3.521 \(\int \frac{\tan ^5(e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\)

Optimal. Leaf size=177 \[ -\frac{8 a^2-8 a b-b^2}{8 f (a+b)^3 \sqrt{a+b \sin ^2(e+f x)}}+\frac{\left (8 a^2-8 a b-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{8 f (a+b)^{7/2}}+\frac{\sec ^4(e+f x)}{4 f (a+b) \sqrt{a+b \sin ^2(e+f x)}}-\frac{(8 a+3 b) \sec ^2(e+f x)}{8 f (a+b)^2 \sqrt{a+b \sin ^2(e+f x)}} \]

[Out]

((8*a^2 - 8*a*b - b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(8*(a + b)^(7/2)*f) - (8*a^2 - 8*a*b -
b^2)/(8*(a + b)^3*f*Sqrt[a + b*Sin[e + f*x]^2]) - ((8*a + 3*b)*Sec[e + f*x]^2)/(8*(a + b)^2*f*Sqrt[a + b*Sin[
e + f*x]^2]) + Sec[e + f*x]^4/(4*(a + b)*f*Sqrt[a + b*Sin[e + f*x]^2])

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Rubi [A]  time = 0.229394, antiderivative size = 177, normalized size of antiderivative = 1., number of steps used = 6, 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} \[ -\frac{8 a^2-8 a b-b^2}{8 f (a+b)^3 \sqrt{a+b \sin ^2(e+f x)}}+\frac{\left (8 a^2-8 a b-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{8 f (a+b)^{7/2}}+\frac{\sec ^4(e+f x)}{4 f (a+b) \sqrt{a+b \sin ^2(e+f x)}}-\frac{(8 a+3 b) \sec ^2(e+f x)}{8 f (a+b)^2 \sqrt{a+b \sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Tan[e + f*x]^5/(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

((8*a^2 - 8*a*b - b^2)*ArcTanh[Sqrt[a + b*Sin[e + f*x]^2]/Sqrt[a + b]])/(8*(a + b)^(7/2)*f) - (8*a^2 - 8*a*b -
b^2)/(8*(a + b)^3*f*Sqrt[a + b*Sin[e + f*x]^2]) - ((8*a + 3*b)*Sec[e + f*x]^2)/(8*(a + b)^2*f*Sqrt[a + b*Sin[
e + f*x]^2]) + Sec[e + f*x]^4/(4*(a + b)*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 )^{3/2}} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{x^2}{(1-x)^3 (a+b x)^{3/2}} \, dx,x,\sin ^2(e+f x)\right )}{2 f}\\ &=\frac{\sec ^4(e+f x)}{4 (a+b) f \sqrt{a+b \sin ^2(e+f x)}}-\frac{\operatorname{Subst}\left (\int \frac{\frac{1}{2} (4 a-b)+2 (a+b) x}{(1-x)^2 (a+b x)^{3/2}} \, dx,x,\sin ^2(e+f x)\right )}{4 (a+b) f}\\ &=-\frac{(8 a+3 b) \sec ^2(e+f x)}{8 (a+b)^2 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\sec ^4(e+f x)}{4 (a+b) f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\left (8 a^2-8 a b-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)^2 f}\\ &=-\frac{8 a^2-8 a b-b^2}{8 (a+b)^3 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{(8 a+3 b) \sec ^2(e+f x)}{8 (a+b)^2 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\sec ^4(e+f x)}{4 (a+b) f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\left (8 a^2-8 a b-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)^3 f}\\ &=-\frac{8 a^2-8 a b-b^2}{8 (a+b)^3 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{(8 a+3 b) \sec ^2(e+f x)}{8 (a+b)^2 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\sec ^4(e+f x)}{4 (a+b) f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\left (8 a^2-8 a b-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)^3 f}\\ &=\frac{\left (8 a^2-8 a b-b^2\right ) \tanh ^{-1}\left (\frac{\sqrt{a+b \sin ^2(e+f x)}}{\sqrt{a+b}}\right )}{8 (a+b)^{7/2} f}-\frac{8 a^2-8 a b-b^2}{8 (a+b)^3 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{(8 a+3 b) \sec ^2(e+f x)}{8 (a+b)^2 f \sqrt{a+b \sin ^2(e+f x)}}+\frac{\sec ^4(e+f x)}{4 (a+b) f \sqrt{a+b \sin ^2(e+f x)}}\\ \end{align*}

