3.242 \(\int \sec ^3(e+f x) (a+b \sec ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=371 \[ \frac{(a+b) (9 a+8 b) \sqrt{\cos ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )} \text{EllipticF}\left (\sin ^{-1}(\sin (e+f x)),\frac{a}{a+b}\right )}{15 f \left (-a \sin ^2(e+f x)+a+b\right )}+\frac{\left (3 a^2+13 a b+8 b^2\right ) \sin (e+f x) \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}{15 b f}-\frac{\left (3 a^2+13 a b+8 b^2\right ) \sqrt{\cos ^2(e+f x)} \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )} E\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right )}{15 b f \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}}}+\frac{b \tan (e+f x) \sec ^3(e+f x) \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}{5 f}+\frac{2 (3 a+2 b) \tan (e+f x) \sec (e+f x) \sqrt{\sec ^2(e+f x) \left (-a \sin ^2(e+f x)+a+b\right )}}{15 f} \]

[Out]

((3*a^2 + 13*a*b + 8*b^2)*Sin[e + f*x]*Sqrt[Sec[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)])/(15*b*f) - ((3*a^2 + 1
3*a*b + 8*b^2)*Sqrt[Cos[e + f*x]^2]*EllipticE[ArcSin[Sin[e + f*x]], a/(a + b)]*Sqrt[Sec[e + f*x]^2*(a + b - a*
Sin[e + f*x]^2)])/(15*b*f*Sqrt[1 - (a*Sin[e + f*x]^2)/(a + b)]) + ((a + b)*(9*a + 8*b)*Sqrt[Cos[e + f*x]^2]*El
lipticF[ArcSin[Sin[e + f*x]], a/(a + b)]*Sqrt[Sec[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)]*Sqrt[1 - (a*Sin[e + f
*x]^2)/(a + b)])/(15*f*(a + b - a*Sin[e + f*x]^2)) + (2*(3*a + 2*b)*Sec[e + f*x]*Sqrt[Sec[e + f*x]^2*(a + b -
a*Sin[e + f*x]^2)]*Tan[e + f*x])/(15*f) + (b*Sec[e + f*x]^3*Sqrt[Sec[e + f*x]^2*(a + b - a*Sin[e + f*x]^2)]*Ta
n[e + f*x])/(5*f)

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Rubi [A]  time = 0.700338, antiderivative size = 470, normalized size of antiderivative = 1.27, number of steps used = 11, number of rules used = 10, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.4, Rules used = {4148, 6722, 1974, 413, 527, 524, 426, 424, 421, 419} \[ \frac{\left (3 a^2+13 a b+8 b^2\right ) \sin (e+f x) \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a+b \sec ^2(e+f x)}}{15 b f \sqrt{a \cos ^2(e+f x)+b}}-\frac{\left (3 a^2+13 a b+8 b^2\right ) \sqrt{\cos ^2(e+f x)} \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a+b \sec ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right )}{15 b f \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \sqrt{a \cos ^2(e+f x)+b}}+\frac{b \tan (e+f x) \sec ^3(e+f x) \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a+b \sec ^2(e+f x)}}{5 f \sqrt{a \cos ^2(e+f x)+b}}+\frac{2 (3 a+2 b) \tan (e+f x) \sec (e+f x) \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a+b \sec ^2(e+f x)}}{15 f \sqrt{a \cos ^2(e+f x)+b}}+\frac{(a+b) (9 a+8 b) \sqrt{\cos ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}} \sqrt{a+b \sec ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right )}{15 f \sqrt{-a \sin ^2(e+f x)+a+b} \sqrt{a \cos ^2(e+f x)+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

