3.92 \(\int \sin ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\)
Optimal. Leaf size=161 \[ -\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{5 f (a-b)}+\frac {2 (5 a-4 b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{15 f (a-b)^2}-\frac {\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{f}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{f} \]
[Out]
2/15*(5*a-4*b)*cos(f*x+e)^3*(a-b+b*sec(f*x+e)^2)^(3/2)/(a-b)^2/f-1/5*cos(f*x+e)^5*(a-b+b*sec(f*x+e)^2)^(3/2)/(
a-b)/f+arctanh(sec(f*x+e)*b^(1/2)/(a-b+b*sec(f*x+e)^2)^(1/2))*b^(1/2)/f-cos(f*x+e)*(a-b+b*sec(f*x+e)^2)^(1/2)/
f
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Rubi [A] time = 0.17, antiderivative size = 161, 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 =
{3664, 462, 451, 277, 217, 206} \[ -\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{5 f (a-b)}+\frac {2 (5 a-4 b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{15 f (a-b)^2}-\frac {\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{f}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{f} \]
Antiderivative was successfully verified.
[In]
Int[Sin[e + f*x]^5*Sqrt[a + b*Tan[e + f*x]^2],x]
[Out]
(Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/f - (Cos[e + f*x]*Sqrt[a - b + b*Sec[
e + f*x]^2])/f + (2*(5*a - 4*b)*Cos[e + f*x]^3*(a - b + b*Sec[e + f*x]^2)^(3/2))/(15*(a - b)^2*f) - (Cos[e + f
*x]^5*(a - b + b*Sec[e + f*x]^2)^(3/2))/(5*(a - b)*f)
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 217
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] && !GtQ[a, 0]
Rule 277
Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[((c*x)^(m + 1)*(a + b*x^n)^p)/(c*(m +
1)), x] - Dist[(b*n*p)/(c^n*(m + 1)), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] && !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]
Rule 451
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[d/e^n, Int[(e*x)^(m + n)*(a + b*x^n)^p, x], x] /; FreeQ[{a,
b, c, d, e, m, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[m + n*(p + 1) + 1, 0] && (IntegerQ[n] || GtQ[e, 0]) && (
(GtQ[n, 0] && LtQ[m, -1]) || (LtQ[n, 0] && GtQ[m + n, -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 3664
Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sec[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a - b + b*ff^2*x^2)^p)/x^(
m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]
Rubi steps
\begin {align*} \int \sin ^5(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (-1+x^2\right )^2 \sqrt {a-b+b x^2}}{x^6} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{5 (a-b) f}+\frac {\operatorname {Subst}\left (\int \frac {\left (-2 (5 a-4 b)+5 (a-b) x^2\right ) \sqrt {a-b+b x^2}}{x^4} \, dx,x,\sec (e+f x)\right )}{5 (a-b) f}\\ &=\frac {2 (5 a-4 b) \cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{15 (a-b)^2 f}-\frac {\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{5 (a-b) f}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {a-b+b x^2}}{x^2} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {\cos (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{f}+\frac {2 (5 a-4 b) \cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{15 (a-b)^2 f}-\frac {\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{5 (a-b) f}+\frac {b \operatorname {Subst}\left (\int \frac {1}{\sqrt {a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {\cos (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{f}+\frac {2 (5 a-4 b) \cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{15 (a-b)^2 f}-\frac {\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{5 (a-b) f}+\frac {b \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{f}\\ &=\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{f}-\frac {\cos (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{f}+\frac {2 (5 a-4 b) \cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{15 (a-b)^2 f}-\frac {\cos ^5(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{5 (a-b) f}\\ \end {align*}
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Mathematica [A] time = 3.29, size = 208, normalized size = 1.29 \[ \frac {\cos (e+f x) \sqrt {\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)} \left (\sqrt {(a-b) \cos (2 (e+f x))+a+b} \left (4 \left (7 a^2-15 a b+8 b^2\right ) \cos (2 (e+f x))-89 a^2-3 (a-b)^2 \cos (4 (e+f x))+254 a b-149 b^2\right )+120 \sqrt {2} \sqrt {b} (a-b)^2 \tanh ^{-1}\left (\frac {\sqrt {(a-b) \cos (2 (e+f x))+a+b}}{\sqrt {2} \sqrt {b}}\right )\right )}{120 \sqrt {2} f (a-b)^2 \sqrt {(a-b) \cos (2 (e+f x))+a+b}} \]
Antiderivative was successfully verified.
