3.4.71 \(\int \frac {\sec ^2(e+f x)}{(a+b \sin ^2(e+f x))^{5/2}} \, dx\) [371]

Optimal. Leaf size=288 \[ -\frac {(3 a-b) b \cos (e+f x) \sin (e+f x)}{3 a (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {b \left (3 a^2-7 a b-2 b^2\right ) \cos (e+f x) \sin (e+f x)}{3 a^2 (a+b)^3 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\left (3 a^2-7 a b-2 b^2\right ) E\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{3 a (a+b)^3 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {(3 a-b) F\left (e+f x\left |-\frac {b}{a}\right .\right ) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{3 a (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\tan (e+f x)}{(a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}} \]

[Out]

-1/3*(3*a-b)*b*cos(f*x+e)*sin(f*x+e)/a/(a+b)^2/f/(a+b*sin(f*x+e)^2)^(3/2)-1/3*b*(3*a^2-7*a*b-2*b^2)*cos(f*x+e)
*sin(f*x+e)/a^2/(a+b)^3/f/(a+b*sin(f*x+e)^2)^(1/2)-1/3*(3*a^2-7*a*b-2*b^2)*(cos(f*x+e)^2)^(1/2)/cos(f*x+e)*Ell
ipticE(sin(f*x+e),(-b/a)^(1/2))*(1+b*sin(f*x+e)^2/a)^(1/2)/a/(a+b)^3/f/(a+b*sin(f*x+e)^2)^(1/2)+1/3*(3*a-b)*(c
os(f*x+e)^2)^(1/2)/cos(f*x+e)*EllipticF(sin(f*x+e),(-b/a)^(1/2))*(1+b*sin(f*x+e)^2/a)^(1/2)/a/(a+b)^2/f/(a+b*s
in(f*x+e)^2)^(1/2)+tan(f*x+e)/(a+b)/f/(a+b*sin(f*x+e)^2)^(3/2)

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Rubi [A]
time = 0.23, antiderivative size = 328, normalized size of antiderivative = 1.14, number of steps used = 9, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, Rules used = {3271, 425, 541, 538, 437, 435, 432, 430} \begin {gather*} -\frac {\left (3 a^2-7 a b-2 b^2\right ) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)} E\left (\text {ArcSin}(\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{3 a^2 f (a+b)^3 \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}-\frac {b \left (3 a^2-7 a b-2 b^2\right ) \sin (e+f x) \cos (e+f x)}{3 a^2 f (a+b)^3 \sqrt {a+b \sin ^2(e+f x)}}+\frac {(3 a-b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {\frac {b \sin ^2(e+f x)}{a}+1} F\left (\text {ArcSin}(\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{3 a f (a+b)^2 \sqrt {a+b \sin ^2(e+f x)}}+\frac {\tan (e+f x)}{f (a+b) \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {b (3 a-b) \sin (e+f x) \cos (e+f x)}{3 a f (a+b)^2 \left (a+b \sin ^2(e+f x)\right )^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/3*((3*a - b)*b*Cos[e + f*x]*Sin[e + f*x])/(a*(a + b)^2*f*(a + b*Sin[e + f*x]^2)^(3/2)) - (b*(3*a^2 - 7*a*b
- 2*b^2)*Cos[e + f*x]*Sin[e + f*x])/(3*a^2*(a + b)^3*f*Sqrt[a + b*Sin[e + f*x]^2]) - ((3*a^2 - 7*a*b - 2*b^2)*
Sqrt[Cos[e + f*x]^2]*EllipticE[ArcSin[Sin[e + f*x]], -(b/a)]*Sec[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(3*a^2*(
a + b)^3*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a]) + ((3*a - b)*Sqrt[Cos[e + f*x]^2]*EllipticF[ArcSin[Sin[e + f*x]], -
(b/a)]*Sec[e + f*x]*Sqrt[1 + (b*Sin[e + f*x]^2)/a])/(3*a*(a + b)^2*f*Sqrt[a + b*Sin[e + f*x]^2]) + Tan[e + f*x
]/((a + b)*f*(a + b*Sin[e + f*x]^2)^(3/2))