Mathematica [C]  time = 0.486389, size = 107, normalized size = 0.6 \[ \frac{\left (-8 a^2+8 a b+b^2\right ) \, _2F_1\left (-\frac{1}{2},1;\frac{1}{2};\frac{b \sin ^2(e+f x)+a}{a+b}\right )-\frac{1}{2} (a+b) \sec ^4(e+f x) ((8 a+3 b) \cos (2 (e+f x))+4 a-b)}{8 f (a+b)^3 \sqrt{a+b \sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Tan[e + f*x]^5/(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

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

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Maple [B]  time = 12.073, size = 3763, normalized size = 21.3 \begin{align*} \text{output 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)^(3/2),x)

[Out]

-1/16/(a^4*b^2*cos(f*x+e)^4+4*a^3*b^3*cos(f*x+e)^4+6*a^2*b^4*cos(f*x+e)^4+4*a*b^5*cos(f*x+e)^4+b^6*cos(f*x+e)^
4-2*a^5*b*cos(f*x+e)^2-10*a^4*b^2*cos(f*x+e)^2-20*a^3*b^3*cos(f*x+e)^2-20*a^2*b^4*cos(f*x+e)^2-10*a*b^5*cos(f*
x+e)^2-2*b^6*cos(f*x+e)^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)/cos(f*x+e)^4/(a+b)^(3/2)*(
20*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^3*b^3*cos(f*x+e)^4+30*ln(2/(
1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^2*b^4*cos(f*x+e)^4+12*ln(2/(1+sin(f*x
+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a*b^5*cos(f*x+e)^4-8*ln(2/(-1+sin(f*x+e))*((a+b)
^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^4*b^2*cos(f*x+e)^8-8*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a
+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^3*b^3*cos(f*x+e)^8+9*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(
f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^2*b^4*cos(f*x+e)^8+10*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2
)^(1/2)+b*sin(f*x+e)+a))*a*b^5*cos(f*x+e)^8-8*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*si
n(f*x+e)+a))*a^4*b^2*cos(f*x+e)^8-8*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a
))*a^3*b^3*cos(f*x+e)^8+9*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^2*b^4
*cos(f*x+e)^8+10*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a*b^5*cos(f*x+e)
^8+16*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^5*b*cos(f*x+e)^6+32*ln(2
/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^4*b^2*cos(f*x+e)^6-2*ln(2/(-1+sin(
f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^3*b^3*cos(f*x+e)^6-38*ln(2/(-1+sin(f*x+e))*
((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^2*b^4*cos(f*x+e)^6-22*ln(2/(-1+sin(f*x+e))*((a+b)^(
1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a*b^5*cos(f*x+e)^6+16*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b
*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^5*b*cos(f*x+e)^6+16*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(3/2)*a^4*cos(f*x
+e)^4-2*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(3/2)*b^4*cos(f*x+e)^4-8*(a+b)^(3/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2
)^(1/2)*a^4*cos(f*x+e)^4+16*(a+b-b*cos(f*x+e)^2)^(3/2)*(a+b)^(3/2)*a^3*cos(f*x+e)^2-32*(a+b-b*cos(f*x+e)^2)^(1
/2)*(a+b)^(3/2)*a^3*b*cos(f*x+e)^6+36*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(3/2)*a*b^3*cos(f*x+e)^6+16*(a+b)^(3/2)
*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a^3*b*cos(f*x+e)^6-64*(a+b)^(3/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(
1/2)*a^2*b^2*cos(f*x+e)^6-80*(a+b)^(3/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a*b^3*cos(f*x+e)^6-32*(a+b-b*
cos(f*x+e)^2)^(3/2)*(a+b)^(3/2)*a^2*b*cos(f*x+e)^4-32*(a+b-b*cos(f*x+e)^2)^(3/2)*(a+b)^(3/2)*a*b^2*cos(f*x+e)^
