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]
1/8*(8*a^2-8*a*b-b^2)*arctanh((a+b*sin(f*x+e)^2)^(1/2)/(a+b)^(1/2))/(a+b)^(7/2)/f+1/8*(-8*a^2+8*a*b+b^2)/(a+b)
^3/f/(a+b*sin(f*x+e)^2)^(1/2)-1/8*(8*a+3*b)*sec(f*x+e)^2/(a+b)^2/f/(a+b*sin(f*x+e)^2)^(1/2)+1/4*sec(f*x+e)^4/(
a+b)/f/(a+b*sin(f*x+e)^2)^(1/2)
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Rubi [A] time = 0.23, antiderivative size = 177, 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 =
{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 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 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 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 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]
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]
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*}
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Mathematica [C] time = 0.47, size = 107, normalized size = 0.60 \[ \frac {\left (-8 a^2+8 a b+b^2\right ) \, _2F_1\left (-\frac {1}{2},1;\frac {1}{2};\frac {b \sin ^2(e+f x)+a}{a+b}\right )-\frac {1}{2} (a+b) \sec ^4(e+f x) ((8 a+3 b) \cos (2 (e+f x))+4 a-b)}{8 f (a+b)^3 \sqrt {a+b \sin ^2(e+f x)}} \]
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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fricas [A] time = 0.78, size = 593, normalized size = 3.35 \[ \left [-\frac {{\left ({\left (8 \, a^{2} b - 8 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{6} - {\left (8 \, a^{3} - 9 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{4}\right )} \sqrt {a + b} \log \left (\frac {b \cos \left (f x + e\right )^{2} + 2 \, \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} - 2 \, a - 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \, {\left ({\left (8 \, a^{3} - 9 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{4} - 2 \, a^{3} - 6 \, a^{2} b - 6 \, a b^{2} - 2 \, b^{3} + {\left (8 \, a^{3} + 19 \, a^{2} b + 14 \, a b^{2} + 3 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{16 \, {\left ({\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{6} - {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{4}\right )}}, -\frac {{\left ({\left (8 \, a^{2} b - 8 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{6} - {\left (8 \, a^{3} - 9 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{4}\right )} \sqrt {-a - b} \arctan \left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{a + b}\right ) - {\left ({\left (8 \, a^{3} - 9 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{4} - 2 \, a^{3} - 6 \, a^{2} b - 6 \, a b^{2} - 2 \, b^{3} + {\left (8 \, a^{3} + 19 \, a^{2} b + 14 \, a b^{2} + 3 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b}}{8 \, {\left ({\left (a^{4} b + 4 \, a^{3} b^{2} + 6 \, a^{2} b^{3} + 4 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{6} - {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} f \cos \left (f x + e\right )^{4}\right )}}\right ] \]
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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giac [B] time = 4.03, size = 2776, normalized size = 15.68 \[ \text {result too large to display} \]
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]
-1/4*(4*((a^6*b + 4*a^5*b^2 + 6*a^4*b^3 + 4*a^3*b^4 + a^2*b^5)*tan(1/2*f*x + 1/2*e)^2/(a^7*b + 7*a^6*b^2 + 21*
a^5*b^3 + 35*a^4*b^4 + 35*a^3*b^5 + 21*a^2*b^6 + 7*a*b^7 + b^8) + (a^6*b + 4*a^5*b^2 + 6*a^4*b^3 + 4*a^3*b^4 +
a^2*b^5)/(a^7*b + 7*a^6*b^2 + 21*a^5*b^3 + 35*a^4*b^4 + 35*a^3*b^5 + 21*a^2*b^6 + 7*a*b^7 + b^8))/sqrt(a*tan(
1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a) + (8*a^2 - 8*a*b - b^2)*arct
an(-1/2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan
(1/2*f*x + 1/2*e)^2 + a) - sqrt(a))/sqrt(-a - b))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt(-a - b)) - 2*(8*(sqrt(
a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2
*e)^2 + a))^7*a^2 - (sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)
^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^7*b^2 - 56*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)
^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^6*a^(5/2) - 32*(sqrt(a)*tan(1/2*f*x + 1/2*e
)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^6*a^(3/2)*
b - 25*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(
1/2*f*x + 1/2*e)^2 + a))^6*sqrt(a)*b^2 - 120*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 +
2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^5*a^3 - 352*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - s
qrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^5*a^2*b - 113*(sq
