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

Optimal. Leaf size=218 \[ -\frac{8 a^2-24 a b+3 b^2}{8 f (a+b)^4 \sqrt{a+b \sin ^2(e+f x)}}-\frac{8 a^2-24 a b+3 b^2}{24 f (a+b)^3 \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\left (8 a^2-24 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{8 f (a+b)^{9/2}}+\frac{\sec ^4(e+f x)}{4 f (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{(8 a+b) \sec ^2(e+f x)}{8 f (a+b)^2 \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

[Out]

((8*a^2 - 24*a*b + 3*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(8*(a + b)^(9/2)*f) - (8*a^2 - 24*a
*b + 3*b^2)/(24*(a + b)^3*f*(a + b*Sin[e + f*x]^2)^(3/2)) - ((8*a + b)*Sec[e + f*x]^2)/(8*(a + b)^2*f*(a + b*S
in[e + f*x]^2)^(3/2)) + Sec[e + f*x]^4/(4*(a + b)*f*(a + b*Sin[e + f*x]^2)^(3/2)) - (8*a^2 - 24*a*b + 3*b^2)/(
8*(a + b)^4*f*Sqrt[a + b*Sin[e + f*x]^2])

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Rubi [A]  time = 0.27721, antiderivative size = 218, normalized size of antiderivative = 1., number of steps used = 7, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3194, 89, 78, 51, 63, 208} \[ -\frac{8 a^2-24 a b+3 b^2}{8 f (a+b)^4 \sqrt{a+b \sin ^2(e+f x)}}-\frac{8 a^2-24 a b+3 b^2}{24 f (a+b)^3 \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\left (8 a^2-24 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{8 f (a+b)^{9/2}}+\frac{\sec ^4(e+f x)}{4 f (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{(8 a+b) \sec ^2(e+f x)}{8 f (a+b)^2 \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((8*a^2 - 24*a*b + 3*b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(8*(a + b)^(9/2)*f) - (8*a^2 - 24*a
*b + 3*b^2)/(24*(a + b)^3*f*(a + b*Sin[e + f*x]^2)^(3/2)) - ((8*a + b)*Sec[e + f*x]^2)/(8*(a + b)^2*f*(a + b*S
in[e + f*x]^2)^(3/2)) + Sec[e + f*x]^4/(4*(a + b)*f*(a + b*Sin[e + f*x]^2)^(3/2)) - (8*a^2 - 24*a*b + 3*b^2)/(
8*(a + b)^4*f*Sqrt[a + b*Sin[e + f*x]^2])

Rule 3194

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.)*tan[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x]^2, x]}, Dist[ff^((m + 1)/2)/(2*f), Subst[Int[(x^((m - 1)/2)*(a + b*ff*x)^p)/(1 - ff*x)^((
m + 1)/2), x], x, Sin[e + f*x]^2/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rule 89

Int[((a_.) + (b_.)*(x_))^2*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> Simp[((b*c - a*
d)^2*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(d^2*(d*e - c*f)*(n + 1)), x] - Dist[1/(d^2*(d*e - c*f)*(n + 1)), In
t[(c + d*x)^(n + 1)*(e + f*x)^p*Simp[a^2*d^2*f*(n + p + 2) + b^2*c*(d*e*(n + 1) + c*f*(p + 1)) - 2*a*b*d*(d*e*
(n + 1) + c*f*(p + 1)) - b^2*d*(d*e - c*f)*(n + 1)*x, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && (LtQ
[n, -1] || (EqQ[n + p + 3, 0] && NeQ[n, -1] && (SumSimplerQ[n, 1] ||  !SumSimplerQ[p, 1])))

Rule 78

Int[((a_.) + (b_.)*(x_))*((c_.) + (d_.)*(x_))^(n_.)*((e_.) + (f_.)*(x_))^(p_.), x_Symbol] :> -Simp[((b*e - a*f
)*(c + d*x)^(n + 1)*(e + f*x)^(p + 1))/(f*(p + 1)*(c*f - d*e)), x] - Dist[(a*d*f*(n + p + 2) - b*(d*e*(n + 1)
+ c*f*(p + 1)))/(f*(p + 1)*(c*f - d*e)), Int[(c + d*x)^n*(e + f*x)^(p + 1), x], x] /; FreeQ[{a, b, c, d, e, f,
n}, x] && LtQ[p, -1] && ( !LtQ[n, -1] || IntegerQ[p] ||  !(IntegerQ[n] ||  !(EqQ[e, 0] ||  !(EqQ[c, 0] || LtQ
[p, n]))))

