3.2.22 \(\int \frac {\sin ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx\) [122]

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

[Out]

3/8*a^2*arctan((a-b)^(1/2)*tan(f*x+e)/(a+b*tan(f*x+e)^2)^(1/2))/(a-b)^(5/2)/f-1/8*(5*a-2*b)*cos(f*x+e)*sin(f*x
+e)*(a+b*tan(f*x+e)^2)^(1/2)/(a-b)^2/f+1/4*cos(f*x+e)^3*sin(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2)/(a-b)/f

________________________________________________________________________________________

Rubi [A]
time = 0.11, antiderivative size = 146, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.240, Rules used = {3744, 481, 541, 12, 385, 209} \begin {gather*} \frac {3 a^2 \text {ArcTan}\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} \end {gather*}

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 12

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

Rule 209

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

Rule 385

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 481

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 541

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 3744

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]

Rubi steps

\begin {align*} \int \frac {\sin ^4(e+f x)}{\sqrt {a+b \tan ^2(e+f x)}} \, dx &=\frac {\text {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 {\text {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 {\text {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 ) \text {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 ) \text {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] Result contains higher order function than in optimal. Order 4 vs. order 3 in optimal.
time = 4.37, size = 314, normalized size = 2.15 \begin {gather*} -\frac {\left ((a-b) \left (7 a^2+8 a b-3 b^2+2 \left (3 a^2-5 a b+2 b^2\right ) \cos (2 (e+f x))-(a-b)^2 \cos (4 (e+f x))\right )+6 \sqrt {2} a^2 (-a+b) \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} F\left (\left .\text {ArcSin}\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 )+6 \sqrt {2} a^3 \sqrt {\frac {(a+b+(a-b) \cos (2 (e+f x))) \csc ^2(e+f x)}{b}} \Pi \left (-\frac {b}{a-b};\left .\text {ArcSin}\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 ) \sec ^2(e+f x) \sin (2 (e+f x))}{32 \sqrt {2} (a-b)^3 f \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/32*(((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)*Co
s[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)])*Cs
c[e + f*x]^2)/b]/Sqrt[2]], 1])*Sec[e + f*x]^2*Sin[2*(e + f*x)])/(Sqrt[2]*(a - b)^3*f*Sqrt[(a + b + (a - b)*Cos
[2*(e + f*x)])*Sec[e + f*x]^2])

________________________________________________________________________________________

Maple [C] Result contains higher order function than in optimal. Order 4 vs. order 3.
time = 0.40, size = 1169, normalized size = 8.01

method	result	size
default	\(\frac {\sin \left (f x +e \right ) \left (2 \sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}\, \left (\cos ^{5}\left (f x +e \right )\right ) a^{2}-4 \sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}\, \left (\cos ^{5}\left (f x +e \right )\right ) a b +2 \sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}\, \left (\cos ^{5}\left (f x +e \right )\right ) b^{2}-2 \sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}\, \left (\cos ^{4}\left (f x +e \right )\right ) a^{2}+4 \sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}\, \left (\cos ^{4}\left (f x +e \right )\right ) a b -2 \sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}\, \left (\cos ^{4}\left (f x +e \right )\right ) b^{2}+6 \sqrt {2}\, \sqrt {\frac {i \cos \left (f x +e \right ) \sqrt {b}\, \sqrt {a -b}-i \sqrt {b}\, \sqrt {a -b}+\cos \left (f x +e \right ) a -b \cos \left (f x +e \right )+b}{\left (\cos \left (f x +e \right )+1\right ) a}}\, \sqrt {-\frac {2 \left (i \cos \left (f x +e \right ) \sqrt {b}\, \sqrt {a -b}-i \sqrt {b}\, \sqrt {a -b}-\cos \left (f x +e \right ) a +b \cos \left (f x +e \right )-b \right )}{\left (\cos \left (f x +e \right )+1\right ) a}}\, \EllipticPi \left (\frac {\left (\cos \left (f x +e \right )-1\right ) \sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}}{\sin \left (f x +e \right )}, -\frac {a}{2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}, \frac {\sqrt {-\frac {2 i \sqrt {b}\, \sqrt {a -b}-a +2 b}{a}}}{\sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}}\right ) a^{2} \sin \left (f x +e \right )-3 \sqrt {2}\, \sqrt {\frac {i \cos \left (f x +e \right ) \sqrt {b}\, \sqrt {a -b}-i \sqrt {b}\, \sqrt {a -b}+\cos \left (f x +e \right ) a -b \cos \left (f x +e \right )+b}{\left (\cos \left (f x +e \right )+1\right ) a}}\, \sqrt {-\frac {2 \left (i \cos \left (f x +e \right ) \sqrt {b}\, \sqrt {a -b}-i \sqrt {b}\, \sqrt {a -b}-\cos \left (f x +e \right ) a +b \cos \left (f x +e \right )-b \right )}{\left (\cos \left (f x +e \right )+1\right ) a}}\, \EllipticF \left (\frac {\left (\cos \left (f x +e \right )-1\right ) \sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}}{\sin \left (f x +e \right )}, \sqrt {\frac {8 i b^{\frac {3}{2}} \sqrt {a -b}-4 i \sqrt {b}\, \sqrt {a -b}\, a +a^{2}-8 a b +8 b^{2}}{a^{2}}}\right ) a^{2} \sin \left (f x +e \right )-5 \sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}\, \left (\cos ^{3}\left (f x +e \right )\right ) a^{2}+9 \sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}\, \left (\cos ^{3}\left (f x +e \right )\right ) a b -4 \sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}\, \left (\cos ^{3}\left (f x +e \right )\right ) b^{2}+5 \sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}\, \left (\cos ^{2}\left (f x +e \right )\right ) a^{2}-9 \sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}\, \left (\cos ^{2}\left (f x +e \right )\right ) a b +4 \sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}\, \left (\cos ^{2}\left (f x +e \right )\right ) b^{2}-5 \sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}\, \cos \left (f x +e \right ) a b +2 \sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}\, \cos \left (f x +e \right ) b^{2}+5 \sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}\, a b -2 \sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}\, b^{2}\right )}{8 f \left (\cos \left (f x +e \right )-1\right ) \sqrt {\frac {a \left (\cos ^{2}\left (f x +e \right )\right )-\left (\cos ^{2}\left (f x +e \right )\right ) b +b}{\cos \left (f x +e \right )^{2}}}\, \cos \left (f x +e \right ) \sqrt {\frac {2 i \sqrt {b}\, \sqrt {a -b}+a -2 b}{a}}\, \left (a -b \right )^{2}}\)	\(1169\)