rt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x +
1/2*e)^2 + a))^5*a*b^2 - 28*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x
+ 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^5*b^3 + 136*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x
+ 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^4*a^(7/2) - 64*(sqrt(a)*tan(1/2*f*
x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^4
*a^(5/2)*b - 561*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2
+ 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^4*a^(3/2)*b^2 - 116*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x +
1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^4*sqrt(a)*b^3 + 344*(sqrt(a)*tan(1/2*
f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))
^3*a^4 + 1088*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4
*b*tan(1/2*f*x + 1/2*e)^2 + a))^3*a^3*b + 597*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4
+ 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^3*a^2*b^2 - 504*(sqrt(a)*tan(1/2*f*x + 1/2*e)^
2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^3*a*b^3 - 11
2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f
*x + 1/2*e)^2 + a))^3*b^4 + 24*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f
*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^2*a^(9/2) + 608*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(
1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^2*a^(7/2)*b + 1565*(sqrt(a)
*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e
)^2 + a))^2*a^(5/2)*b^2 + 952*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*
x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^2*a^(3/2)*b^3 - 176*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*t
an(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^2*sqrt(a)*b^4 - 232*(sqr
t(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1
/2*e)^2 + a))*a^5 - 736*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/
2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a^4*b - 483*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1
/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a^3*b^2 + 532*(sqrt(a)*tan(1/2*f*x + 1
/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a^2*b^
3 + 496*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan
(1/2*f*x + 1/2*e)^2 + a))*a*b^4 - 64*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan
(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*b^5 - 104*a^(11/2) - 512*a^(9/2)*b - 979*a^(7/2)*b^2 -
836*a^(5/2)*b^3 - 208*a^(3/2)*b^4 + 64*sqrt(a)*b^5)/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*((sqrt(a)*tan(1/2*f*x + 1
/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^2 - 2*
(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 + 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x
+ 1/2*e)^2 + a))*sqrt(a) - 3*a - 4*b)^4))/f
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maple [B] time = 13.20, size = 3763, normalized size = 21.26 \[ \text {output too large to display} \]
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)*(
-4*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(3/2)*a^3-4*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(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/(sin(f*x+e)-1)*((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/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*s
in(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/(sin(f*x+e)-1)*((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/(sin(f*x
+e)-1)*((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)^4-2*(a+b)^(3/2)*(a+b-b*cos(f*x+
e)^2)^(1/2)*b^4*cos(f*x+e)^8-2*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(3/2)*b^3*cos(f*x+e)^6+4*(a+b)^(3/2)*(a+b-b*co
s(f*x+e)^2)^(1/2)*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)^4-12*(
a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(3/2)*a^2*b-12*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(3/2)*a*b^2-8*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/(sin(f*x+e)-1)*((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/(sin(f*x+e)-1)*((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/(sin(f*x+e)-1)*((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)^8-2*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*co
s(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+10*ln(2/(sin(f*x+e)-1)*((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)^6-8*(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)^(3/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*a^4*cos(f*x+e)^4-2*(a+b)^(
3/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*b^4*cos(f*x+e)^4+16*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(3/2)*a^3*cos(f*x+e)^2+6*
