*x + e)^2)*b*cos(f*x + e))/a^4 + (3*(a + b/cos(f*x + e)^2)^(5/2)*cos(f*x + e)^5 - 20*(a + b/cos(
f*x + e)^2)^(3/2)*b*cos(f*x + e)^3 + 90*sqrt(a + b/cos(f*x + e)^2)*b^2*cos(f*x + e))/a^5 + 5*(6*(a + b/cos(f*x
+ e)^2)*b*cos(f*x + e)^2 - b^2)/((a + b/cos(f*x + e)^2)^(3/2)*a^3*cos(f*x + e)^3) + 10*(9*(a + b/cos(f*x + e)
^2)*b^2*cos(f*x + e)^2 - b^3)/((a + b/cos(f*x + e)^2)^(3/2)*a^4*cos(f*x + e)^3) + 5*(12*(a + b/cos(f*x + e)^2)
*b^3*cos(f*x + e)^2 - b^4)/((a + b/cos(f*x + e)^2)^(3/2)*a^5*cos(f*x + e)^3))/f

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^5/(a + b/cos(e + f*x)^2)^(5/2),x)

[Out]

int(sin(e + f*x)^5/(a + b/cos(e + f*x)^2)^(5/2), 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(sin(f*x+e)**5/(a+b*sec(f*x+e)**2)**(5/2),x)

[Out]

Timed out

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