((3*a^2 + 13*a*b + 8*b^2)*Sqrt[a + b*Sec[e + f*x]^2]*Sin[e + f*x]*Sqrt[a + b - a*Sin[e + f*x]^2])/(15*b*f*Sqrt
[b + a*Cos[e + f*x]^2]) - ((3*a^2 + 13*a*b + 8*b^2)*Sqrt[Cos[e + f*x]^2]*EllipticE[ArcSin[Sin[e + f*x]], a/(a
+ b)]*Sqrt[a + b*Sec[e + f*x]^2]*Sqrt[a + b - a*Sin[e + f*x]^2])/(15*b*f*Sqrt[b + a*Cos[e + f*x]^2]*Sqrt[1 - (
a*Sin[e + f*x]^2)/(a + b)]) + ((a + b)*(9*a + 8*b)*Sqrt[Cos[e + f*x]^2]*EllipticF[ArcSin[Sin[e + f*x]], a/(a +
b)]*Sqrt[a + b*Sec[e + f*x]^2]*Sqrt[1 - (a*Sin[e + f*x]^2)/(a + b)])/(15*f*Sqrt[b + a*Cos[e + f*x]^2]*Sqrt[a
+ b - a*Sin[e + f*x]^2]) + (2*(3*a + 2*b)*Sec[e + f*x]*Sqrt[a + b*Sec[e + f*x]^2]*Sqrt[a + b - a*Sin[e + f*x]^
2]*Tan[e + f*x])/(15*f*Sqrt[b + a*Cos[e + f*x]^2]) + (b*Sec[e + f*x]^3*Sqrt[a + b*Sec[e + f*x]^2]*Sqrt[a + b -
a*Sin[e + f*x]^2]*Tan[e + f*x])/(5*f*Sqrt[b + a*Cos[e + f*x]^2])

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]

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 413

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

Rule 527

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 524

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 426

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

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 421

Int[1/(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[1/(Sqrt[a + b*x^2]*Sqrt[1 + (d*x^2)/c]), x], x] /; FreeQ[{a, b, c, d}, x] &&  !GtQ[c, 0]

