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3.258 \(\int \frac {\sec ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx\)

Optimal. Leaf size=170 \[ \frac {\tan (e+f x) \sec (e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}{b f \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}-\frac {\sqrt {a} \sqrt {a+b} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}} E\left (\sin ^{-1}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )|\frac {a+b}{a}\right )}{b f \sqrt {\cos ^2(e+f x)} \sqrt {\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}} \]

[Out]

-EllipticE(sin(f*x+e)*a^(1/2)/(a+b)^(1/2),((a+b)/a)^(1/2))*a^(1/2)*(a+b)^(1/2)*(1-a*sin(f*x+e)^2/(a+b))^(1/2)/
b/f/(cos(f*x+e)^2)^(1/2)/(sec(f*x+e)^2*(a+b-a*sin(f*x+e)^2))^(1/2)+sec(f*x+e)*(a+b-a*sin(f*x+e)^2)*tan(f*x+e)/
b/f/(sec(f*x+e)^2*(a+b-a*sin(f*x+e)^2))^(1/2)

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Rubi [A]  time = 0.37, antiderivative size = 202, normalized size of antiderivative = 1.19, number of steps used = 7, number of rules used = 7, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {4148, 6722, 1974, 414, 21, 427, 424} \[ \frac {\tan (e+f x) \sec (e+f x) \sqrt {-a \sin ^2(e+f x)+a+b} \sqrt {a \cos ^2(e+f x)+b}}{b f \sqrt {a+b \sec ^2(e+f x)}}-\frac {\sqrt {a} \sqrt {a+b} \sqrt {1-\frac {a \sin ^2(e+f x)}{a+b}} \sqrt {a \cos ^2(e+f x)+b} E\left (\sin ^{-1}\left (\frac {\sqrt {a} \sin (e+f x)}{\sqrt {a+b}}\right )|\frac {a+b}{a}\right )}{b f \sqrt {\cos ^2(e+f x)} \sqrt {-a \sin ^2(e+f x)+a+b} \sqrt {a+b \sec ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^3/Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

-((Sqrt[a]*Sqrt[a + b]*Sqrt[b + a*Cos[e + f*x]^2]*EllipticE[ArcSin[(Sqrt[a]*Sin[e + f*x])/Sqrt[a + b]], (a + b
)/a]*Sqrt[1 - (a*Sin[e + f*x]^2)/(a + b)])/(b*f*Sqrt[Cos[e + f*x]^2]*Sqrt[a + b*Sec[e + f*x]^2]*Sqrt[a + b - a
*Sin[e + f*x]^2])) + (Sqrt[b + a*Cos[e + f*x]^2]*Sec[e + f*x]*Sqrt[a + b - a*Sin[e + f*x]^2]*Tan[e + f*x])/(b*
f*Sqrt[a + b*Sec[e + f*x]^2])

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
 n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
 d*x, a + b*x])

Rule 414

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[1/(a*n*(p + 1)*(b*c - a*d)), Int[(a + b*x^n)^(p + 1)*
(c + d*x^n)^q*Simp[b*c + n*(p + 1)*(b*c - a*d) + d*b*(n*(p + q + 2) + 1)*x^n, x], x], x] /; FreeQ[{a, b, c, d,
 n, q}, x] && NeQ[b*c - a*d, 0] && LtQ[p, -1] &&  !( !IntegerQ[p] && IntegerQ[q] && LtQ[q, -1]) && IntBinomial
Q[a, b, c, d, n, p, q, x]

Rule 424

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Simp[(Sqrt[a]*EllipticE[ArcSin[Rt[-(d/c)
, 2]*x], (b*c)/(a*d)])/(Sqrt[c]*Rt[-(d/c), 2]), x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] && GtQ[c, 0] && GtQ[
a, 0]

Rule 427

Int[Sqrt[(a_) + (b_.)*(x_)^2]/Sqrt[(c_) + (d_.)*(x_)^2], x_Symbol] :> Dist[Sqrt[1 + (d*x^2)/c]/Sqrt[c + d*x^2]
, Int[Sqrt[a + b*x^2]/Sqrt[1 + (d*x^2)/c], x], x] /; FreeQ[{a, b, c, d}, x] && NegQ[d/c] &&  !GtQ[c, 0]

Rule 1974

Int[(u_)^(p_.)*(v_)^(q_.), x_Symbol] :> Int[ExpandToSum[u, x]^p*ExpandToSum[v, x]^q, x] /; FreeQ[{p, q}, x] &&
 BinomialQ[{u, v}, x] && EqQ[BinomialDegree[u, x] - BinomialDegree[v, x], 0] &&  !BinomialMatchQ[{u, v}, x]

Rule 4148

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

Rule 6722

Int[(u_.)*((a_.) + (b_.)*(v_)^(n_))^(p_), x_Symbol] :> Dist[(a + b*v^n)^FracPart[p]/(v^(n*FracPart[p])*(b + a/
v^n)^FracPart[p]), Int[u*v^(n*p)*(b + a/v^n)^p, x], x] /; FreeQ[{a, b, p}, x] &&  !IntegerQ[p] && ILtQ[n, 0] &
& BinomialQ[v, x] &&  !LinearQ[v, x]

Rubi steps

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

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Mathematica [F]  time = 10.63, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^3(e+f x)}{\sqrt {a+b \sec ^2(e+f x)}} \, dx \]

Verification is Not applicable to the result.

