3.122 \(\int \frac{\sin ^4(e+f x)}{\sqrt{a+b \tan ^2(e+f x)}} \, dx\)

Optimal. Leaf size=146 \[ \frac{3 a^2 \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{8 f (a-b)^{5/2}}+\frac{\sin (e+f x) \cos ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{4 f (a-b)}-\frac{(5 a-2 b) \sin (e+f x) \cos (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{8 f (a-b)^2} \]

[Out]

(3*a^2*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(8*(a - b)^(5/2)*f) - ((5*a - 2*b)*Cos[e
+ f*x]*Sin[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(8*(a - b)^2*f) + (Cos[e + f*x]^3*Sin[e + f*x]*Sqrt[a + b*Tan
[e + f*x]^2])/(4*(a - b)*f)

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Rubi [A]  time = 0.166636, antiderivative size = 146, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.24, Rules used = {3663, 470, 527, 12, 377, 203} \[ \frac{3 a^2 \tan ^{-1}\left (\frac{\sqrt{a-b} \tan (e+f x)}{\sqrt{a+b \tan ^2(e+f x)}}\right )}{8 f (a-b)^{5/2}}+\frac{\sin (e+f x) \cos ^3(e+f x) \sqrt{a+b \tan ^2(e+f x)}}{4 f (a-b)}-\frac{(5 a-2 b) \sin (e+f x) \cos (e+f x) \sqrt{a+b \tan ^2(e+f x)}}{8 f (a-b)^2} \]

Antiderivative was successfully verified.

[In]

Int[Sin[e + f*x]^4/Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

(3*a^2*ArcTan[(Sqrt[a - b]*Tan[e + f*x])/Sqrt[a + b*Tan[e + f*x]^2]])/(8*(a - b)^(5/2)*f) - ((5*a - 2*b)*Cos[e
+ f*x]*Sin[e + f*x]*Sqrt[a + b*Tan[e + f*x]^2])/(8*(a - b)^2*f) + (Cos[e + f*x]^3*Sin[e + f*x]*Sqrt[a + b*Tan
[e + f*x]^2])/(4*(a - b)*f)

Rule 3663

Int[sin[(e_.) + (f_.)*(x_)]^(m_)*((a_) + (b_.)*((c_.)*tan[(e_.) + (f_.)*(x_)])^(n_))^(p_.), x_Symbol] :> With[
{ff = FreeFactors[Tan[e + f*x], x]}, Dist[(c*ff^(m + 1))/f, Subst[Int[(x^m*(a + b*(ff*x)^n)^p)/(c^2 + ff^2*x^2
)^(m/2 + 1), x], x, (c*Tan[e + f*x])/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x] && IntegerQ[m/2]

Rule 470

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> -Simp[(a*e^(2
*n - 1)*(e*x)^(m - 2*n + 1)*(a + b*x^n)^(p + 1)*(c + d*x^n)^(q + 1))/(b*n*(b*c - a*d)*(p + 1)), x] + Dist[e^(2
*n)/(b*n*(b*c - a*d)*(p + 1)), Int[(e*x)^(m - 2*n)*(a + b*x^n)^(p + 1)*(c + d*x^n)^q*Simp[a*c*(m - 2*n + 1) +
(a*d*(m - n + n*q + 1) + b*c*n*(p + 1))*x^n, x], x], x] /; FreeQ[{a, b, c, d, e, q}, x] && NeQ[b*c - a*d, 0] &
& IGtQ[n, 0] && LtQ[p, -1] && GtQ[m - n + 1, n] && IntBinomialQ[a, b, c, d, e, m, 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 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

Rule 377

Int[((a_) + (b_.)*(x_)^(n_))^(p_)/((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Subst[Int[1/(c - (b*c - a*d)*x^n), x]
, x, x/(a + b*x^n)^(1/n)] /; FreeQ[{a, b, c, d}, x] && NeQ[b*c - a*d, 0] && EqQ[n*p + 1, 0] && IntegerQ[n]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rubi steps

