3.4.59 \(\int \frac {\cos ^4(e+f x)}{(a+b \sin ^2(e+f x))^{3/2}} \, dx\) [359]

Optimal. Leaf size=202 \[ \frac {(a+b) \cos (e+f x) \sin (e+f x)}{a b f \sqrt {a+b \sin ^2(e+f x)}}+\frac {(2 a+b) \sqrt {\cos ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{a b^2 f \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}-\frac {2 (a+b) \sqrt {\cos ^2(e+f x)} F\left (\sin ^{-1}(\sin (e+f x))|-\frac {b}{a}\right ) \sec (e+f x) \sqrt {1+\frac {b \sin ^2(e+f x)}{a}}}{b^2 f \sqrt {a+b \sin ^2(e+f x)}} \]

[Out]

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

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Rubi [A]
time = 0.13, antiderivative size = 202, normalized size of antiderivative = 1.00, 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 = {3271, 424, 538, 437, 435, 432, 430} \begin {gather*} -\frac {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 (\text {ArcSin}(\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{b^2 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 (\text {ArcSin}(\sin (e+f x))\left |-\frac {b}{a}\right .\right )}{a b^2 f \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}+\frac {(a+b) \sin (e+f x) \cos (e+f x)}{a b f \sqrt {a+b \sin ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 424

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

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Mathematica [A]
time = 0.41, size = 139, normalized size = 0.69 \begin {gather*} \frac {2 a (2 a+b) \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} E\left (e+f x\left |-\frac {b}{a}\right .\right )-(a+b) \left (4 a \sqrt {\frac {2 a+b-b \cos (2 (e+f x))}{a}} F\left (e+f x\left |-\frac {b}{a}\right .\right )-\sqrt {2} b \sin (2 (e+f x))\right )}{2 a b^2 f \sqrt {2 a+b-b \cos (2 (e+f x))}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*a*(2*a + b)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, -(b/a)] - (a + b)*(4*a*Sqrt[(2*a + b
- b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)] - Sqrt[2]*b*Sin[2*(e + f*x)]))/(2*a*b^2*f*Sqrt[2*a + b - b
*Cos[2*(e + f*x)]])

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Maple [A]
time = 8.31, size = 269, normalized size = 1.33

method	result	size
default	\(\frac {\left (a b +b^{2}\right ) \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right )+2 \EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, a^{2}+\EllipticE \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, a b -2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a^{2}-2 \sqrt {\frac {\cos \left (2 f x +2 e \right )}{2}+\frac {1}{2}}\, \sqrt {-\frac {b \left (\cos ^{2}\left (f x +e \right )\right )}{a}+\frac {a +b}{a}}\, \EllipticF \left (\sin \left (f x +e \right ), \sqrt {-\frac {b}{a}}\right ) a b}{a \,b^{2} \cos \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\, f}\)	\(269\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

((a*b+b^2)*sin(f*x+e)*cos(f*x+e)^2+2*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e
)^2+(a+b)/a)^(1/2)*a^2+EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*(cos(f*x+e)^2)^(1/2)*(-b/a*cos(f*x+e)^2+(a+b)/a)^(
1/2)*a*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))*a^2-2*(
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))*a*b)/a/b^2/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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Fricas [F]
time = 0.11, size = 69, normalized size = 0.34 \begin {gather*} {\rm integral}\left (\frac {\sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \cos \left (f x + e\right )^{4}}{b^{2} \cos \left (f x + e\right )^{4} - 2 \, {\left (a b + b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 2 \, a b + b^{2}}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integral(sqrt(-b*cos(f*x + e)^2 + a + b)*cos(f*x + e)^4/(b^2*cos(f*x + e)^4 - 2*(a*b + b^2)*cos(f*x + e)^2 + a
^2 + 2*a*b + b^2), x)

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Sympy [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: SystemError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: SystemError >> excessive stack use: stack is 4847 deep

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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