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

Optimal. Leaf size=143 \[ \frac {a (3 a+4 b) \tanh ^{-1}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{8 f (a+b)^{3/2}}+\frac {\tan (e+f x) \sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 f (a+b)}+\frac {(3 a+4 b) \tan (e+f x) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{8 f (a+b)} \]

[Out]

1/8*a*(3*a+4*b)*arctanh(sin(f*x+e)*(a+b)^(1/2)/(a+b*sin(f*x+e)^2)^(1/2))/(a+b)^(3/2)/f+1/4*sec(f*x+e)^3*(a+b*s
in(f*x+e)^2)^(3/2)*tan(f*x+e)/(a+b)/f+1/8*(3*a+4*b)*sec(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)*tan(f*x+e)/(a+b)/f

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Rubi [A]  time = 0.14, antiderivative size = 143, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3190, 382, 378, 377, 206} \[ \frac {a (3 a+4 b) \tanh ^{-1}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{8 f (a+b)^{3/2}}+\frac {\tan (e+f x) \sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 f (a+b)}+\frac {(3 a+4 b) \tan (e+f x) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{8 f (a+b)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a*(3*a + 4*b)*ArcTanh[(Sqrt[a + b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]])/(8*(a + b)^(3/2)*f) + ((3*a + 4
*b)*Sec[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2]*Tan[e + f*x])/(8*(a + b)*f) + (Sec[e + f*x]^3*(a + b*Sin[e + f*x]^
2)^(3/2)*Tan[e + f*x])/(4*(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 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 378

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_.), x_Symbol] :> -Simp[(x*(a + b*x^n)^(p + 1)*(c
+ d*x^n)^q)/(a*n*(p + 1)), x] - Dist[(c*q)/(a*(p + 1)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^(q - 1), x], x] /
; FreeQ[{a, b, c, d, n, p}, x] && NeQ[b*c - a*d, 0] && EqQ[n*(p + q + 1) + 1, 0] && GtQ[q, 0] && NeQ[p, -1]

Rule 382

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(b*x*(a + b*x^n)^(p + 1)*(
c + d*x^n)^(q + 1))/(a*n*(p + 1)*(b*c - a*d)), x] + Dist[(b*c + n*(p + 1)*(b*c - a*d))/(a*n*(p + 1)*(b*c - a*d
)), Int[(a + b*x^n)^(p + 1)*(c + d*x^n)^q, x], x] /; FreeQ[{a, b, c, d, n, q}, x] && NeQ[b*c - a*d, 0] && EqQ[
n*(p + q + 2) + 1, 0] && (LtQ[p, -1] ||  !LtQ[q, -1]) && NeQ[p, -1]

Rule 3190

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \sec ^5(e+f x) \sqrt {a+b \sin ^2(e+f x)} \, dx &=\frac {\operatorname {Subst}\left (\int \frac {\sqrt {a+b x^2}}{\left (1-x^2\right )^3} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x)}{4 (a+b) f}+\frac {(3 a+4 b) \operatorname {Subst}\left (\int \frac {\sqrt {a+b x^2}}{\left (1-x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{4 (a+b) f}\\ &=\frac {(3 a+4 b) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{8 (a+b) f}+\frac {\sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x)}{4 (a+b) f}+\frac {(a (3 a+4 b)) \operatorname {Subst}\left (\int \frac {1}{\left (1-x^2\right ) \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{8 (a+b) f}\\ &=\frac {(3 a+4 b) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{8 (a+b) f}+\frac {\sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x)}{4 (a+b) f}+\frac {(a (3 a+4 b)) \operatorname {Subst}\left (\int \frac {1}{1-(a+b) x^2} \, dx,x,\frac {\sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{8 (a+b) f}\\ &=\frac {a (3 a+4 b) \tanh ^{-1}\left (\frac {\sqrt {a+b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{8 (a+b)^{3/2} f}+\frac {(3 a+4 b) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} \tan (e+f x)}{8 (a+b) f}+\frac {\sec ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \tan (e+f x)}{4 (a+b) f}\\ \end {align*}