[In]

Integrate[Sec[e + f*x]^3/Sqrt[a + b*Sec[e + f*x]^2],x]

[Out]

Integrate[Sec[e + f*x]^3/Sqrt[a + b*Sec[e + f*x]^2], x]

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fricas [F]  time = 0.48, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sec \left (f x + e\right )^{3}}{\sqrt {b \sec \left (f x + e\right )^{2} + a}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^3/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

integral(sec(f*x + e)^3/sqrt(b*sec(f*x + e)^2 + a), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (f x + e\right )^{3}}{\sqrt {b \sec \left (f x + e\right )^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^3/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

integrate(sec(f*x + e)^3/sqrt(b*sec(f*x + e)^2 + a), x)

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maple [C]  time = 1.72, size = 3029, normalized size = 17.82 \[ \text {output too large to display} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^3/(a+b*sec(f*x+e)^2)^(1/2),x)

[Out]

-1/2/f*(4*I*cos(f*x+e)^3*a^(3/2)*b^(1/2)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)+4*I*cos(f*x+e)*a^(1/2)*b^(3/2
)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)+2*I*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I
*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/
2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/
2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a^(1/2)*b^(3/2)-2*I*co
s(f*x+e)*sin(f*x+e)*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a
+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*Ell
ipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b
^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a^(3/2)*b^(1/2)-2*I*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)*((I*a^(1/2)*b^(1/2)*
cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a
^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b
)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a^(3/2)*b^
(1/2)-4*cos(f*x+e)^2*sin(f*x+e)*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+co
s(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b
))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4
*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a*b+cos(f*x+e)^2*sin(f*x+e)*EllipticE((-1+cos(f*x+e))*((2*I*
a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2
)^(1/2))*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*
(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*a^2+2*cos(f*x+
e)^2*sin(f*x+e)*EllipticE((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^
(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1
/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+
e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*a*b+cos(f*x+e)^2*sin(f*x+e)*EllipticE((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+
a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*2^(1/2)
*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b
^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*b^2-4*I*cos(f*x+e)^2*a^(3/2)*b
^(1/2)*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)-4*sin(f*x+e)*cos(f*x+e)*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-
I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1
/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1
/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a*b+sin(f*x+e)*cos(f*
x+e)*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*
(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*EllipticE((-1+cos(
f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*
b-b^2)/(a+b)^2)^(1/2))*a^2+2*sin(f*x+e)*cos(f*x+e)*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*
cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/
(1+cos(f*x+e))/(a+b))^(1/2)*EllipticE((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*
I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))*a*b+sin(f*x+e)*cos(f*x+e)*2^(1/2)*((I*a^(
1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*cos(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*c
os(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1+cos(f*x+e))/(a+b))^(1/2)*EllipticE((-1+cos(f*x+e))*((2*I*a^(1/2
)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2
))*b^2+2*I*sin(f*x+e)*cos(f*x+e)*a^(1/2)*b^(3/2)*2^(1/2)*((I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)+a*co
s(f*x+e)+b)/(1+cos(f*x+e))/(a+b))^(1/2)*(-2*(I*a^(1/2)*b^(1/2)*cos(f*x+e)-I*a^(1/2)*b^(1/2)-a*cos(f*x+e)-b)/(1
+cos(f*x+e))/(a+b))^(1/2)*EllipticF((-1+cos(f*x+e))*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)/sin(f*x+e),(-(4*I*
a^(3/2)*b^(1/2)-4*I*a^(1/2)*b^(3/2)-a^2+6*a*b-b^2)/(a+b)^2)^(1/2))-4*I*a^(1/2)*b^(3/2)*((2*I*a^(1/2)*b^(1/2)+a
-b)/(a+b))^(1/2)-2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^3*a^2+2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b)
)^(1/2)*cos(f*x+e)^3*a*b+2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)^2*a^2-2*((2*I*a^(1/2)*b^(1/2)+a-
b)/(a+b))^(1/2)*cos(f*x+e)^2*a*b-2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)*a*b+2*((2*I*a^(1/2)*b^(1
/2)+a-b)/(a+b))^(1/2)*cos(f*x+e)*b^2+2*((2*I*a^(1/2)*b^(1/2)+a-b)/(a+b))^(1/2)*a*b-2*((2*I*a^(1/2)*b^(1/2)+a-b
)/(a+b))^(1/2)*b^2)/((b+a*cos(f*x+e)^2)/cos(f*x+e)^2)^(1/2)/cos(f*x+e)^2/sin(f*x+e)/((2*I*a^(1/2)*b^(1/2)+a-b)
/(a+b))^(1/2)/b/(2*I*a^(1/2)*b^(1/2)-a+b)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec \left (f x + e\right )^{3}}{\sqrt {b \sec \left (f x + e\right )^{2} + a}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^3/(a+b*sec(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

integrate(sec(f*x + e)^3/sqrt(b*sec(f*x + e)^2 + a), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {1}{{\cos \left (e+f\,x\right )}^3\,\sqrt {a+\frac {b}{{\cos \left (e+f\,x\right )}^2}}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(e + f*x)^3*(a + b/cos(e + f*x)^2)^(1/2)),x)

[Out]

int(1/(cos(e + f*x)^3*(a + b/cos(e + f*x)^2)^(1/2)), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {\sec ^{3}{\left (e + f x \right )}}{\sqrt {a + b \sec ^{2}{\left (e + f x \right )}}}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)**3/(a+b*sec(f*x+e)**2)**(1/2),x)

[Out]

Integral(sec(e + f*x)**3/sqrt(a + b*sec(e + f*x)**2), x)

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