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,method=_RETURNVERBOSE)

[Out]

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

________________________________________________________________________________________

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

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]

Exception raised: RuntimeError >> ECL says: Error executing code in Maxima: sign: argument cannot be imaginary
; found %i

________________________________________________________________________________________

Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 326 vs. \(2 (138) = 276\).
time = 5.70, size = 817, normalized size = 5.60 \begin {gather*} \left [-\frac {3 \, a^{2} \sqrt {-a + b} \log \left (128 \, {\left (a^{4} - 4 \, a^{3} b + 6 \, a^{2} b^{2} - 4 \, a b^{3} + b^{4}\right )} \cos \left (f x + e\right )^{8} - 256 \, {\left (a^{4} - 5 \, a^{3} b + 9 \, a^{2} b^{2} - 7 \, a b^{3} + 2 \, b^{4}\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{4} - 34 \, a^{3} b + 77 \, a^{2} b^{2} - 72 \, a b^{3} + 24 \, b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} - 32 \, a^{3} b + 160 \, a^{2} b^{2} - 256 \, a b^{3} + 128 \, b^{4} - 32 \, {\left (a^{4} - 11 \, a^{3} b + 34 \, a^{2} b^{2} - 40 \, a b^{3} + 16 \, b^{4}\right )} \cos \left (f x + e\right )^{2} + 8 \, {\left (16 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{7} - 24 \, {\left (a^{3} - 4 \, a^{2} b + 5 \, a b^{2} - 2 \, b^{3}\right )} \cos \left (f x + e\right )^{5} + 2 \, {\left (5 \, a^{3} - 29 \, a^{2} b + 48 \, a b^{2} - 24 \, b^{3}\right )} \cos \left (f x + e\right )^{3} - {\left (a^{3} - 10 \, a^{2} b + 24 \, a b^{2} - 16 \, b^{3}\right )} \cos \left (f x + e\right )\right )} \sqrt {-a + b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )\right ) - 8 \, {\left (2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (5 \, a^{2} - 7 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{64 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} f}, \frac {3 \, \sqrt {a - b} a^{2} \arctan \left (-\frac {{\left (8 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{5} - 8 \, {\left (a^{2} - 3 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{3} + {\left (a^{2} - 8 \, a b + 8 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {a - b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, {\left (2 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} \cos \left (f x + e\right )^{4} - a^{2} b + 3 \, a b^{2} - 2 \, b^{3} - {\left (a^{3} - 6 \, a^{2} b + 9 \, a b^{2} - 4 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ) + 4 \, {\left (2 \, {\left (a^{2} - 2 \, a b + b^{2}\right )} \cos \left (f x + e\right )^{3} - {\left (5 \, a^{2} - 7 \, a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \sin \left (f x + e\right )}{32 \, {\left (a^{3} - 3 \, a^{2} b + 3 \, a b^{2} - b^{3}\right )} f}\right ] \end {gather*}

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)]

________________________________________________________________________________________

Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sin ^{4}{\left (e + f x \right )}}{\sqrt {a + b \tan ^{2}{\left (e + f x \right )}}}\, dx \end {gather*}

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)

________________________________________________________________________________________

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(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)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\sin \left (e+f\,x\right )}^4}{\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

________________________________________________________________________________________