3.341 \(\int \cos ^2(e+f x) (a+b \sin ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=259 \[ -\frac{\left (3 a^2-7 a b-2 b^2\right ) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{15 b f \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}}-\frac{b \sin (e+f x) \cos ^3(e+f x) \sqrt{a+b \sin ^2(e+f x)}}{5 f}+\frac{2 (3 a+b) \sin (e+f x) \cos (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{15 f}+\frac{a (3 a-b) (a+b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{15 b f \sqrt{a+b \sin ^2(e+f x)}} \]

[Out]

(2*(3*a + b)*Cos[e + f*x]*Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(15*f) - (b*Cos[e + f*x]^3*Sin[e + f*x]*Sqr
t[a + b*Sin[e + f*x]^2])/(5*f) - ((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])/(15*b*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a]) + (a*(3*a - b)*(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])/(
15*b*f*Sqrt[a + b*Sin[e + f*x]^2])

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Rubi [A]  time = 0.284961, antiderivative size = 259, 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, 416, 528, 524, 426, 424, 421, 419} \[ -\frac{\left (3 a^2-7 a b-2 b^2\right ) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{a+b \sin ^2(e+f x)} E\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{15 b f \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}}-\frac{b \sin (e+f x) \cos ^3(e+f x) \sqrt{a+b \sin ^2(e+f x)}}{5 f}+\frac{2 (3 a+b) \sin (e+f x) \cos (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{15 f}+\frac{a (3 a-b) (a+b) \sqrt{\cos ^2(e+f x)} \sec (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} F\left (\sin ^{-1}(\sin (e+f x))|-\frac{b}{a}\right )}{15 b f \sqrt{a+b \sin ^2(e+f x)}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(2*(3*a + b)*Cos[e + f*x]*Sin[e + f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(15*f) - (b*Cos[e + f*x]^3*Sin[e + f*x]*Sqr
t[a + b*Sin[e + f*x]^2])/(5*f) - ((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])/(15*b*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a]) + (a*(3*a - b)*(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])/(
15*b*f*Sqrt[a + b*Sin[e + f*x]^2])

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 416

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

Rule 528

Int[((a_) + (b_.)*(x_)^(n_))^(p_.)*((c_) + (d_.)*(x_)^(n_))^(q_.)*((e_) + (f_.)*(x_)^(n_)), x_Symbol] :> Simp[
(f*x*(a + b*x^n)^(p + 1)*(c + d*x^n)^q)/(b*(n*(p + q + 1) + 1)), x] + Dist[1/(b*(n*(p + q + 1) + 1)), Int[(a +
b*x^n)^p*(c + d*x^n)^(q - 1)*Simp[c*(b*e - a*f + b*e*n*(p + q + 1)) + (d*(b*e - a*f) + f*n*q*(b*c - a*d) + b*
d*e*n*(p + q + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, f, n, p}, x] && GtQ[q, 0] && NeQ[n*(p + q + 1) + 1
, 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 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 \cos ^2(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 \sqrt{1-x^2} \left (a+b x^2\right )^{3/2} \, dx,x,\sin (e+f x)\right )}{f}\\ &=-\frac{b \cos ^3(e+f x) \sin (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{5 f}-\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{1-x^2} \left (-a (5 a+b)-2 b (3 a+b) x^2\right )}{\sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{5 f}\\ &=\frac{2 (3 a+b) \cos (e+f x) \sin (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{15 f}-\frac{b \cos ^3(e+f x) \sin (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{5 f}-\frac{\left (\sqrt{\cos ^2(e+f x)} \sec (e+f x)\right ) \operatorname{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 )}{15 b f}\\ &=\frac{2 (3 a+b) \cos (e+f x) \sin (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{15 f}-\frac{b \cos ^3(e+f x) \sin (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{5 f}+\frac{\left (a (3 a-b) (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 )}{15 b 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 ) \operatorname{Subst}\left (\int \frac{\sqrt{a+b x^2}}{\sqrt{1-x^2}} \, dx,x,\sin (e+f x)\right )}{15 b f}\\ &=\frac{2 (3 a+b) \cos (e+f x) \sin (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{15 f}-\frac{b \cos ^3(e+f x) \sin (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{5 f}-\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 ) \operatorname{Subst}\left (\int \frac{\sqrt{1+\frac{b x^2}{a}}}{\sqrt{1-x^2}} \, dx,x,\sin (e+f x)\right )}{15 b f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}+\frac{\left (a (3 a-b) (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 )}{15 b f \sqrt{a+b \sin ^2(e+f x)}}\\ &=\frac{2 (3 a+b) \cos (e+f x) \sin (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{15 f}-\frac{b \cos ^3(e+f x) \sin (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{5 f}-\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)}}{15 b f \sqrt{1+\frac{b \sin ^2(e+f x)}{a}}}+\frac{a (3 a-b) (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}}}{15 b f \sqrt{a+b \sin ^2(e+f x)}}\\ \end{align*}