Rule 51

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

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int \frac{\tan ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{(1-x)^3 (a+b x)^{5/2}} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=\frac{\sec ^4(e+f x)}{4 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} (4 a-3 b)+2 (a+b) x}{(1-x)^2 (a+b x)^{5/2}} \, dx,x,\sin ^2(e+f x)\right )}{4 (a+b) f}\\ &=-\frac{(8 a+b) \sec ^2(e+f x)}{8 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\sec ^4(e+f x)}{4 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\left (8 a^2-24 a b+3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{(1-x) (a+b x)^{5/2}} \, dx,x,\sin ^2(e+f x)\right )}{16 (a+b)^2 f}\\ &=-\frac{8 a^2-24 a b+3 b^2}{24 (a+b)^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{(8 a+b) \sec ^2(e+f x)}{8 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\sec ^4(e+f x)}{4 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\left (8 a^2-24 a b+3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{(1-x) (a+b x)^{3/2}} \, dx,x,\sin ^2(e+f x)\right )}{16 (a+b)^3 f}\\ &=-\frac{8 a^2-24 a b+3 b^2}{24 (a+b)^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{(8 a+b) \sec ^2(e+f x)}{8 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\sec ^4(e+f x)}{4 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{8 a^2-24 a b+3 b^2}{8 (a+b)^4 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\left (8 a^2-24 a b+3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{(1-x) \sqrt{a+b x}} \, dx,x,\sin ^2(e+f x)\right )}{16 (a+b)^4 f}\\ &=-\frac{8 a^2-24 a b+3 b^2}{24 (a+b)^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{(8 a+b) \sec ^2(e+f x)}{8 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\sec ^4(e+f x)}{4 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{8 a^2-24 a b+3 b^2}{8 (a+b)^4 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\left (8 a^2-24 a b+3 b^2\right ) \operatorname{Subst}\left (\int \frac{1}{1+\frac{a}{b}-\frac{x^2}{b}} \, dx,x,\sqrt{a+b \sin ^2(e+f x)}\right )}{8 b (a+b)^4 f}\\ &=\frac{\left (8 a^2-24 a b+3 b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{8 (a+b)^{9/2} f}-\frac{8 a^2-24 a b+3 b^2}{24 (a+b)^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{(8 a+b) \sec ^2(e+f x)}{8 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac{\sec ^4(e+f x)}{4 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac{8 a^2-24 a b+3 b^2}{8 (a+b)^4 f \sqrt{a+b \sin ^2(e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.484806, size = 107, normalized size = 0.49 \[ \frac{\left (-8 a^2+24 a b-3 b^2\right ) \, _2F_1\left (-\frac{3}{2},1;-\frac{1}{2};\frac{b \sin ^2(e+f x)+a}{a+b}\right )-\frac{3}{2} (a+b) \sec ^4(e+f x) ((8 a+b) \cos (2 (e+f x))+4 a-3 b)}{24 f (a+b)^3 \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((-8*a^2 + 24*a*b - 3*b^2)*Hypergeometric2F1[-3/2, 1, -1/2, (a + b*Sin[e + f*x]^2)/(a + b)] - (3*(a + b)*(4*a
- 3*b + (8*a + b)*Cos[2*(e + f*x)])*Sec[e + f*x]^4)/2)/(24*(a + b)^3*f*(a + b*Sin[e + f*x]^2)^(3/2))