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

Mathematica [C]  time = 4.01079, size = 314, normalized size = 2.15 \[ -\frac{\sin (2 (e+f x)) \sec ^2(e+f x) \left (6 \sqrt{2} a^2 (b-a) \sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}}}{\sqrt{2}}\right ),1\right )+(a-b) \left (2 \left (3 a^2-5 a b+2 b^2\right ) \cos (2 (e+f x))+7 a^2-(a-b)^2 \cos (4 (e+f x))+8 a b-3 b^2\right )+6 \sqrt{2} a^3 \sqrt{\frac{\csc ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}{b}} \Pi \left (-\frac{b}{a-b};\left .\sin ^{-1}\left (\frac{\sqrt{\frac{(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}}}{\sqrt{2}}\right )\right |1\right )\right )}{32 \sqrt{2} f (a-b)^3 \sqrt{\sec ^2(e+f x) ((a-b) \cos (2 (e+f x))+a+b)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sin[e + f*x]^4/Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

-(((a - b)*(7*a^2 + 8*a*b - 3*b^2 + 2*(3*a^2 - 5*a*b + 2*b^2)*Cos[2*(e + f*x)] - (a - b)^2*Cos[4*(e + f*x)]) +
6*Sqrt[2]*a^2*(-a + b)*Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]*EllipticF[ArcSin[Sqrt[((a
+ b + (a - b)*Cos[2*(e + f*x)])*Csc[e + f*x]^2)/b]/Sqrt[2]], 1] + 6*Sqrt[2]*a^3*Sqrt[((a + b + (a - b)*Cos[2*(
e + f*x)])*Csc[e + f*x]^2)/b]*EllipticPi[-(b/(a - b)), ArcSin[Sqrt[((a + b + (a - b)*Cos[2*(e + f*x)])*Csc[e +
f*x]^2)/b]/Sqrt[2]], 1])*Sec[e + f*x]^2*Sin[2*(e + f*x)])/(32*Sqrt[2]*(a - b)^3*f*Sqrt[(a + b + (a - b)*Cos[2
*(e + f*x)])*Sec[e + f*x]^2])

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Maple [C]  time = 0.277, size = 1169, normalized size = 8. \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sin(f*x+e)^4/(a+b*tan(f*x+e)^2)^(1/2),x)

[Out]

1/8/f/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/(a-b)^2*sin(f*x+e)*(2*cos(f*x+e)^5*((2*I*b^(1/2)*(a-b)^(1/2)+a
-2*b)/a)^(1/2)*a^2-4*cos(f*x+e)^5*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a*b+2*cos(f*x+e)^5*((2*I*b^(1/2)*(
a-b)^(1/2)+a-2*b)/a)^(1/2)*b^2-2*cos(f*x+e)^4*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^2+4*cos(f*x+e)^4*((2
*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a*b-2*cos(f*x+e)^4*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*b^2-3*2^(1
/2)*(1/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1))^
(1/2)*(-2/a*(I*cos(f*x+e)*b^(1/2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1
))^(1/2)*EllipticF((cos(f*x+e)-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),((8*I*b^(3/2)*(a-b)^(1/
2)-4*I*b^(1/2)*(a-b)^(1/2)*a+a^2-8*a*b+8*b^2)/a^2)^(1/2))*a^2*sin(f*x+e)+6*2^(1/2)*(1/a*(I*cos(f*x+e)*b^(1/2)*
(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)+cos(f*x+e)*a-b*cos(f*x+e)+b)/(cos(f*x+e)+1))^(1/2)*(-2/a*(I*cos(f*x+e)*b^(1/
2)*(a-b)^(1/2)-I*b^(1/2)*(a-b)^(1/2)-cos(f*x+e)*a+b*cos(f*x+e)-b)/(cos(f*x+e)+1))^(1/2)*EllipticPi((cos(f*x+e)
-1)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)/sin(f*x+e),-1/(2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)*a,(-(2*I*b^(1/2)*(
a-b)^(1/2)-a+2*b)/a)^(1/2)/((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2))*a^2*sin(f*x+e)-5*cos(f*x+e)^3*((2*I*b^(1
/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^2+9*cos(f*x+e)^3*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a*b-4*cos(f*x+e)^
3*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*b^2+5*cos(f*x+e)^2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a^2-9
*cos(f*x+e)^2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a*b+4*cos(f*x+e)^2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)
^(1/2)*b^2-5*cos(f*x+e)*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a*b+2*cos(f*x+e)*((2*I*b^(1/2)*(a-b)^(1/2)+a
-2*b)/a)^(1/2)*b^2+5*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)*a*b-2*((2*I*b^(1/2)*(a-b)^(1/2)+a-2*b)/a)^(1/2)
*b^2)/(cos(f*x+e)-1)/((a*cos(f*x+e)^2-cos(f*x+e)^2*b+b)/cos(f*x+e)^2)^(1/2)/cos(f*x+e)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