Rule 425

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]) && IntBinomi
alQ[a, b, c, d, n, p, q, x]

Rule 430

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

Rule 432

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

Rule 435

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

Rule 437

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

Rule 538

Int[((e_) + (f_.)*(x_)^(n_))/(Sqrt[(a_) + (b_.)*(x_)^(n_)]*Sqrt[(c_) + (d_.)*(x_)^(n_)]), x_Symbol] :> Dist[f/
b, Int[Sqrt[a + b*x^n]/Sqrt[c + d*x^n], x], x] + Dist[(b*e - a*f)/b, Int[1/(Sqrt[a + b*x^n]*Sqrt[c + d*x^n]),
x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] &&  !(EqQ[n, 2] && ((PosQ[b/a] && PosQ[d/c]) || (NegQ[b/a] && (PosQ[
d/c] || (GtQ[a, 0] && ( !GtQ[c, 0] || SimplerSqrtQ[-b/a, -d/c]))))))

Rule 541

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

Rule 3271

Int[cos[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = FreeF
actors[Sin[e + f*x], x]}, Dist[ff*(Sqrt[Cos[e + f*x]^2]/(f*Cos[e + f*x])), 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/2] &&  !IntegerQ
[p]

Rubi steps

\begin {align*} \int \frac {\sec ^2(e+f x)}{\left (a+b \sin ^2(e+f x)\right )^{5/2}} \, dx &=\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {1}{\left (1-x^2\right )^{3/2} \left (a+b x^2\right )^{5/2}} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\tan (e+f x)}{(a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {b+3 b x^2}{\sqrt {1-x^2} \left (a+b x^2\right )^{5/2}} \, dx,x,\sin (e+f x)\right )}{(a+b) f}\\ &=-\frac {(3 a-b) b \cos (e+f x) \sin (e+f x)}{3 a (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\tan (e+f x)}{(a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {-2 b (3 a+b)-(3 a-b) b x^2}{\sqrt {1-x^2} \left (a+b x^2\right )^{3/2}} \, dx,x,\sin (e+f x)\right )}{3 a (a+b)^2 f}\\ &=-\frac {(3 a-b) b \cos (e+f x) \sin (e+f x)}{3 a (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {b \left (3 a^2-7 a b-2 b^2\right ) \cos (e+f x) \sin (e+f x)}{3 a^2 (a+b)^3 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\tan (e+f x)}{(a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\left (\sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {a b (9 a+b)-b \left (3 a^2-7 a b-2 b^2\right ) x^2}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{3 a^2 (a+b)^3 f}\\ &=-\frac {(3 a-b) b \cos (e+f x) \sin (e+f x)}{3 a (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {b \left (3 a^2-7 a b-2 b^2\right ) \cos (e+f x) \sin (e+f x)}{3 a^2 (a+b)^3 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\tan (e+f x)}{(a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\left ((3 a-b) \sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{3 a (a+b)^2 f}+\frac {\left (\left (-3 a^2+7 a b+2 b^2\right ) \sqrt {\cos ^2(e+f x)} \sec (e+f x)\right ) \text {Subst}\left (\int \frac {\sqrt {a+b x^2}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{3 a^2 (a+b)^3 f}\\ &=-\frac {(3 a-b) b \cos (e+f x) \sin (e+f x)}{3 a (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {b \left (3 a^2-7 a b-2 b^2\right ) \cos (e+f x) \sin (e+f x)}{3 a^2 (a+b)^3 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\tan (e+f x)}{(a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}+\frac {\left (\left (-3 a^2+7 a b+2 b^2\right ) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}\right ) \text {Subst}\left (\int \frac {\sqrt {1+\frac {b x^2}{a}}}{\sqrt {1-x^2}} \, dx,x,\sin (e+f x)\right )}{3 a^2 (a+b)^3 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {\left ((3 a-b) \sqrt {\cos ^2(e+f x)} \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}\right ) \text {Subst}\left (\int \frac {1}{\sqrt {1-x^2} \sqrt {1+\frac {b x^2}{a}}} \, dx,x,\sin (e+f x)\right )}{3 a (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}\\ &=-\frac {(3 a-b) b \cos (e+f x) \sin (e+f x)}{3 a (a+b)^2 f \left (a+b \sin ^2(e+f x)\right )^{3/2}}-\frac {b \left (3 a^2-7 a b-2 b^2\right ) \cos (e+f x) \sin (e+f x)}{3 a^2 (a+b)^3 f \sqrt {a+b \sin ^2(e+f x)}}-\frac {\left (3 a^2-7 a b-2 b^2\right ) \sqrt {\cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{3 a^2 (a+b)^3 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}+\frac {(3 a-b) \sqrt {\cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{3 a (a+b)^2 f \sqrt {a+b \sin ^2(e+f x)}}+\frac {\tan (e+f x)}{(a+b) f \left (a+b \sin ^2(e+f x)\right )^{3/2}}\\ \end {align*}