[In]
Integrate[Sin[e + f*x]^5*Sqrt[a + b*Tan[e + f*x]^2],x]
[Out]
(Cos[e + f*x]*(120*Sqrt[2]*(a - b)^2*Sqrt[b]*ArcTanh[Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]]/(Sqrt[2]*Sqrt[b])]
+ Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]]*(-89*a^2 + 254*a*b - 149*b^2 + 4*(7*a^2 - 15*a*b + 8*b^2)*Cos[2*(e +
f*x)] - 3*(a - b)^2*Cos[4*(e + f*x)]))*Sqrt[(a + b + (a - b)*Cos[2*(e + f*x)])*Sec[e + f*x]^2])/(120*Sqrt[2]*
(a - b)^2*f*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]])
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fricas [A] time = 0.66, size = 378, normalized size = 2.35 \[ \left [\frac {15 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {b} \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \, {\left (3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (10 \, a^{2} - 21 \, a b + 11 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} - 40 \, a b + 23 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{30 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} f}, -\frac {15 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right ) + {\left (3 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - {\left (10 \, a^{2} - 21 \, a b + 11 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} - 40 \, a b + 23 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} f}\right ] \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(f*x+e)^5*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")
[Out]
[1/30*(15*(a^2 - 2*a*b + b^2)*sqrt(b)*log(-((a - b)*cos(f*x + e)^2 + 2*sqrt(b)*sqrt(((a - b)*cos(f*x + e)^2 +
b)/cos(f*x + e)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^2) - 2*(3*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 - (10*a^2 - 2
1*a*b + 11*b^2)*cos(f*x + e)^3 + (15*a^2 - 40*a*b + 23*b^2)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/co
s(f*x + e)^2))/((a^2 - 2*a*b + b^2)*f), -1/15*(15*(a^2 - 2*a*b + b^2)*sqrt(-b)*arctan(sqrt(-b)*sqrt(((a - b)*c
os(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b) + (3*(a^2 - 2*a*b + b^2)*cos(f*x + e)^5 - (10*a^2 - 21*a*b
+ 11*b^2)*cos(f*x + e)^3 + (15*a^2 - 40*a*b + 23*b^2)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x
+ e)^2))/((a^2 - 2*a*b + 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*tan(f*x+e)^2)^(1/2),x, algorithm="giac")
[Out]
Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (4*pi/x/2)>(-4*pi/
x/2)2/f*32*(1/240*(15*b*(-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(a*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2
+4*b*tan((f*x+exp(1))/2)^2+a))^9-165*sqrt(a)*b*(-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(a*tan((f*x+exp(1))/2)^4-2*
a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))^8-320*a^2*(-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(a*tan((f*
x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))^7+320*b^2*(-sqrt(a)*tan((f*x+exp(1))/2)
^2+sqrt(a*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))^7+540*a*b*(-sqrt(a)*ta
n((f*x+exp(1))/2)^2+sqrt(a*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))^7-640
*sqrt(a)*a^2*(-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(a*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+4*b*tan((f
*x+exp(1))/2)^2+a))^6-2960*sqrt(a)*b^2*(-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(a*tan((f*x+exp(1))/2)^4-2*a*tan((f
*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))^6+2940*sqrt(a)*a*b*(-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(a*tan((f
*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))^6+832*a^3*(-sqrt(a)*tan((f*x+exp(1))/2
)^2+sqrt(a*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))^5+2848*b^3*(-sqrt(a)*
tan((f*x+exp(1))/2)^2+sqrt(a*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))^5-2
464*a*b^2*(-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(a*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+
exp(1))/2)^2+a))^5-1246*a^2*b*(-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(a*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1)
)/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))^5-2880*a^5*(-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(a*tan((f*x+exp(1))/2)^4-2
*a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))+3840*b^5*(-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(a*tan((f*
x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))+2560*sqrt(a)*a^3*(-sqrt(a)*tan((f*x+exp
(1))/2)^2+sqrt(a*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))^4-23040*a*b^4*(
-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(a*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2)^
2+a))+46240*a^2*b^3*(-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(a*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+4*b