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

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Fricas [B]  time = 23.7585, size = 1864, normalized size = 12.77 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[-1/64*(3*a^2*sqrt(-a + b)*log(128*(a^4 - 4*a^3*b + 6*a^2*b^2 - 4*a*b^3 + b^4)*cos(f*x + e)^8 - 256*(a^4 - 5*a
^3*b + 9*a^2*b^2 - 7*a*b^3 + 2*b^4)*cos(f*x + e)^6 + 32*(5*a^4 - 34*a^3*b + 77*a^2*b^2 - 72*a*b^3 + 24*b^4)*co
s(f*x + e)^4 + a^4 - 32*a^3*b + 160*a^2*b^2 - 256*a*b^3 + 128*b^4 - 32*(a^4 - 11*a^3*b + 34*a^2*b^2 - 40*a*b^3
+ 16*b^4)*cos(f*x + e)^2 + 8*(16*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^7 - 24*(a^3 - 4*a^2*b + 5*a*b^2
- 2*b^3)*cos(f*x + e)^5 + 2*(5*a^3 - 29*a^2*b + 48*a*b^2 - 24*b^3)*cos(f*x + e)^3 - (a^3 - 10*a^2*b + 24*a*b^
2 - 16*b^3)*cos(f*x + e))*sqrt(-a + b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*sin(f*x + e)) - 8*(2*
(a^2 - 2*a*b + b^2)*cos(f*x + e)^3 - (5*a^2 - 7*a*b + 2*b^2)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/c
os(f*x + e)^2)*sin(f*x + e))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*f), 1/32*(3*sqrt(a - b)*a^2*arctan(-1/4*(8*(a^2
- 2*a*b + b^2)*cos(f*x + e)^5 - 8*(a^2 - 3*a*b + 2*b^2)*cos(f*x + e)^3 + (a^2 - 8*a*b + 8*b^2)*cos(f*x + e))*s
qrt(a - b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)/((2*(a^3 - 3*a^2*b + 3*a*b^2 - b^3)*cos(f*x + e)^
4 - a^2*b + 3*a*b^2 - 2*b^3 - (a^3 - 6*a^2*b + 9*a*b^2 - 4*b^3)*cos(f*x + e)^2)*sin(f*x + e))) + 4*(2*(a^2 - 2
*a*b + b^2)*cos(f*x + e)^3 - (5*a^2 - 7*a*b + 2*b^2)*cos(f*x + e))*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x +
e)^2)*sin(f*x + e))/((a^3 - 3*a^2*b + 3*a*b^2 - b^3)*f)]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)**4/(a+b*tan(f*x+e)**2)**(1/2),x)

[Out]

Integral(sin(e + f*x)**4/sqrt(a + b*tan(e + f*x)**2), x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sin(f*x+e)^4/(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

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