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Mathematica [A]
time = 2.33, size = 245, normalized size = 0.85 \begin {gather*} \frac {-2 a^2 \left (3 a^2-7 a b-2 b^2\right ) \left (\frac {2 a+b-b \cos (2 (e+f x))}{a}\right )^{3/2} E\left (e+f x\left |-\frac {b}{a}\right .\right )+2 a^2 \left (3 a^2+2 a b-b^2\right ) \left (\frac {2 a+b-b \cos (2 (e+f x))}{a}\right )^{3/2} F\left (e+f x\left |-\frac {b}{a}\right .\right )+\frac {\left (24 a^4+24 a^3 b+41 a^2 b^2+19 a b^3+2 b^4-4 a b \left (6 a^2-5 a b-3 b^2\right ) \cos (2 (e+f x))+b^2 \left (3 a^2-7 a b-2 b^2\right ) \cos (4 (e+f x))\right ) \tan (e+f x)}{\sqrt {2}}}{6 a^2 (a+b)^3 f (2 a+b-b \cos (2 (e+f x)))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*a^2*(3*a^2 - 7*a*b - 2*b^2)*((2*a + b - b*Cos[2*(e + f*x)])/a)^(3/2)*EllipticE[e + f*x, -(b/a)] + 2*a^2*(3
*a^2 + 2*a*b - b^2)*((2*a + b - b*Cos[2*(e + f*x)])/a)^(3/2)*EllipticF[e + f*x, -(b/a)] + ((24*a^4 + 24*a^3*b
+ 41*a^2*b^2 + 19*a*b^3 + 2*b^4 - 4*a*b*(6*a^2 - 5*a*b - 3*b^2)*Cos[2*(e + f*x)] + b^2*(3*a^2 - 7*a*b - 2*b^2)
*Cos[4*(e + f*x)])*Tan[e + f*x])/Sqrt[2])/(6*a^2*(a + b)^3*f*(2*a + b - b*Cos[2*(e + f*x)])^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1081\) vs. \(2(308)=616\).
time = 17.94, size = 1082, normalized size = 3.76