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Maple [B]  time = 7.612, size = 2139, normalized size = 9.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/12/f*b^2*a/(b+(-a*b)^(1/2))^3/(-b+(-a*b)^(1/2))^3/(sin(f*x+e)+(-a*b)^(1/2)/b)^2*(-b*cos(f*x+e)^2+(a*b+b^2)/
b)^(1/2)+1/12/f*b^2*(-a*b)^(1/2)/(b+(-a*b)^(1/2))^3/(-b+(-a*b)^(1/2))^3/(sin(f*x+e)+(-a*b)^(1/2)/b)*(-b*cos(f*
x+e)^2+(a*b+b^2)/b)^(1/2)+1/16/f*b^2/(b+(-a*b)^(1/2))^2/(-b+(-a*b)^(1/2))^2/(a+b)/(-1+sin(f*x+e))^2*(a+b-b*cos
(f*x+e)^2)^(1/2)-3/16/f*b^3/(b+(-a*b)^(1/2))^2/(-b+(-a*b)^(1/2))^2/(a+b)^2/(-1+sin(f*x+e))*(a+b-b*cos(f*x+e)^2
)^(1/2)+1/16/f*b^4/(b+(-a*b)^(1/2))^3/(-b+(-a*b)^(1/2))^3/(a+b)/(-1+sin(f*x+e))*(a+b-b*cos(f*x+e)^2)^(1/2)-1/f
*b^5*a/(b+(-a*b)^(1/2))^4/(-b+(-a*b)^(1/2))^4/(a+b)^(1/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+2*b*sin
(f*x+e)+2*a)/(-1+sin(f*x+e)))+1/2/f*b^4*a^2/(b+(-a*b)^(1/2))^4/(-b+(-a*b)^(1/2))^4/(a+b)^(1/2)*ln((2*(a+b)^(1/
2)*(a+b-b*cos(f*x+e)^2)^(1/2)+2*b*sin(f*x+e)+2*a)/(-1+sin(f*x+e)))+1/16/f*b^2/(b+(-a*b)^(1/2))^2/(-b+(-a*b)^(1
/2))^2/(a+b)/(1+sin(f*x+e))^2*(a+b-b*cos(f*x+e)^2)^(1/2)-1/12/f*b^2*a/(b+(-a*b)^(1/2))^3/(-b+(-a*b)^(1/2))^3/(
sin(f*x+e)-(-a*b)^(1/2)/b)^2*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)-1/12/f*b^2*(-a*b)^(1/2)/(b+(-a*b)^(1/2))^3/(-
b+(-a*b)^(1/2))^3/(sin(f*x+e)-(-a*b)^(1/2)/b)*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)-1/16/f*b^4/(b+(-a*b)^(1/2))^
3/(-b+(-a*b)^(1/2))^3/(a+b)/(1+sin(f*x+e))*(a+b-b*cos(f*x+e)^2)^(1/2)+7/16/f*b^4/(b+(-a*b)^(1/2))^3/(-b+(-a*b)
^(1/2))^3*a/(a+b)^(3/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-2*b*sin(f*x+e)+2*a)/(1+sin(f*x+e)))+7/16/
f*b^4/(b+(-a*b)^(1/2))^3/(-b+(-a*b)^(1/2))^3*a/(a+b)^(3/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+2*b*si
n(f*x+e)+2*a)/(-1+sin(f*x+e)))+1/2/f*b^4*a^2/(b+(-a*b)^(1/2))^4/(-b+(-a*b)^(1/2))^4/(a+b)^(1/2)*ln((2*(a+b)^(1
/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-2*b*sin(f*x+e)+2*a)/(1+sin(f*x+e)))-1/f*b^5*a/(b+(-a*b)^(1/2))^4/(-b+(-a*b)^(1/
2))^4/(a+b)^(1/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-2*b*sin(f*x+e)+2*a)/(1+sin(f*x+e)))+3/16/f*b^3/
(b+(-a*b)^(1/2))^2/(-b+(-a*b)^(1/2))^2/(a+b)^2/(1+sin(f*x+e))*(a+b-b*cos(f*x+e)^2)^(1/2)+1/2/f*b^4*a^2/(b+(-a*
b)^(1/2))^4/(-b+(-a*b)^(1/2))^4/(-a*b)^(1/2)/(sin(f*x+e)+(-a*b)^(1/2)/b)*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)-1
/2/f*b^4*a^2/(b+(-a*b)^(1/2))^4/(-b+(-a*b)^(1/2))^4/(-a*b)^(1/2)/(sin(f*x+e)-(-a*b)^(1/2)/b)*(-b*cos(f*x+e)^2+
(a*b+b^2)/b)^(1/2)+1/f*b^5*a/(b+(-a*b)^(1/2))^4/(-b+(-a*b)^(1/2))^4/(-a*b)^(1/2)/(sin(f*x+e)-(-a*b)^(1/2)/b)*(
-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)-7/16/f*b^3/(b+(-a*b)^(1/2))^3/(-b+(-a*b)^(1/2))^3*a/(a+b)/(-1+sin(f*x+e))*(
a+b-b*cos(f*x+e)^2)^(1/2)-1/f*b^5*a/(b+(-a*b)^(1/2))^4/(-b+(-a*b)^(1/2))^4/(-a*b)^(1/2)/(sin(f*x+e)+(-a*b)^(1/
2)/b)*(-b*cos(f*x+e)^2+(a*b+b^2)/b)^(1/2)+7/16/f*b^3/(b+(-a*b)^(1/2))^3/(-b+(-a*b)^(1/2))^3*a/(a+b)/(1+sin(f*x
+e))*(a+b-b*cos(f*x+e)^2)^(1/2)+3/16/f*b^4/(b+(-a*b)^(1/2))^2/(-b+(-a*b)^(1/2))^2/(a+b)^(5/2)*ln((2*(a+b)^(1/2
