3.67 \(\int \sqrt {a+b \sec ^2(e+f x)} \sin ^5(e+f x) \, dx\)

Optimal. Leaf size=139 \[ \frac {2 (5 a+b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{15 a^2 f}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{5 a f}-\frac {\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{f}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{f} \]

[Out]

2/15*(5*a+b)*cos(f*x+e)^3*(a+b*sec(f*x+e)^2)^(3/2)/a^2/f-1/5*cos(f*x+e)^5*(a+b*sec(f*x+e)^2)^(3/2)/a/f+arctanh
(sec(f*x+e)*b^(1/2)/(a+b*sec(f*x+e)^2)^(1/2))*b^(1/2)/f-cos(f*x+e)*(a+b*sec(f*x+e)^2)^(1/2)/f

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Rubi [A]  time = 0.15, antiderivative size = 139, 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 = {4134, 462, 451, 277, 217, 206} \[ \frac {2 (5 a+b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{15 a^2 f}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{5 a f}-\frac {\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{f}+\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{f} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + b*Sec[e + f*x]^2]*Sin[e + f*x]^5,x]

[Out]

(Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a + b*Sec[e + f*x]^2]])/f - (Cos[e + f*x]*Sqrt[a + b*Sec[e + f*x]
^2])/f + (2*(5*a + b)*Cos[e + f*x]^3*(a + b*Sec[e + f*x]^2)^(3/2))/(15*a^2*f) - (Cos[e + f*x]^5*(a + b*Sec[e +
f*x]^2)^(3/2))/(5*a*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 4134

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + (f_.)*(x_)]^(m_.), x_Symbol] :> With
[{ff = FreeFactors[Cos[e + f*x], x]}, Dist[1/(f*ff^m), Subst[Int[((-1 + ff^2*x^2)^((m - 1)/2)*(a + b*(c*ff*x)^
n)^p)/x^(m + 1), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[(m - 1)/2] && (Gt
Q[m, 0] || EqQ[n, 2] || EqQ[n, 4])

Rubi steps

\begin {align*} \int \sqrt {a+b \sec ^2(e+f x)} \sin ^5(e+f x) \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\left (-1+x^2\right )^2 \sqrt {a+b x^2}}{x^6} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{5 a f}+\frac {\operatorname {Subst}\left (\int \frac {\left (-2 (5 a+b)+5 a x^2\right ) \sqrt {a+b x^2}}{x^4} \, dx,x,\sec (e+f x)\right )}{5 a f}\\ &=\frac {2 (5 a+b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{15 a^2 f}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{5 a f}+\frac {\operatorname {Subst}\left (\int \frac {\sqrt {a+b x^2}}{x^2} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{f}+\frac {2 (5 a+b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{15 a^2 f}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{5 a f}+\frac {b \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sec (e+f x)\right )}{f}\\ &=-\frac {\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{f}+\frac {2 (5 a+b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{15 a^2 f}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{5 a f}+\frac {b \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{f}\\ &=\frac {\sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{f}-\frac {\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{f}+\frac {2 (5 a+b) \cos ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{15 a^2 f}-\frac {\cos ^5(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{5 a f}\\ \end {align*}

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Mathematica [A]  time = 0.85, size = 152, normalized size = 1.09 \[ -\frac {\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)} \left (\frac {2 \left (a \cos ^2(e+f x)+b\right )^{5/2}}{5 a^2}-\frac {2 (2 a+b) \left (a \cos ^2(e+f x)+b\right )^{3/2}}{3 a^2}+2 \sqrt {a \cos ^2(e+f x)+b}-2 \sqrt {b} \tanh ^{-1}\left (\frac {\sqrt {a \cos ^2(e+f x)+b}}{\sqrt {b}}\right )\right )}{\sqrt {2} f \sqrt {a \cos (2 e+2 f x)+a+2 b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + b*Sec[e + f*x]^2]*Sin[e + f*x]^5,x]

[Out]