method	result	size
default	\(\text {Expression too large to display}\)	\(1082\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/3*((-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*b^2*(3*a^2-7*a*b-2*b^2)*sin(f*x+e)*cos(f*x+e)^4-2*(-b*cos(f*x+
e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*b*(3*a^3-a^2*b-5*a*b^2-b^3)*cos(f*x+e)^2*sin(f*x+e)+3*(-b*cos(f*x+e)^4+(a+b)*co
s(f*x+e)^2)^(1/2)*a^2*(a^2+2*a*b+b^2)*sin(f*x+e)-(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*(cos(f*x+e)^2)^(1/2)*(-b*co
s(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*a*b*(3*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a^2+2*EllipticF(sin(f*x+e),(-
1/a*b)^(1/2))*a*b-EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*b^2-3*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^2+7*Ellipt
icE(sin(f*x+e),(-1/a*b)^(1/2))*a*b+2*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*b^2)*cos(f*x+e)^2+3*(cos(f*x+e)^2)^(
1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^
2)^(1/2)*a^4+5*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*(-b
*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*a^3*b+(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticF
(sin(f*x+e),(-1/a*b)^(1/2))*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*a^2*b^2-(cos(f*x+e)^2)^(1/2)*(-b/a*cos(
f*x+e)^2+(a+b)/a)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*a*b^3-
3*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*(-b*cos(f*x+e)^4
+(a+b)*cos(f*x+e)^2)^(1/2)*a^4+4*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*EllipticE(sin(f*x+e),(
-1/a*b)^(1/2))*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*a^3*b+9*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b
)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*a^2*b^2+2*(cos(f*x+
e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(1/2)*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*EllipticE(sin(f*x+e),
(-1/a*b)^(1/2))*a*b^3)/(-(a+b*sin(f*x+e)^2)*(sin(f*x+e)-1)*(1+sin(f*x+e)))^(1/2)/(a+b*sin(f*x+e)^2)^(3/2)/a^2/
(a+b)^3/cos(f*x+e)/f