)*(a+b-b*cos(f*x+e)^2)^(1/2)+2*b*sin(f*x+e)+2*a)/(-1+sin(f*x+e)))-1/16/f*b^3/(b+(-a*b)^(1/2))^2/(-b+(-a*b)^(1/
2))^2/(a+b)^(3/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+2*b*sin(f*x+e)+2*a)/(-1+sin(f*x+e)))-1/16/f*b^3
/(b+(-a*b)^(1/2))^2/(-b+(-a*b)^(1/2))^2/(a+b)^(3/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-2*b*sin(f*x+e
)+2*a)/(1+sin(f*x+e)))-1/16/f*b^5/(b+(-a*b)^(1/2))^3/(-b+(-a*b)^(1/2))^3/(a+b)^(3/2)*ln((2*(a+b)^(1/2)*(a+b-b*
cos(f*x+e)^2)^(1/2)+2*b*sin(f*x+e)+2*a)/(-1+sin(f*x+e)))+3/16/f*b^4/(b+(-a*b)^(1/2))^2/(-b+(-a*b)^(1/2))^2/(a+
b)^(5/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-2*b*sin(f*x+e)+2*a)/(1+sin(f*x+e)))-1/16/f*b^5/(b+(-a*b)
^(1/2))^3/(-b+(-a*b)^(1/2))^3/(a+b)^(3/2)*ln((2*(a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-2*b*sin(f*x+e)+2*a)/(1+
sin(f*x+e)))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [B]  time = 4.63943, size = 2287, normalized size = 10.49 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/48*(3*((8*a^2*b^2 - 24*a*b^3 + 3*b^4)*cos(f*x + e)^8 - 2*(8*a^3*b - 16*a^2*b^2 - 21*a*b^3 + 3*b^4)*cos(f*x
+ e)^6 + (8*a^4 - 8*a^3*b - 37*a^2*b^2 - 18*a*b^3 + 3*b^4)*cos(f*x + e)^4)*sqrt(a + b)*log((b*cos(f*x + e)^2 -
2*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a + b) - 2*a - 2*b)/cos(f*x + e)^2) + 2*(3*(8*a^3*b - 16*a^2*b^2 - 21*
a*b^3 + 3*b^4)*cos(f*x + e)^6 - 4*(8*a^4 - 8*a^3*b - 37*a^2*b^2 - 18*a*b^3 + 3*b^4)*cos(f*x + e)^4 + 6*a^4 + 2
4*a^3*b + 36*a^2*b^2 + 24*a*b^3 + 6*b^4 - 3*(8*a^4 + 25*a^3*b + 27*a^2*b^2 + 11*a*b^3 + b^4)*cos(f*x + e)^2)*s
qrt(-b*cos(f*x + e)^2 + a + b))/((a^5*b^2 + 5*a^4*b^3 + 10*a^3*b^4 + 10*a^2*b^5 + 5*a*b^6 + b^7)*f*cos(f*x + e
)^8 - 2*(a^6*b + 6*a^5*b^2 + 15*a^4*b^3 + 20*a^3*b^4 + 15*a^2*b^5 + 6*a*b^6 + b^7)*f*cos(f*x + e)^6 + (a^7 + 7
*a^6*b + 21*a^5*b^2 + 35*a^4*b^3 + 35*a^3*b^4 + 21*a^2*b^5 + 7*a*b^6 + b^7)*f*cos(f*x + e)^4), -1/24*(3*((8*a^
2*b^2 - 24*a*b^3 + 3*b^4)*cos(f*x + e)^8 - 2*(8*a^3*b - 16*a^2*b^2 - 21*a*b^3 + 3*b^4)*cos(f*x + e)^6 + (8*a^4
- 8*a^3*b - 37*a^2*b^2 - 18*a*b^3 + 3*b^4)*cos(f*x + e)^4)*sqrt(-a - b)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b
)*sqrt(-a - b)/(a + b)) - (3*(8*a^3*b - 16*a^2*b^2 - 21*a*b^3 + 3*b^4)*cos(f*x + e)^6 - 4*(8*a^4 - 8*a^3*b - 3
7*a^2*b^2 - 18*a*b^3 + 3*b^4)*cos(f*x + e)^4 + 6*a^4 + 24*a^3*b + 36*a^2*b^2 + 24*a*b^3 + 6*b^4 - 3*(8*a^4 + 2
5*a^3*b + 27*a^2*b^2 + 11*a*b^3 + b^4)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^5*b^2 + 5*a^4*b^3
+ 10*a^3*b^4 + 10*a^2*b^5 + 5*a*b^6 + b^7)*f*cos(f*x + e)^8 - 2*(a^6*b + 6*a^5*b^2 + 15*a^4*b^3 + 20*a^3*b^4 +
15*a^2*b^5 + 6*a*b^6 + b^7)*f*cos(f*x + e)^6 + (a^7 + 7*a^6*b + 21*a^5*b^2 + 35*a^4*b^3 + 35*a^3*b^4 + 21*a^2
*b^5 + 7*a*b^6 + b^7)*f*cos(f*x + e)^4)]

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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(tan(f*x+e)**5/(a+b*sin(f*x+e)**2)**(5/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan \left (f x + e\right )^{5}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(tan(f*x + e)^5/(b*sin(f*x + e)^2 + a)^(5/2), x)