\(\int \frac {\tan ^5(e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\) [532]

Optimal result

Integrand size = 25, antiderivative size = 218 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\left (8 a^2-24 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{8 (a+b)^{9/2} f}-\frac {8 a^2-24 a b+3 b^2}{24 (a+b)^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {(8 a+b) \sec ^2(e+f x)}{8 (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\sec ^4(e+f x)}{4 (a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {8 a^2-24 a b+3 b^2}{8 (a+b)^4 f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

1/8*(8*a^2-24*a*b+3*b^2)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(9/2)/f+1/24*(-8*a^2+24*a*b-3*b^2
)/(a+b)^3/f/(a+b*sin(f*x+e)^2)^(3/2)-1/8*(8*a+b)*sec(f*x+e)^2/(a+b)^2/f/(a+b*sin(f*x+e)^2)^(3/2)+1/4*sec(f*x+e
)^4/(a+b)/f/(a+b*sin(f*x+e)^2)^(3/2)+1/8*(-8*a^2+24*a*b-3*b^2)/(a+b)^4/f/(a+b*sin(f*x+e)^2)^(1/2)

Rubi [A] (verified)

Time = 0.34 (sec) , antiderivative size = 218, normalized size of antiderivative = 1.00, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3273, 91, 79, 53, 65, 214} \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\left (8 a^2-24 a b+3 b^2\right ) \text {arctanh}\left (\frac {\sqrt {a+b \sin ^2(e+f x)}}{\sqrt {a+b}}\right )}{8 f (a+b)^{9/2}}-\frac {8 a^2-24 a b+3 b^2}{8 f (a+b)^4 \sqrt {a+b \sin ^2(e+f x)}}-\frac {8 a^2-24 a b+3 b^2}{24 f (a+b)^3 \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\sec ^4(e+f x)}{4 f (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {(8 a+b) \sec ^2(e+f x)}{8 f (a+b)^2 \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

[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 53

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 65

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 79

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] || L
tQ[p, n]))))

Rule 91

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 214

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

Rule 3273

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]

Rubi steps \begin{align*} \text {integral}& = \frac {\text {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 {\text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {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 ) \text {arctanh}\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] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.50 (sec) , antiderivative size = 107, normalized size of antiderivative = 0.49 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\frac {\left (-8 a^2+24 a b-3 b^2\right ) \operatorname {Hypergeometric2F1}\left (-\frac {3}{2},1,-\frac {1}{2},\frac {a+b \sin ^2(e+f x)}{a+b}\right )-\frac {3}{2} (a+b) (4 a-3 b+(8 a+b) \cos (2 (e+f x))) \sec ^4(e+f x)}{24 (a+b)^3 f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

[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))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1908\) vs. \(2(194)=388\).

Time = 4.23 (sec) , antiderivative size = 1909, normalized size of antiderivative = 8.76

method	result	size
default	\(\text {Expression too large to display}\)	\(1909\)

[In]

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

[Out]

(-1/16*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*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)/(sin(f*x+e)-1))+1/16*b^4/(b+(-a*b)^(1/2))^3/(-b+(-a*b)^(1/2))^3/(a+b)/(sin(f*x+e)-1)*(a+b-b*cos(f*x+e)^2)
^(1/2)-1/12*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)+1/16*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*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*b^2/(b
+(-a*b)^(1/2))^2/(-b+(-a*b)^(1/2))^2/(a+b)/(sin(f*x+e)-1)^2*(a+b-b*cos(f*x+e)^2)^(1/2)-3/16*b^3/(b+(-a*b)^(1/2
))^2/(-b+(-a*b)^(1/2))^2/(a+b)^2/(sin(f*x+e)-1)*(a+b-b*cos(f*x+e)^2)^(1/2)+1/12*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)-1/12*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)-1/12*
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)
+1/2*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)/(sin(f*x+e)-1))-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)/(sin(f*x+e)-1))+1/2*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)))-1/2*b^4*a*(a-
2*b)/(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*b^4*a*(a-2*b)/(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)-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)))+7/16*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*b^3/(b+(-
a*b)^(1/2))^3/(-b+(-a*b)^(1/2))^3*a/(a+b)/(sin(f*x+e)-1)*(a+b-b*cos(f*x+e)^2)^(1/2)+7/16*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*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)/(sin(f*x+e)-1))-1/16*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)/(sin(f*x+e)-1))-1/16*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*c
os(f*x+e)^2)^(1/2)+2*b*sin(f*x+e)+2*a)/(sin(f*x+e)-1))-1/16*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*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*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))))/f

Fricas [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 490 vs. \(2 (194) = 388\).