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [C] Result contains complex when optimal does not.
time = 0.27, size = 1746, normalized size = 6.06 \begin {gather*} \text {Too large to display} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/6*((2*((-3*I*a^2*b^3 + 7*I*a*b^4 + 2*I*b^5)*cos(f*x + e)^5 - 2*(-3*I*a^3*b^2 + 4*I*a^2*b^3 + 9*I*a*b^4 + 2*I
*b^5)*cos(f*x + e)^3 + (-3*I*a^4*b + I*a^3*b^2 + 13*I*a^2*b^3 + 11*I*a*b^4 + 2*I*b^5)*cos(f*x + e))*sqrt(-b)*s
qrt((a^2 + a*b)/b^2) - ((6*I*a^3*b^2 - 11*I*a^2*b^3 - 11*I*a*b^4 - 2*I*b^5)*cos(f*x + e)^5 + 2*(-6*I*a^4*b + 5
*I*a^3*b^2 + 22*I*a^2*b^3 + 13*I*a*b^4 + 2*I*b^5)*cos(f*x + e)^3 + (6*I*a^5 + I*a^4*b - 27*I*a^3*b^2 - 35*I*a^
2*b^3 - 15*I*a*b^4 - 2*I*b^5)*cos(f*x + e))*sqrt(-b))*sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_e
(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*(cos(f*x + e) + I*sin(f*x + e))), (8*a^2 + 8*a*b + b^2 -
4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))/b^2) + (2*((3*I*a^2*b^3 - 7*I*a*b^4 - 2*I*b^5)*cos(f*x + e)^5 - 2*(3*I
*a^3*b^2 - 4*I*a^2*b^3 - 9*I*a*b^4 - 2*I*b^5)*cos(f*x + e)^3 + (3*I*a^4*b - I*a^3*b^2 - 13*I*a^2*b^3 - 11*I*a*
b^4 - 2*I*b^5)*cos(f*x + e))*sqrt(-b)*sqrt((a^2 + a*b)/b^2) - ((-6*I*a^3*b^2 + 11*I*a^2*b^3 + 11*I*a*b^4 + 2*I
*b^5)*cos(f*x + e)^5 + 2*(6*I*a^4*b - 5*I*a^3*b^2 - 22*I*a^2*b^3 - 13*I*a*b^4 - 2*I*b^5)*cos(f*x + e)^3 + (-6*
I*a^5 - I*a^4*b + 27*I*a^3*b^2 + 35*I*a^2*b^3 + 15*I*a*b^4 + 2*I*b^5)*cos(f*x + e))*sqrt(-b))*sqrt((2*b*sqrt((
a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_e(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*(cos(f*x + e) -
I*sin(f*x + e))), (8*a^2 + 8*a*b + b^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))/b^2) - 2*(4*((3*I*a^2*b^3 + 4*
I*a*b^4 + I*b^5)*cos(f*x + e)^5 + 2*(-3*I*a^3*b^2 - 7*I*a^2*b^3 - 5*I*a*b^4 - I*b^5)*cos(f*x + e)^3 + (3*I*a^4
*b + 10*I*a^3*b^2 + 12*I*a^2*b^3 + 6*I*a*b^4 + I*b^5)*cos(f*x + e))*sqrt(-b)*sqrt((a^2 + a*b)/b^2) + ((-18*I*a
^3*b^2 - 11*I*a^2*b^3 - I*a*b^4)*cos(f*x + e)^5 + 2*(18*I*a^4*b + 29*I*a^3*b^2 + 12*I*a^2*b^3 + I*a*b^4)*cos(f
*x + e)^3 + (-18*I*a^5 - 47*I*a^4*b - 41*I*a^3*b^2 - 13*I*a^2*b^3 - I*a*b^4)*cos(f*x + e))*sqrt(-b))*sqrt((2*b
*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_f(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*(cos(f*x
+ e) + I*sin(f*x + e))), (8*a^2 + 8*a*b + b^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))/b^2) - 2*(4*((-3*I*a^2*
b^3 - 4*I*a*b^4 - I*b^5)*cos(f*x + e)^5 + 2*(3*I*a^3*b^2 + 7*I*a^2*b^3 + 5*I*a*b^4 + I*b^5)*cos(f*x + e)^3 + (
-3*I*a^4*b - 10*I*a^3*b^2 - 12*I*a^2*b^3 - 6*I*a*b^4 - I*b^5)*cos(f*x + e))*sqrt(-b)*sqrt((a^2 + a*b)/b^2) + (
(18*I*a^3*b^2 + 11*I*a^2*b^3 + I*a*b^4)*cos(f*x + e)^5 + 2*(-18*I*a^4*b - 29*I*a^3*b^2 - 12*I*a^2*b^3 - I*a*b^
4)*cos(f*x + e)^3 + (18*I*a^5 + 47*I*a^4*b + 41*I*a^3*b^2 + 13*I*a^2*b^3 + I*a*b^4)*cos(f*x + e))*sqrt(-b))*sq
rt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*elliptic_f(arcsin(sqrt((2*b*sqrt((a^2 + a*b)/b^2) + 2*a + b)/b)*(c
os(f*x + e) - I*sin(f*x + e))), (8*a^2 + 8*a*b + b^2 - 4*(2*a*b + b^2)*sqrt((a^2 + a*b)/b^2))/b^2) + 2*(3*a^4*
b + 6*a^3*b^2 + 3*a^2*b^3 + (3*a^2*b^3 - 7*a*b^4 - 2*b^5)*cos(f*x + e)^4 - 2*(3*a^3*b^2 - a^2*b^3 - 5*a*b^4 -
b^5)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e))/((a^5*b^3 + 3*a^4*b^4 + 3*a^3*b^5 + a^2*b^6
)*f*cos(f*x + e)^5 - 2*(a^6*b^2 + 4*a^5*b^3 + 6*a^4*b^4 + 4*a^3*b^5 + a^2*b^6)*f*cos(f*x + e)^3 + (a^7*b + 5*a
^6*b^2 + 10*a^5*b^3 + 10*a^4*b^4 + 5*a^3*b^5 + a^2*b^6)*f*cos(f*x + e))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{2}{\left (e + f x \right )}}{\left (a + b \sin ^{2}{\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral(sec(e + f*x)**2/(a + b*sin(e + f*x)**2)**(5/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\cos \left (e+f\,x\right )}^2\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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