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]

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

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Rubi [A]  time = 0.26, antiderivative size = 236, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.320, 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 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 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)])

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 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 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 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 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 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]

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*}

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Mathematica [A]  time = 2.03, 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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fricas [F]  time = 0.45, size = 0, normalized size = 0.00 \[ {\rm integral}\left (-{\left (b \cos \left (f x + e\right )^{2} - a - b\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sec \left (f x + e\right )^{4}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sec \left (f x + e\right )^{4}\,{d x} \]

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)

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maple [A]  time = 2.69, size = 375, normalized size = 1.59 \[ \frac {2 \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, b \left (a -b \right ) \sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right )-\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (2 a^{2}-a b -3 b^{2}\right ) \left (\cos ^{2}\left (f x +e \right )\right ) \sin \left (f x +e \right )-\sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \left (a^{2}+2 a b +b^{2}\right ) \sin \left (f x +e \right )-\sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-b \left (\cos ^{4}\left (f x +e \right )\right )+\left (a +b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, a \left (2 \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a -\EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b -2 \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a +2 \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) b \right ) \left (\cos ^{2}\left (f x +e \right )\right )}{3 \sqrt {-\left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right ) \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right )}\, \left (\sin \left (f x +e \right )-1\right ) \left (1+\sin \left (f x +e \right )\right ) \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f} \]

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)-(cos(f*x+e)^2)^(1/2)*(-b*cos(f*x+e)^4+(a+b)*cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(
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)*(sin(f*x+e)-1)
*(1+sin(f*x+e)))^(1/2)/(sin(f*x+e)-1)/(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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sec \left (f x + e\right )^{4}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}}{{\cos \left (e+f\,x\right )}^4} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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