-((Cos[e + f*x]*(-2*Sqrt[b]*ArcTanh[Sqrt[b + a*Cos[e + f*x]^2]/Sqrt[b]] + 2*Sqrt[b + a*Cos[e + f*x]^2] - (2*(2
*a + b)*(b + a*Cos[e + f*x]^2)^(3/2))/(3*a^2) + (2*(b + a*Cos[e + f*x]^2)^(5/2))/(5*a^2))*Sqrt[a + b*Sec[e + f
*x]^2])/(Sqrt[2]*f*Sqrt[a + 2*b + a*Cos[2*e + 2*f*x]]))

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fricas [A]  time = 0.88, size = 295, normalized size = 2.12 \[ \left [\frac {15 \, a^{2} \sqrt {b} \log \left (\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \, {\left (3 \, a^{2} \cos \left (f x + e\right )^{5} - {\left (10 \, a^{2} - a b\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{30 \, a^{2} f}, -\frac {15 \, a^{2} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right ) + {\left (3 \, a^{2} \cos \left (f x + e\right )^{5} - {\left (10 \, a^{2} - a b\right )} \cos \left (f x + e\right )^{3} + {\left (15 \, a^{2} - 10 \, a b - 2 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{15 \, a^{2} f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/30*(15*a^2*sqrt(b)*log((a*cos(f*x + e)^2 + 2*sqrt(b)*sqrt((a*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*cos(f*x + e)^5 - (10*a^2 - a*b)*cos(f*x + e)^3 + (15*a^2 - 10*a*b - 2*b^2
)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(a^2*f), -1/15*(15*a^2*sqrt(-b)*arctan(sqrt(-b)*s
qrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b) + (3*a^2*cos(f*x + e)^5 - (10*a^2 - a*b)*cos(f*x +
e)^3 + (15*a^2 - 10*a*b - 2*b^2)*cos(f*x + e))*sqrt((a*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/(a^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*sec(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)-64*(1/240*(-15*b*(-sqrt(a+b)*tan(1/2*(f*x+exp(1)))^2+sqrt(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)
))^4-2*a*tan(1/2*(f*x+exp(1)))^2+2*b*tan(1/2*(f*x+exp(1)))^2+a+b))^9+165*b*sqrt(a+b)*(-sqrt(a+b)*tan(1/2*(f*x+
exp(1)))^2+sqrt(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+exp(1)))^2+2*b*tan(1/2*(f
*x+exp(1)))^2+a+b))^8-(-sqrt(a+b)*tan(1/2*(f*x+exp(1)))^2+sqrt(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)
))^4-2*a*tan(1/2*(f*x+exp(1)))^2+2*b*tan(1/2*(f*x+exp(1)))^2+a+b))^7*(-320*a^2+540*b^2-100*a*b)-sqrt(a+b)*(-sq
rt(a+b)*tan(1/2*(f*x+exp(1)))^2+sqrt(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+exp(
1)))^2+2*b*tan(1/2*(f*x+exp(1)))^2+a+b))^6*(-640*a^2-660*b^2+1660*a*b)-(-sqrt(a+b)*tan(1/2*(f*x+exp(1)))^2+sqr
t(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+exp(1)))^2+2*b*tan(1/2*(f*x+exp(1)))^2+
a+b))^5*(832*a^3-30*b^3-2460*a*b^2+1250*a^2*b)-(-sqrt(a+b)*tan(1/2*(f*x+exp(1)))^2+sqrt(a*tan(1/2*(f*x+exp(1))
)^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+exp(1)))^2+2*b*tan(1/2*(f*x+exp(1)))^2+a+b))*(-2880*a^5+135*b^5
+700*a*b^4-1430*a^2*b^3+380*a^3*b^2+3335*a^4*b)-sqrt(a+b)*(-sqrt(a+b)*tan(1/2*(f*x+exp(1)))^2+sqrt(a*tan(1/2*(
f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+exp(1)))^2+2*b*tan(1/2*(f*x+exp(1)))^2+a+b))^4*(2560
*a^3+810*b^3+900*a*b^2-3910*a^2*b)-(-sqrt(a+b)*tan(1/2*(f*x+exp(1)))^2+sqrt(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/
2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+exp(1)))^2+2*b*tan(1/2*(f*x+exp(1)))^2+a+b))^3*(320*a^4-660*b^4+900*a*b^3+3
140*a^2*b^2-3220*a^3*b)-sqrt(a+b)*(-sqrt(a+b)*tan(1/2*(f*x+exp(1)))^2+sqrt(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/2
*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+exp(1)))^2+2*b*tan(1/2*(f*x+exp(1)))^2+a+b))^2*(-3200*a^4+60*b^4-1500*a*b^3+
980*a^2*b^2+3180*a^3*b)-sqrt(a+b)*(-768*a^5-45*b^5-100*a*b^4+450*a^2*b^3-1028*a^3*b^2+1539*a^4*b))/(2*sqrt(a+b
)*(-sqrt(a+b)*tan(1/2*(f*x+exp(1)))^2+sqrt(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*
x+exp(1)))^2+2*b*tan(1/2*(f*x+exp(1)))^2+a+b))-(-sqrt(a+b)*tan(1/2*(f*x+exp(1)))^2+sqrt(a*tan(1/2*(f*x+exp(1))
)^4+b*tan(1/2*(f*x+exp(1)))^4-2*a*tan(1/2*(f*x+exp(1)))^2+2*b*tan(1/2*(f*x+exp(1)))^2+a+b))^2+3*a-b)^5+1/32*b*
atan(1/2*(-sqrt(a+b)*tan(1/2*(f*x+exp(1)))^2+sqrt(a+b)+sqrt(a*tan(1/2*(f*x+exp(1)))^4+b*tan(1/2*(f*x+exp(1)))^
4-2*a*tan(1/2*(f*x+exp(1)))^2+2*b*tan(1/2*(f*x+exp(1)))^2+a+b))/sqrt(-b))/sqrt(-b))*sign(cos(f*x+exp(1)))/f