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Mathematica [C]  time = 13.30, size = 669, normalized size = 4.68 \[ -\frac {\tan (e+f x) \sec ^3(e+f x) \left (\frac {b \sin ^2(e+f x)}{a}+1\right ) \left (10 b \sin ^2(e+f x) \sqrt {-\frac {(a+b) \tan ^2(e+f x) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a^2}}+15 a \sqrt {-\frac {(a+b) \tan ^2(e+f x) \sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a^2}}+32 b \sin ^2(e+f x) \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{7/2} \, _2F_1\left (2,4;\frac {7}{2};-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}}+32 a \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{7/2} \, _2F_1\left (2,4;\frac {7}{2};-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}}-32 b \sin ^2(e+f x) \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{5/2} \, _2F_1\left (2,4;\frac {7}{2};-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}}-32 a \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{5/2} \, _2F_1\left (2,4;\frac {7}{2};-\frac {(a+b) \tan ^2(e+f x)}{a}\right ) \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}}-15 a \sin ^{-1}\left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right )-10 b \sin ^2(e+f x) \sin ^{-1}\left (\sqrt {-\frac {(a+b) \tan ^2(e+f x)}{a}}\right )-20 b \sin ^2(e+f x) \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{3/2} \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}}-30 a \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{3/2} \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}}\right )}{40 f \sqrt {a+b \sin ^2(e+f x)} \left (-\frac {(a+b) \tan ^2(e+f x)}{a}\right )^{3/2} \sqrt {\frac {\sec ^2(e+f x) \left (a+b \sin ^2(e+f x)\right )}{a}}} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

-1/40*(Sec[e + f*x]^3*(1 + (b*Sin[e + f*x]^2)/a)*Tan[e + f*x]*(-15*a*ArcSin[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a)
]] - 10*b*ArcSin[Sqrt[-(((a + b)*Tan[e + f*x]^2)/a)]]*Sin[e + f*x]^2 - 30*a*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e
+ f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(3/2) - 20*b*Sin[e + f*x]^2*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e +
f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(3/2) - 32*a*Hypergeometric2F1[2, 4, 7/2, -(((a + b)*Tan[e + f*x]^
2)/a)]*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(5/2) - 32*b*Hypergeome
tric2F1[2, 4, 7/2, -(((a + b)*Tan[e + f*x]^2)/a)]*Sin[e + f*x]^2*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/
a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(5/2) + 32*a*Hypergeometric2F1[2, 4, 7/2, -(((a + b)*Tan[e + f*x]^2)/a)]*Sq
rt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(7/2) + 32*b*Hypergeometric2F1[2
, 4, 7/2, -(((a + b)*Tan[e + f*x]^2)/a)]*Sin[e + f*x]^2*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a
+ b)*Tan[e + f*x]^2)/a))^(7/2) + 15*a*Sqrt[-(((a + b)*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2)*Tan[e + f*x]^2)/a
^2)] + 10*b*Sin[e + f*x]^2*Sqrt[-(((a + b)*Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2)*Tan[e + f*x]^2)/a^2)]))/(f*Sq
rt[a + b*Sin[e + f*x]^2]*Sqrt[(Sec[e + f*x]^2*(a + b*Sin[e + f*x]^2))/a]*(-(((a + b)*Tan[e + f*x]^2)/a))^(3/2)
)