Time = 0.58 (sec) , antiderivative size = 995, normalized size of antiderivative = 4.56 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\text {Too large to display} \]

[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)]

Sympy [F]

\[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int \frac {\tan ^{5}{\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \]

[In]

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

[Out]

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

Maxima [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 424 vs. \(2 (194) = 388\).

Time = 0.38 (sec) , antiderivative size = 424, normalized size of antiderivative = 1.94 \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=-\frac {\frac {3 \, {\left (8 \, a^{2} b^{3} - 24 \, a b^{4} + 3 \, b^{5}\right )} \log \left (\frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} - \sqrt {a + b}}{\sqrt {b \sin \left (f x + e\right )^{2} + a} + \sqrt {a + b}}\right )}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} \sqrt {a + b}} + \frac {2 \, {\left (8 \, a^{5} b^{3} + 24 \, a^{4} b^{4} + 24 \, a^{3} b^{5} + 8 \, a^{2} b^{6} + 3 \, {\left (8 \, a^{2} b^{3} - 24 \, a b^{4} + 3 \, b^{5}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{3} - 5 \, {\left (8 \, a^{3} b^{3} - 16 \, a^{2} b^{4} - 21 \, a b^{5} + 3 \, b^{6}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{2} + 8 \, {\left (a^{4} b^{3} - 4 \, a^{3} b^{4} - 11 \, a^{2} b^{5} - 6 \, a b^{6}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}\right )}}{{\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {7}{2}} - 2 \, {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}} + {\left (a^{6} + 6 \, a^{5} b + 15 \, a^{4} b^{2} + 20 \, a^{3} b^{3} + 15 \, a^{2} b^{4} + 6 \, a b^{5} + b^{6}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}}}}{48 \, b^{3} f} \]

[In]

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

[Out]

-1/48*(3*(8*a^2*b^3 - 24*a*b^4 + 3*b^5)*log((sqrt(b*sin(f*x + e)^2 + a) - sqrt(a + b))/(sqrt(b*sin(f*x + e)^2
+ a) + sqrt(a + b)))/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*sqrt(a + b)) + 2*(8*a^5*b^3 + 24*a^4*b^4 + 2
4*a^3*b^5 + 8*a^2*b^6 + 3*(8*a^2*b^3 - 24*a*b^4 + 3*b^5)*(b*sin(f*x + e)^2 + a)^3 - 5*(8*a^3*b^3 - 16*a^2*b^4
- 21*a*b^5 + 3*b^6)*(b*sin(f*x + e)^2 + a)^2 + 8*(a^4*b^3 - 4*a^3*b^4 - 11*a^2*b^5 - 6*a*b^6)*(b*sin(f*x + e)^
2 + a))/((a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b^4)*(b*sin(f*x + e)^2 + a)^(7/2) - 2*(a^5 + 5*a^4*b + 10*a^3*
b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*(b*sin(f*x + e)^2 + a)^(5/2) + (a^6 + 6*a^5*b + 15*a^4*b^2 + 20*a^3*b^3 + 15
*a^2*b^4 + 6*a*b^5 + b^6)*(b*sin(f*x + e)^2 + a)^(3/2)))/(b^3*f)

Giac [F]

\[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\int { \frac {\tan \left (f x + e\right )^{5}}{{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

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

[Out]

sage0*x

Mupad [F(-1)]

Timed out. \[ \int \frac {\tan ^5(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx=\text {Hanged} \]

[In]

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

[Out]

\text{Hanged}