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maple [B]  time = 4.58, size = 1840, normalized size = 13.24 \[ \text {Expression too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^5*(a+b*sec(f*x+e)^2)^(1/2),x)

[Out]

-1/30/f*(-1+cos(f*x+e))^2*(-4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)*b^(3/2)*(a+b)^(3/2)+30*((b+a*cos(f*x
+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*b^(1/2)*(a+b)^(3/2)*4^(1/2)*a^2+15*ln(-4*(-1+cos(f*x+e))*(((b+a*cos(
f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*(a+b)^(1/2)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/
2)-a*cos(f*x+e)+b)/sin(f*x+e)^2/(a+b)^(1/2))*b^(5/2)*4^(1/2)*a^2-15*ln(-2*(-1+cos(f*x+e))*(((b+a*cos(f*x+e)^2)
/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*(a+b)^(1/2)+((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(
f*x+e)+b)/sin(f*x+e)^2/(a+b)^(1/2))*b^(5/2)*4^(1/2)*a^2+10*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)*b^(1/2)
*(a+b)^(3/2)*a+15*ln(-4*(-1+cos(f*x+e))*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*(a+b)^(1/2)+((
b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+b)/sin(f*x+e)^2/(a+b)^(1/2))*b^(3/2)*4^(1/2
)*a^3-15*ln(-2*(-1+cos(f*x+e))*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*(a+b)^(1/2)+((b+a*cos(f
*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+b)/sin(f*x+e)^2/(a+b)^(1/2))*b^(3/2)*4^(1/2)*a^3-15*
cos(f*x+e)*arctanh(1/8*(-1+cos(f*x+e))*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)-2)/sin(f*x+e)^2/((b+a*cos(f*x+
e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2)*4^(1/2))*(a+b)^(3/2)*4^(1/2)*a^2*b-4*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^
2)^(3/2)*cos(f*x+e)^4*b^(3/2)*(a+b)^(3/2)-16*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)*cos(f*x+e)^3*b^(3/2)*
(a+b)^(3/2)-24*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)*cos(f*x+e)^2*b^(3/2)*(a+b)^(3/2)-16*((b+a*cos(f*x+e
)^2)/(1+cos(f*x+e))^2)^(3/2)*cos(f*x+e)*b^(3/2)*(a+b)^(3/2)+16*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)*cos
(f*x+e)^4*b^(1/2)*(a+b)^(3/2)*a-56*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)*cos(f*x+e)^3*b^(1/2)*(a+b)^(3/2
)*a-84*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)*cos(f*x+e)^2*b^(1/2)*(a+b)^(3/2)*a+15*cos(f*x+e)*ln(-4*(-1+
cos(f*x+e))*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*(a+b)^(1/2)+((b+a*cos(f*x+e)^2)/(1+cos(f*x
+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+b)/sin(f*x+e)^2/(a+b)^(1/2))*b^(5/2)*4^(1/2)*a^2-15*cos(f*x+e)*ln(-2*(-
1+cos(f*x+e))*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*(a+b)^(1/2)+((b+a*cos(f*x+e)^2)/(1+cos(f
*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+b)/sin(f*x+e)^2/(a+b)^(1/2))*b^(5/2)*4^(1/2)*a^2-15*((b+a*cos(f*x+e)^