Mathematica [A]  time = 1.32096, size = 200, normalized size = 0.77 \[ \frac{\sqrt{2} b \sin (2 (e+f x)) \left (48 a^2-4 b (9 a+2 b) \cos (2 (e+f x))+28 a b+3 b^2 \cos (4 (e+f x))+5 b^2\right )+16 a \left (3 a^2+2 a b-b^2\right ) \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} F\left (e+f x\left |-\frac{b}{a}\right .\right )-16 a \left (3 a^2-7 a b-2 b^2\right ) \sqrt{\frac{2 a-b \cos (2 (e+f x))+b}{a}} E\left (e+f x\left |-\frac{b}{a}\right .\right )}{240 b f \sqrt{2 a-b \cos (2 (e+f x))+b}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(-16*a*(3*a^2 - 7*a*b - 2*b^2)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticE[e + f*x, -(b/a)] + 16*a*(3*a^2
+ 2*a*b - b^2)*Sqrt[(2*a + b - b*Cos[2*(e + f*x)])/a]*EllipticF[e + f*x, -(b/a)] + Sqrt[2]*b*(48*a^2 + 28*a*b
+ 5*b^2 - 4*b*(9*a + 2*b)*Cos[2*(e + f*x)] + 3*b^2*Cos[4*(e + f*x)])*Sin[2*(e + f*x)])/(240*b*f*Sqrt[2*a + b
- b*Cos[2*(e + f*x)]])

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Maple [A]  time = 1.421, size = 429, normalized size = 1.7 \begin{align*}{\frac{1}{15\,b\cos \left ( fx+e \right ) f} \left ( -3\,{b}^{3} \left ( \sin \left ( fx+e \right ) \right ) ^{7}-9\,a{b}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{5}+4\,{b}^{3} \left ( \sin \left ( fx+e \right ) \right ) ^{5}+3\,\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{a}}}{\it EllipticF} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ){a}^{3}+2\,{a}^{2}\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{a}}}{\it EllipticF} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) b-a\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{a}}}{\it EllipticF} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ){b}^{2}-3\,\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{a}}}{\it EllipticE} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ){a}^{3}+7\,\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{a}}}{\it EllipticE} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ){a}^{2}b+2\,\sqrt{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}}\sqrt{{\frac{a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2}}{a}}}{\it EllipticE} \left ( \sin \left ( fx+e \right ) ,\sqrt{-{\frac{b}{a}}} \right ) a{b}^{2}-6\,{a}^{2}b \left ( \sin \left ( fx+e \right ) \right ) ^{3}+10\,a{b}^{2} \left ( \sin \left ( fx+e \right ) \right ) ^{3}- \left ( \sin \left ( fx+e \right ) \right ) ^{3}{b}^{3}+6\,\sin \left ( fx+e \right ){a}^{2}b-a{b}^{2}\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(cos(f*x+e)^2*(a+b*sin(f*x+e)^2)^(3/2),x)

[Out]

1/15*(-3*b^3*sin(f*x+e)^7-9*a*b^2*sin(f*x+e)^5+4*b^3*sin(f*x+e)^5+3*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a
)^(1/2)*EllipticF(sin(f*x+e),(-1/a*b)^(1/2))*a^3+2*a^2*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*Ellip
ticF(sin(f*x+e),(-1/a*b)^(1/2))*b-a*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticF(sin(f*x+e),(-1
/a*b)^(1/2))*b^2-3*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^3+
7*(cos(f*x+e)^2)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a^2*b+2*(cos(f*x+e)^2
)^(1/2)*((a+b*sin(f*x+e)^2)/a)^(1/2)*EllipticE(sin(f*x+e),(-1/a*b)^(1/2))*a*b^2-6*a^2*b*sin(f*x+e)^3+10*a*b^2*
sin(f*x+e)^3-sin(f*x+e)^3*b^3+6*sin(f*x+e)*a^2*b-a*b^2*sin(f*x+e))/b/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}} \cos \left (f x + e\right )^{2}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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