3.344 \(\int \sec ^4(e+f x) (a+b \sin ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=236 \[ \frac{2 (a-b) \tan (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 f}+\frac{(a+b) \tan (e+f x) \sec ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 f}+\frac{a (2 a-b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{2 (a-b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 f \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}} \]

[Out]

(-2*(a - b)*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*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a]) + (a*(2*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*f*Sqrt[a + b*Sin[e + f*x]^2]) + (2*(a - b)*Sqrt[a + b*S
in[e + f*x]^2]*Tan[e + f*x])/(3*f) + ((a + b)*Sec[e + f*x]^2*Sqrt[a + b*Sin[e + f*x]^2]*Tan[e + f*x])/(3*f)

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Rubi [A]  time = 0.256409, antiderivative size = 236, normalized size of antiderivative = 1., number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.32, Rules used = {3192, 413, 527, 524, 426, 424, 421, 419} \[ \frac{2 (a-b) \tan (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 f}+\frac{(a+b) \tan (e+f x) \sec ^2(e+f x) \sqrt{a+b \sin ^2(e+f x)}}{3 f}+\frac{a (2 a-b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 f \sqrt{a+b \sin ^2(e+f x)}}-\frac{2 (a-b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{3 f \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[e + f*x]^4*(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

(-2*(a - b)*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*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a]) + (a*(2*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*f*Sqrt[a + b*Sin[e + f*x]^2]) + (2*(a - b)*Sqrt[a + b*S
in[e + f*x]^2]*Tan[e + f*x])/(3*f) + ((a + b)*Sec[e + f*x]^2*Sqrt[a + b*Sin[e + f*x]^2]*Tan[e + f*x])/(3*f)

Rule 3192

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]

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

Mathematica [A]  time = 2.03666, size = 190, normalized size = 0.81 \[ \frac{\frac{\tan (e+f x) \sec ^2(e+f x) \left (\left (4 a^2-6 a b-2 b^2\right ) \cos (2 (e+f x))+8 a^2+b (b-a) \cos (4 (e+f x))+3 a b+b^2\right )}{\sqrt{2}}+2 a (2 a-b) \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} F\left (e+f x\left |-\frac{b}{a}\right .\right )-4 a (a-b) \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{6 f \sqrt{2 a-b \cos (2 (e+f x))+b}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[e + f*x]^4*(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

(-4*a*(a - b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, -(b/a)] + 2*a*(2*a - b)*Sqrt[(2*a + b
- b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)] + ((8*a^2 + 3*a*b + b^2 + (4*a^2 - 6*a*b - 2*b^2)*Cos[2*(e
+ f*x)] + b*(-a + b)*Cos[4*(e + f*x)])*Sec[e + f*x]^2*Tan[e + f*x])/Sqrt[2])/(6*f*Sqrt[2*a + b - b*Cos[2*(e +
f*x)]])

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Maple [A]  time = 2.043, size = 375, normalized size = 1.6 \begin{align*}{\frac{1}{ \left ( -3+3\,\sin \left ( fx+e \right ) \right ) \left ( 1+\sin \left ( fx+e \right ) \right ) \cos \left ( fx+e \right ) f} \left ( 2\,\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}b \left ( a-b \right ) \sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}-\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ( 2\,{a}^{2}-ab-3\,{b}^{2} \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) -\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \left ({a}^{2}+2\,ab+{b}^{2} \right ) \sin \left ( fx+e \right ) -\sqrt{-{\frac{b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}{a}}+{\frac{a+b}{a}}}\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{-b \left ( \cos \left ( fx+e \right ) \right ) ^{4}+ \left ( a+b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2}}a \left ( 2\,{\it EllipticF} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) a-{\it EllipticF} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) b-2\,{\it EllipticE} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) a+2\,{\it EllipticE} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) b \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{2} \right ){\frac{1}{\sqrt{- \left ( a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) \left ( -1+\sin \left ( fx+e \right ) \right ) \left ( 1+\sin \left ( fx+e \right ) \right ) }}}{\frac{1}{\sqrt{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(f*x+e)^4*(a+b*sin(f*x+e)^2)^(3/2),x)

[Out]

1/3*(2*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*b*(a-b)*sin(f*x+e)*cos(f*x+e)^4-(-b*cos(f*x+e)^4+(a+b)*cos(f
*x+e)^2)^(1/2)*(2*a^2-a*b-3*b^2)*cos(f*x+e)^2*sin(f*x+e)-(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*(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*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(
1/2)*a*(2*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a-EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*b-2*EllipticE(sin(f*x+e)
,(-1/a*b)^(1/2))*a+2*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*b)*cos(f*x+e)^2)/(-(a+b*sin(f*x+e)^2)*(-1+sin(f*x+e)
)*(1+sin(f*x+e)))^(1/2)/(-1+sin(f*x+e))/(1+sin(f*x+e))/cos(f*x+e)/(a+b*sin(f*x+e)^2)^(1/2)/f

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \sec \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

integrate((b*sin(f*x + e)^2 + a)^(3/2)*sec(f*x + e)^4, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (-{\left (b \cos \left (f x + e\right )^{2} - a - b\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sec \left (f x + e\right )^{4}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

integral(-(b*cos(f*x + e)^2 - a - b)*sqrt(-b*cos(f*x + e)^2 + a + b)*sec(f*x + e)^4, 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)**4*(a+b*sin(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 \sin \left (f x + e\right )^{2} + a\right )}^{\frac{3}{2}} \sec \left (f x + e\right )^{4}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(f*x+e)^4*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

integrate((b*sin(f*x + e)^2 + a)^(3/2)*sec(f*x + e)^4, x)