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fricas [A]  time = 0.86, size = 443, normalized size = 3.10 \[ \left [\frac {{\left (3 \, a^{2} + 4 \, a b\right )} \sqrt {a + b} \cos \left (f x + e\right )^{4} \log \left (\frac {{\left (a^{2} + 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )^{4} - 8 \, {\left (a^{2} + 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} - 4 \, {\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {a + b} \sin \left (f x + e\right ) + 8 \, a^{2} + 16 \, a b + 8 \, b^{2}}{\cos \left (f x + e\right )^{4}}\right ) + 4 \, {\left ({\left (3 \, a^{2} + 5 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, a^{2} + 4 \, a b + 2 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{32 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{4}}, -\frac {{\left (3 \, a^{2} + 4 \, a b\right )} \sqrt {-a - b} \arctan \left (\frac {{\left ({\left (a + 2 \, b\right )} \cos \left (f x + e\right )^{2} - 2 \, a - 2 \, b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-a - b}}{2 \, {\left ({\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} - a^{2} - 2 \, a b - b^{2}\right )} \sin \left (f x + e\right )}\right ) \cos \left (f x + e\right )^{4} - 2 \, {\left ({\left (3 \, a^{2} + 5 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + 2 \, a^{2} + 4 \, a b + 2 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{16 \, {\left (a^{2} + 2 \, a b + b^{2}\right )} f \cos \left (f x + e\right )^{4}}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/32*((3*a^2 + 4*a*b)*sqrt(a + b)*cos(f*x + e)^4*log(((a^2 + 8*a*b + 8*b^2)*cos(f*x + e)^4 - 8*(a^2 + 3*a*b +
2*b^2)*cos(f*x + e)^2 - 4*((a + 2*b)*cos(f*x + e)^2 - 2*a - 2*b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(a + b)*
sin(f*x + e) + 8*a^2 + 16*a*b + 8*b^2)/cos(f*x + e)^4) + 4*((3*a^2 + 5*a*b + 2*b^2)*cos(f*x + e)^2 + 2*a^2 + 4
*a*b + 2*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e))/((a^2 + 2*a*b + b^2)*f*cos(f*x + e)^4), -1/16*((3*
a^2 + 4*a*b)*sqrt(-a - b)*arctan(1/2*((a + 2*b)*cos(f*x + e)^2 - 2*a - 2*b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sq
rt(-a - b)/(((a*b + b^2)*cos(f*x + e)^2 - a^2 - 2*a*b - b^2)*sin(f*x + e)))*cos(f*x + e)^4 - 2*((3*a^2 + 5*a*b
+ 2*b^2)*cos(f*x + e)^2 + 2*a^2 + 4*a*b + 2*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e))/((a^2 + 2*a*b
+ b^2)*f*cos(f*x + e)^4)]

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \sin \left (f x + e\right )^{2} + a} \sec \left (f x + e\right )^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*sin(f*x + e)^2 + a)*sec(f*x + e)^5, x)

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maple [B]  time = 4.85, size = 570, normalized size = 3.99 \[ \frac {2 \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a +b \right )^{\frac {3}{2}} b \left (3 a +4 b \right ) \sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right )+2 \left (a +b -b \left (\cos ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}} \left (a +b \right )^{\frac {3}{2}} \left (3 a +4 b \right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )+4 \left (a +b -b \left (\cos ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}} \left (a +b \right )^{\frac {5}{2}} \sin \left (f x +e \right )+a \left (3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{3}+10 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a^{2} b +11 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) a \,b^{2}+4 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}+2 b \sin \left (f x +e \right )+2 a}{\sin \left (f x +e \right )-1}\right ) b^{3}-3 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{3}-10 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a^{2} b -11 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) a \,b^{2}-4 \ln \left (\frac {2 \sqrt {a +b}\, \sqrt {a +b -b \left (\cos ^{2}\left (f x +e \right )\right )}-2 b \sin \left (f x +e \right )+2 a}{1+\sin \left (f x +e \right )}\right ) b^{3}\right ) \left (\cos ^{4}\left (f x +e \right )\right )}{16 \left (a +b \right )^{\frac {3}{2}} \cos \left (f x +e \right )^{4} \left (a^{2}+2 a b +b^{2}\right ) f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/16*(2*(a+b-b*cos(f*x+e)^2)^(1/2)*(a+b)^(3/2)*b*(3*a+4*b)*sin(f*x+e)*cos(f*x+e)^4+2*(a+b-b*cos(f*x+e)^2)^(3/2
)*(a+b)^(3/2)*(3*a+4*b)*cos(f*x+e)^2*sin(f*x+e)+4*(a+b-b*cos(f*x+e)^2)^(3/2)*(a+b)^(5/2)*sin(f*x+e)+a*(3*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+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^2*b+11*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^2+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^3
-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))*a^3-10*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^2*b-11*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^2-4*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^3)*cos(f*x+e)^4)/(a+b)^(3/2)/cos(f*x+e)^4/(a^2+2*a*b+b^2)/f

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \sqrt {b \sin \left (f x + e\right )^{2} + a} \sec \left (f x + e\right )^{5}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate(sqrt(b*sin(f*x + e)^2 + a)*sec(f*x + e)^5, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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