4-4*(a+b-b*cos(f*x+e)^2)^(3/2)*(a+b)^(3/2)*a^3-4*(a+b-b*cos(f*x+e)^2)^(3/2)*(a+b)^(3/2)*b^3+ln(2/(-1+sin(f*x+e
))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*b^6*cos(f*x+e)^8+ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*
(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*b^6*cos(f*x+e)^8-2*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*
x+e)^2)^(1/2)+b*sin(f*x+e)+a))*b^6*cos(f*x+e)^6-2*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-
b*sin(f*x+e)+a))*b^6*cos(f*x+e)^6-8*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+
a))*a^6*cos(f*x+e)^4+ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*b^6*cos(f*x
+e)^4-8*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^6*cos(f*x+e)^4+ln(2/(1+
sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*b^6*cos(f*x+e)^4+6*(a+b-b*cos(f*x+e)^2)^(
3/2)*(a+b)^(3/2)*b^3*cos(f*x+e)^2-12*(a+b-b*cos(f*x+e)^2)^(3/2)*(a+b)^(3/2)*a^2*b-12*(a+b-b*cos(f*x+e)^2)^(3/2
)*(a+b)^(3/2)*a*b^2-2*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(3/2)*b^4*cos(f*x+e)^8-2*(a+b-b*cos(f*x+e)^2)^(3/2)*(a+
b)^(3/2)*b^3*cos(f*x+e)^6+4*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(3/2)*b^4*cos(f*x+e)^6+8*(a+b)^(3/2)*(-b*cos(f*x+
e)^2+(a*b^2+b^3)/b^2)^(3/2)*a^3*cos(f*x+e)^4+32*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*
sin(f*x+e)+a))*a^4*b^2*cos(f*x+e)^6-2*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)
+a))*a^3*b^3*cos(f*x+e)^6-38*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^2*
b^4*cos(f*x+e)^6-22*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a*b^5*cos(f*x
+e)^6-24*ln(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^5*b*cos(f*x+e)^4-15*l
n(2/(-1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^4*b^2*cos(f*x+e)^4+20*ln(2/(-1+
sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^3*b^3*cos(f*x+e)^4+30*ln(2/(-1+sin(f*x+
e))*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^2*b^4*cos(f*x+e)^4+12*ln(2/(-1+sin(f*x+e))*((a+
b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a*b^5*cos(f*x+e)^4-24*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a
+b-b*cos(f*x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^5*b*cos(f*x+e)^4-15*ln(2/(1+sin(f*x+e))*((a+b)^(1/2)*(a+b-b*cos(f*
x+e)^2)^(1/2)-b*sin(f*x+e)+a))*a^4*b^2*cos(f*x+e)^4-8*(a+b)^(3/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(3/2)*a*b^
2*cos(f*x+e)^4+16*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(3/2)*a^3*b*cos(f*x+e)^4-18*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b
)^(3/2)*a^2*b^2*cos(f*x+e)^4-20*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(3/2)*a*b^3*cos(f*x+e)^4+24*(a+b)^(3/2)*(-b*c
os(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a^3*b*cos(f*x+e)^4+72*(a+b)^(3/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a
^2*b^2*cos(f*x+e)^4+40*(a+b)^(3/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a*b^3*cos(f*x+e)^4+38*(a+b-b*cos(f*
x+e)^2)^(3/2)*(a+b)^(3/2)*a^2*b*cos(f*x+e)^2+28*(a+b-b*cos(f*x+e)^2)^(3/2)*(a+b)^(3/2)*a*b^2*cos(f*x+e)^2+16*(
a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(3/2)*a^2*b^2*cos(f*x+e)^8-16*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(3/2)*a*b^3*cos
(f*x+e)^8-8*(a+b)^(3/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a^2*b^2*cos(f*x+e)^8+40*(a+b)^(3/2)*(-b*cos(f*
x+e)^2+(a*b^2+b^3)/b^2)^(1/2)*a*b^3*cos(f*x+e)^8+16*(a+b-b*cos(f*x+e)^2)^(3/2)*(a+b)^(3/2)*a*b^2*cos(f*x+e)^6+
8*(a+b)^(3/2)*(-b*cos(f*x+e)^2+(a*b^2+b^3)/b^2)^(3/2)*a^2*b*cos(f*x+e)^6+8*(a+b)^(3/2)*(-b*cos(f*x+e)^2+(a*b^2
+b^3)/b^2)^(3/2)*a*b^2*cos(f*x+e)^6)/f