(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(3/2)*b^3*cos(f*x+e)^2-22*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/(sin(f*x+e)-1)*((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+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/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*cos(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^5*b*c
os(f*x+e)^4-15*ln(2/(sin(f*x+e)-1)*((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/(sin(f*x+e)-1)*((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/(sin(f*x+e)-1)*((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+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/(sin(f*x+e)-1)*(
(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+32*ln(2/(sin(f*x+e)-1)*((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/(sin(f*x+e)-1)*((a+b)^(1/2)*(a+b-b*co
s(f*x+e)^2)^(1/2)+b*sin(f*x+e)+a))*a^3*b^3*cos(f*x+e)^6-38*ln(2/(sin(f*x+e)-1)*((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/(sin(f*x+e)-1)*((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*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+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+16*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(3/2)*a*b^2*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)^(3/2)*(a+b-b*cos(f*x+e)^2)^(1
/2)*a^3*b*cos(f*x+e)^6+36*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*a*b^3*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)^4-32*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(3/2)*a^2*b*cos(f*x+e)^4-32
*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(3/2)*a*b^2*cos(f*x+e)^4+24*(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)^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+16*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*a^3*b
*cos(f*x+e)^4-18*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*a^2*b^2*cos(f*x+e)^4-20*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^
2)^(1/2)*a*b^3*cos(f*x+e)^4+38*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(3/2)*a^2*b*cos(f*x+e)^2+28*(a+b)^(3/2)*(a+b-b
*cos(f*x+e)^2)^(3/2)*a*b^2*cos(f*x+e)^2-8*(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)^(3/2)*(a+b-b*cos(f*x+e
)^2)^(1/2)*a^2*b^2*cos(f*x+e)^8-16*(a+b)^(3/2)*(a+b-b*cos(f*x+e)^2)^(1/2)*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)^(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 [B] time = 0.79, size = 334, normalized size = 1.89 \[ -\frac {\frac {{\left (8 \, a^{2} b^{3} - 8 \, a b^{4} - b^{5}\right )} \log \left (\frac {\sqrt {b \sin \left (f x + e\right )^{2} + a} - \sqrt {a + b}}{\sqrt {b \sin \left (f x + e\right )^{2} + a} + \sqrt {a + b}}\right )}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} \sqrt {a + b}} + \frac {2 \, {\left (8 \, a^{4} b^{3} + 16 \, a^{3} b^{4} + 8 \, a^{2} b^{5} + {\left (8 \, a^{2} b^{3} - 8 \, a b^{4} - b^{5}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{2} - {\left (16 \, a^{3} b^{3} + 8 \, a^{2} b^{4} - 7 \, a b^{5} + b^{6}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}\right )}}{{\left (a^{3} + 3 \, a^{2} b + 3 \, a b^{2} + b^{3}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {5}{2}} - 2 \, {\left (a^{4} + 4 \, a^{3} b + 6 \, a^{2} b^{2} + 4 \, a b^{3} + b^{4}\right )} {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} + {\left (a^{5} + 5 \, a^{4} b + 10 \, a^{3} b^{2} + 10 \, a^{2} b^{3} + 5 \, a b^{4} + b^{5}\right )} \sqrt {b \sin \left (f x + e\right )^{2} + a}}}{16 \, b^{3} f} \]
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]
-1/16*((8*a^2*b^3 - 8*a*b^4 - 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^3 + 3*a^2*b + 3*a*b^2 + b^3)*sqrt(a + b)) + 2*(8*a^4*b^3 + 16*a^3*b^4 + 8*a^2*b^5 + (8*a^2
*b^3 - 8*a*b^4 - b^5)*(b*sin(f*x + e)^2 + a)^2 - (16*a^3*b^3 + 8*a^2*b^4 - 7*a*b^5 + b^6)*(b*sin(f*x + e)^2 +
a))/((a^3 + 3*a^2*b + 3*a*b^2 + b^3)*(b*sin(f*x + e)^2 + a)^(5/2) - 2*(a^4 + 4*a^3*b + 6*a^2*b^2 + 4*a*b^3 + b
^4)*(b*sin(f*x + e)^2 + a)^(3/2) + (a^5 + 5*a^4*b + 10*a^3*b^2 + 10*a^2*b^3 + 5*a*b^4 + b^5)*sqrt(b*sin(f*x +
e)^2 + a)))/(b^3*f)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\mathrm {tan}\left (e+f\,x\right )}^5}{{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(tan(e + f*x)^5/(a + b*sin(e + f*x)^2)^(3/2),x)
[Out]
int(tan(e + f*x)^5/(a + b*sin(e + f*x)^2)^(3/2), x)
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\tan ^{5}{\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {3}{2}}}\, dx \]
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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