Rule 419

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

Rubi steps

\begin{align*} \int \sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \frac{\left (a+\frac{b}{1-x^2}\right )^{3/2}}{\left (1-x^2\right )^2} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac{\left (\sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{\left (b+a \left (1-x^2\right )\right )^{3/2}}{\left (1-x^2\right )^{7/2}} \, dx,x,\sin (e+f x)\right )}{f \sqrt{b+a \cos ^2(e+f x)}}\\ &=\frac{\left (\sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{\left (a+b-a x^2\right )^{3/2}}{\left (1-x^2\right )^{7/2}} \, dx,x,\sin (e+f x)\right )}{f \sqrt{b+a \cos ^2(e+f x)}}\\ &=\frac{b \sec ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)} \tan (e+f x)}{5 f \sqrt{b+a \cos ^2(e+f x)}}-\frac{\left (\sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{-(a+b) (5 a+4 b)+a (5 a+3 b) x^2}{\left (1-x^2\right )^{5/2} \sqrt{a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{5 f \sqrt{b+a \cos ^2(e+f x)}}\\ &=\frac{2 (3 a+2 b) \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)} \tan (e+f x)}{15 f \sqrt{b+a \cos ^2(e+f x)}}+\frac{b \sec ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)} \tan (e+f x)}{5 f \sqrt{b+a \cos ^2(e+f x)}}-\frac{\left (\sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{-b (a+b) (9 a+8 b)+2 a b (3 a+2 b) x^2}{\left (1-x^2\right )^{3/2} \sqrt{a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{15 b f \sqrt{b+a \cos ^2(e+f x)}}\\ &=\frac{\left (3 a^2+13 a b+8 b^2\right ) \sqrt{a+b \sec ^2(e+f x)} \sin (e+f x) \sqrt{a+b-a \sin ^2(e+f x)}}{15 b f \sqrt{b+a \cos ^2(e+f x)}}+\frac{2 (3 a+2 b) \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)} \tan (e+f x)}{15 f \sqrt{b+a \cos ^2(e+f x)}}+\frac{b \sec ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)} \tan (e+f x)}{5 f \sqrt{b+a \cos ^2(e+f x)}}-\frac{\left (\sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{a b (a+b) (3 a+4 b)-a b \left (3 a^2+13 a b+8 b^2\right ) x^2}{\sqrt{1-x^2} \sqrt{a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{15 b^2 f \sqrt{b+a \cos ^2(e+f x)}}\\ &=\frac{\left (3 a^2+13 a b+8 b^2\right ) \sqrt{a+b \sec ^2(e+f x)} \sin (e+f x) \sqrt{a+b-a \sin ^2(e+f x)}}{15 b f \sqrt{b+a \cos ^2(e+f x)}}+\frac{2 (3 a+2 b) \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)} \tan (e+f x)}{15 f \sqrt{b+a \cos ^2(e+f x)}}+\frac{b \sec ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)} \tan (e+f x)}{5 f \sqrt{b+a \cos ^2(e+f x)}}+\frac{\left ((a+b) (9 a+8 b) \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{a+b-a x^2}} \, dx,x,\sin (e+f x)\right )}{15 f \sqrt{b+a \cos ^2(e+f x)}}-\frac{\left (\left (3 a^2+13 a b+8 b^2\right ) \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b-a x^2}}{\sqrt{1-x^2}} \, dx,x,\sin (e+f x)\right )}{15 b f \sqrt{b+a \cos ^2(e+f x)}}\\ &=\frac{\left (3 a^2+13 a b+8 b^2\right ) \sqrt{a+b \sec ^2(e+f x)} \sin (e+f x) \sqrt{a+b-a \sin ^2(e+f x)}}{15 b f \sqrt{b+a \cos ^2(e+f x)}}+\frac{2 (3 a+2 b) \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)} \tan (e+f x)}{15 f \sqrt{b+a \cos ^2(e+f x)}}+\frac{b \sec ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)} \tan (e+f x)}{5 f \sqrt{b+a \cos ^2(e+f x)}}-\frac{\left (\left (3 a^2+13 a b+8 b^2\right ) \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1-\frac{a x^2}{a+b}}}{\sqrt{1-x^2}} \, dx,x,\sin (e+f x)\right )}{15 b f \sqrt{b+a \cos ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}}}+\frac{\left ((a+b) (9 a+8 b) \sqrt{\cos ^2(e+f x)} \sqrt{a+b \sec ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}}\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1-x^2} \sqrt{1-\frac{a x^2}{a+b}}} \, dx,x,\sin (e+f x)\right )}{15 f \sqrt{b+a \cos ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}\\ &=\frac{\left (3 a^2+13 a b+8 b^2\right ) \sqrt{a+b \sec ^2(e+f x)} \sin (e+f x) \sqrt{a+b-a \sin ^2(e+f x)}}{15 b f \sqrt{b+a \cos ^2(e+f x)}}-\frac{\left (3 a^2+13 a b+8 b^2\right ) \sqrt{\cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right ) \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}{15 b f \sqrt{b+a \cos ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}}}+\frac{(a+b) (9 a+8 b) \sqrt{\cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|\frac{a}{a+b}\right ) \sqrt{a+b \sec ^2(e+f x)} \sqrt{1-\frac{a \sin ^2(e+f x)}{a+b}}}{15 f \sqrt{b+a \cos ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)}}+\frac{2 (3 a+2 b) \sec (e+f x) \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)} \tan (e+f x)}{15 f \sqrt{b+a \cos ^2(e+f x)}}+\frac{b \sec ^3(e+f x) \sqrt{a+b \sec ^2(e+f x)} \sqrt{a+b-a \sin ^2(e+f x)} \tan (e+f x)}{5 f \sqrt{b+a \cos ^2(e+f x)}}\\ \end{align*}

Mathematica [F]  time = 16.4127, size = 0, normalized size = 0. \[ \int \sec ^3(e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

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Maple [C]  time = 0.654, size = 6562, normalized size = 17.7 \begin{align*} \text{output too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

result too large to display

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \sec \left (f x + e\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (b \sec \left (f x + e\right )^{5} + a \sec \left (f x + e\right )^{3}\right )} \sqrt{b \sec \left (f x + e\right )^{2} + a}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral((b*sec(f*x + e)^5 + a*sec(f*x + e)^3)*sqrt(b*sec(f*x + e)^2 + a), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sec \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \sec \left (f x + e\right )^{3}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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