*tan((f*x+exp(1))/2)^2+a))-41760*a^3*b^2*(-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(a*tan((f*x+exp(1))/2)^4-2*a*tan(
(f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))+17735*a^4*b*(-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(a*tan((f*x+ex
p(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))-6560*sqrt(a)*b^3*(-sqrt(a)*tan((f*x+exp(1))
/2)^2+sqrt(a*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))^4+16400*sqrt(a)*a*b
^2*(-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(a*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))
/2)^2+a))^4-11590*sqrt(a)*a^2*b*(-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(a*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(
1))/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))^4+320*a^4*(-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(a*tan((f*x+exp(1))/2)^4-
2*a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))^3+5120*b^4*(-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(a*tan(
(f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))^3-16320*a*b^3*(-sqrt(a)*tan((f*x+exp
(1))/2)^2+sqrt(a*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))^3+14720*a^2*b^2
*(-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(a*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2
)^2+a))^3-4500*a^3*b*(-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(a*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+4*
b*tan((f*x+exp(1))/2)^2+a))^3-3200*sqrt(a)*a^4*(-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(a*tan((f*x+exp(1))/2)^4-2*
a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))^2-3840*sqrt(a)*b^4*(-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(
a*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))^2+18880*sqrt(a)*a*b^3*(-sqrt(a
)*tan((f*x+exp(1))/2)^2+sqrt(a*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))^2
-27760*sqrt(a)*a^2*b^2*(-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(a*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+
4*b*tan((f*x+exp(1))/2)^2+a))^2+15980*sqrt(a)*a^3*b*(-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(a*tan((f*x+exp(1))/2)
^4-2*a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))^2-768*sqrt(a)*a^5+3840*sqrt(a)*b^5-14080*sqrt(a)*a*
b^4+20448*sqrt(a)*a^2*b^3-14864*sqrt(a)*a^3*b^2+5379*sqrt(a)*a^4*b)/(-2*sqrt(a)*(-sqrt(a)*tan((f*x+exp(1))/2)^
2+sqrt(a*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))+(-sqrt(a)*tan((f*x+exp(
1))/2)^2+sqrt(a*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+4*b*tan((f*x+exp(1))/2)^2+a))^2-3*a+4*b)^5+1/3
2*b*atan(1/2*(-sqrt(a)*tan((f*x+exp(1))/2)^2+sqrt(a)+sqrt(a*tan((f*x+exp(1))/2)^4-2*a*tan((f*x+exp(1))/2)^2+4*
b*tan((f*x+exp(1))/2)^2+a))/sqrt(-b))/sqrt(-b))*sign(tan((f*x+exp(1))/2)^2-1)
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maple [B] time = 5.18, size = 7044, normalized size = 43.75 \[ \text {output too large to display} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sin(f*x+e)^5*(a+b*tan(f*x+e)^2)^(1/2),x)
[Out]
result too large to display
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maxima [A] time = 0.81, size = 203, normalized size = 1.26 \[ \frac {\frac {20 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3}}{a - b} - 30 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - 15 \, \sqrt {b} \log \left (\frac {\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right ) - \frac {2 \, {\left (3 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{5} - 5 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} b \cos \left (f x + e\right )^{3}\right )}}{a^{2} - 2 \, a b + b^{2}}}{30 \, f} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sin(f*x+e)^5*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")
[Out]
1/30*(20*(a - b + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3/(a - b) - 30*sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x +
e) - 15*sqrt(b)*log((sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e) - sqrt(b))/(sqrt(a - b + b/cos(f*x + e)^2)*c
os(f*x + e) + sqrt(b))) - 2*(3*(a - b + b/cos(f*x + e)^2)^(5/2)*cos(f*x + e)^5 - 5*(a - b + b/cos(f*x + e)^2)^
(3/2)*b*cos(f*x + e)^3)/(a^2 - 2*a*b + b^2))/f
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\sin \left (e+f\,x\right )}^5\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a} \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sin(e + f*x)^5*(a + b*tan(e + f*x)^2)^(1/2),x)
[Out]
int(sin(e + f*x)^5*(a + b*tan(e + f*x)^2)^(1/2), 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*tan(f*x+e)**2)**(1/2),x)
[Out]
Timed out
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