2)/(1+cos(f*x+e))^2)^(1/2)*b^(3/2)*(a+b)^(3/2)*4^(1/2)*a+15*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*b^(1/2
)*(a+b)^(3/2)*4^(1/2)*a^2-20*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)*cos(f*x+e)*b^(1/2)*(a+b)^(3/2)*a+15*c
os(f*x+e)*ln(-4*(-1+cos(f*x+e))*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*(a+b)^(1/2)+((b+a*cos(
f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+b)/sin(f*x+e)^2/(a+b)^(1/2))*b^(3/2)*4^(1/2)*a^3-15
*cos(f*x+e)*ln(-2*(-1+cos(f*x+e))*(((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*cos(f*x+e)*(a+b)^(1/2)+((b+a*co
s(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)*(a+b)^(1/2)-a*cos(f*x+e)+b)/sin(f*x+e)^2/(a+b)^(1/2))*b^(3/2)*4^(1/2)*a^3-
15*arctanh(1/8*(-1+cos(f*x+e))*(cos(f*x+e)*4^(1/2)-2*cos(f*x+e)-4^(1/2)-2)/sin(f*x+e)^2/((b+a*cos(f*x+e)^2)/(1
+cos(f*x+e))^2)^(1/2)*b^(1/2)*4^(1/2))*(a+b)^(3/2)*4^(1/2)*a^2*b+6*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)
*cos(f*x+e)^6*b^(1/2)*(a+b)^(3/2)*a+24*((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(3/2)*cos(f*x+e)^5*b^(1/2)*(a+b)^
(3/2)*a)*cos(f*x+e)*((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(1/2)/((b+a*cos(f*x+e)^2)/(1+cos(f*x+e))^2)^(1/2)/sin(f*
x+e)^4/b^(1/2)/(a+b)^(3/2)/a^2

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maxima [A]  time = 0.43, size = 171, normalized size = 1.23 \[ \frac {\frac {20 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3}}{a} - 30 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - 15 \, \sqrt {b} \log \left (\frac {\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right ) - \frac {2 \, {\left (3 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {5}{2}} \cos \left (f x + e\right )^{5} - 5 \, {\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} b \cos \left (f x + e\right )^{3}\right )}}{a^{2}}}{30 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^5*(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

1/30*(20*(a + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3/a - 30*sqrt(a + b/cos(f*x + e)^2)*cos(f*x + e) - 15*sqrt(
b)*log((sqrt(a + b/cos(f*x + e)^2)*cos(f*x + e) - sqrt(b))/(sqrt(a + b/cos(f*x + e)^2)*cos(f*x + e) + sqrt(b))
) - 2*(3*(a + b/cos(f*x + e)^2)^(5/2)*cos(f*x + e)^5 - 5*(a + b/cos(f*x + e)^2)^(3/2)*b*cos(f*x + e)^3)/a^2)/f

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\sin \left (e+f\,x\right )}^5\,\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(e + f*x)^5*(a + b/cos(e + f*x)^2)^(1/2),x)

[Out]

int(sin(e + f*x)^5*(a + b/cos(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*sec(f*x+e)**2)**(1/2),x)

[Out]

Timed out

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