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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)^(3/2),x, algorithm="maxima")

[Out]

Exception raised: ValueError

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Fricas [A]  time = 4.16715, size = 1345, normalized size = 7.6 \begin{align*} \left [-\frac{{\left ({\left (8 \, a^{2} b - 8 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{6} -{\left (8 \, a^{3} - 9 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{4}\right )} \sqrt{a + b} \log \left (\frac{b \cos \left (f x + e\right )^{2} + 2 \, \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \,{\left ({\left (8 \, a^{3} - 9 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{4} - 2 \, a^{3} - 6 \, a^{2} b - 6 \, a b^{2} - 2 \, b^{3} +{\left (8 \, a^{3} + 19 \, a^{2} b + 14 \, a b^{2} + 3 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{16 \,{\left ({\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{6} -{\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{4}\right )}}, -\frac{{\left ({\left (8 \, a^{2} b - 8 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{6} -{\left (8 \, a^{3} - 9 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{4}\right )} \sqrt{-a - b} \arctan \left (\frac{\sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sqrt{-a - b}}{a + b}\right ) -{\left ({\left (8 \, a^{3} - 9 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{4} - 2 \, a^{3} - 6 \, a^{2} b - 6 \, a b^{2} - 2 \, b^{3} +{\left (8 \, a^{3} + 19 \, a^{2} b + 14 \, a b^{2} + 3 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b}}{8 \,{\left ({\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{6} -{\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{4}\right )}}\right ] \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)^(3/2),x, algorithm="fricas")

[Out]

[-1/16*(((8*a^2*b - 8*a*b^2 - b^3)*cos(f*x + e)^6 - (8*a^3 - 9*a*b^2 - b^3)*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*((8*a^3 - 9*a
*b^2 - b^3)*cos(f*x + e)^4 - 2*a^3 - 6*a^2*b - 6*a*b^2 - 2*b^3 + (8*a^3 + 19*a^2*b + 14*a*b^2 + 3*b^3)*cos(f*x
+ e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^4*b + 4*a^3*b^2 + 6*a^2*b^3 + 4*a*b^4 + b^5)*f*cos(f*x + e)^6 -
(a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*f*cos(f*x + e)^4), -1/8*(((8*a^2*b - 8*a*b^2 - b^3)*
cos(f*x + e)^6 - (8*a^3 - 9*a*b^2 - b^3)*cos(f*x + e)^4)*sqrt(-a - b)*arctan(sqrt(-b*cos(f*x + e)^2 + a + b)*s
qrt(-a - b)/(a + b)) - ((8*a^3 - 9*a*b^2 - b^3)*cos(f*x + e)^4 - 2*a^3 - 6*a^2*b - 6*a*b^2 - 2*b^3 + (8*a^3 +
19*a^2*b + 14*a*b^2 + 3*b^3)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b))/((a^4*b + 4*a^3*b^2 + 6*a^2*b^3
+ 4*a*b^4 + b^5)*f*cos(f*x + e)^6 - (a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*f*cos(f*x + e)^4
)]

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\tan ^{5}{\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac{3}{2}}}\, dx \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)**(3/2),x)

[Out]

Integral(tan(e + f*x)**5/(a + b*sin(e + f*x)**2)**(3/2), x)

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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{3}{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)^(3/2),x, algorithm="giac")

[Out]

integrate(tan(f*x + e)^5/(b*sin(f*x + e)^2 + a)^(3/2), x)