\(\int \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx\) [93]

Optimal result

Integrand size = 25, antiderivative size = 113 \[ \int \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{f}-\frac {\cos (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{f}+\frac {\cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{3 (a-b) f} \]

[Out]

1/3*cos(f*x+e)^3*(a-b+b*sec(f*x+e)^2)^(3/2)/(a-b)/f+arctanh(sec(f*x+e)*b^(1/2)/(a-b+b*sec(f*x+e)^2)^(1/2))*b^(
1/2)/f-cos(f*x+e)*(a-b+b*sec(f*x+e)^2)^(1/2)/f

Rubi [A] (verified)

Time = 0.12 (sec) , antiderivative size = 113, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3745, 462, 283, 223, 212} \[ \int \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )}{f}+\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{3 f (a-b)}-\frac {\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{f} \]

[In]

Int[Sin[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

(Sqrt[b]*ArcTanh[(Sqrt[b]*Sec[e + f*x])/Sqrt[a - b + b*Sec[e + f*x]^2]])/f - (Cos[e + f*x]*Sqrt[a - b + b*Sec[
e + f*x]^2])/f + (Cos[e + f*x]^3*(a - b + b*Sec[e + f*x]^2)^(3/2))/(3*(a - b)*f)

Rule 212

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

Rule 223

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Subst[Int[1/(1 - b*x^2), x], x, x/Sqrt[a + b*x^2]] /; FreeQ[{a,
b}, x] &&  !GtQ[a, 0]

Rule 283

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(c*x)^(m + 1)*((a + b*x^n)^p/(c*(m + 1
))), x] - Dist[b*n*(p/(c^n*(m + 1))), Int[(c*x)^(m + n)*(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] &&
IGtQ[n, 0] && GtQ[p, 0] && LtQ[m, -1] &&  !ILtQ[(m + n*p + n + 1)/n, 0] && IntBinomialQ[a, b, c, n, m, p, x]

Rule 462

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

Rule 3745

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

Rubi steps \begin{align*} \text {integral}& = \frac {\text {Subst}\left (\int \frac {\left (-1+x^2\right ) \sqrt {a-b+b x^2}}{x^4} \, dx,x,\sec (e+f x)\right )}{f} \\ & = \frac {\cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{3 (a-b) f}+\frac {\text {Subst}\left (\int \frac {\sqrt {a-b+b x^2}}{x^2} \, dx,x,\sec (e+f x)\right )}{f} \\ & = -\frac {\cos (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{f}+\frac {\cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{3 (a-b) f}+\frac {b \text {Subst}\left (\int \frac {1}{\sqrt {a-b+b x^2}} \, dx,x,\sec (e+f x)\right )}{f} \\ & = -\frac {\cos (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{f}+\frac {\cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{3 (a-b) f}+\frac {b \text {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{f} \\ & = \frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a-b+b \sec ^2(e+f x)}}\right )}{f}-\frac {\cos (e+f x) \sqrt {a-b+b \sec ^2(e+f x)}}{f}+\frac {\cos ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^{3/2}}{3 (a-b) f} \\ \end{align*}

Mathematica [A] (verified)

Time = 1.47 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.50 \[ \int \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {\cos (e+f x) \left (6 \sqrt {2} (a-b) \sqrt {b} \text {arctanh}\left (\frac {\sqrt {a+b+(a-b) \cos (2 (e+f x))}}{\sqrt {2} \sqrt {b}}\right )+\sqrt {a+b+(a-b) \cos (2 (e+f x))} (-5 a+7 b+(a-b) \cos (2 (e+f x)))\right ) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{6 \sqrt {2} (a-b) f \sqrt {a+b+(a-b) \cos (2 (e+f x))}} \]

[In]

Integrate[Sin[e + f*x]^3*Sqrt[a + b*Tan[e + f*x]^2],x]

[Out]

(Cos[e + f*x]*(6*Sqrt[2]*(a - b)*Sqrt[b]*ArcTanh[Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]]/(Sqrt[2]*Sqrt[b])] + S
qrt[a + b + (a - b)*Cos[2*(e + f*x)]]*(-5*a + 7*b + (a - b)*Cos[2*(e + f*x)]))*Sqrt[(a + b + (a - b)*Cos[2*(e
+ f*x)])*Sec[e + f*x]^2])/(6*Sqrt[2]*(a - b)*f*Sqrt[a + b + (a - b)*Cos[2*(e + f*x)]])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(642\) vs. \(2(101)=202\).

Time = 0.50 (sec) , antiderivative size = 643, normalized size of antiderivative = 5.69

method	result	size
default	\(\frac {\left (\sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )^{3} a -\sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, b \cos \left (f x +e \right )^{3}+\sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )^{2} a -\sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right )^{2} b -3 \ln \left (-4 \sqrt {b}\, \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-4 \sqrt {b}\, \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sec \left (f x +e \right )-4 b \sec \left (f x +e \right )\right ) b^{\frac {3}{2}}-3 \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \cos \left (f x +e \right ) a +4 \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, b \cos \left (f x +e \right )+3 \ln \left (-4 \sqrt {b}\, \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}-4 \sqrt {b}\, \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, \sec \left (f x +e \right )-4 b \sec \left (f x +e \right )\right ) \sqrt {b}\, a -3 \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, a +4 \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, b \right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}\, \cos \left (f x +e \right )}{3 f \left (a -b \right ) \left (\cos \left (f x +e \right )+1\right ) \sqrt {\frac {a \cos \left (f x +e \right )^{2}-b \cos \left (f x +e \right )^{2}+b}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\)	\(643\)

[In]

int(sin(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2),x,method=_RETURNVERBOSE)

[Out]

1/3/f/(a-b)*(((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)^3*a-((a*cos(f*x+e)^2-b*cos(
f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*b*cos(f*x+e)^3+((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*
cos(f*x+e)^2*a-((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*cos(f*x+e)^2*b-3*ln(-4*b^(1/2)*((a*c
os(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)-4*b^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)
+1)^2)^(1/2)*sec(f*x+e)-4*b*sec(f*x+e))*b^(3/2)-3*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*c
os(f*x+e)*a+4*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*b*cos(f*x+e)+3*ln(-4*b^(1/2)*((a*cos(
f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)-4*b^(1/2)*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)
^2)^(1/2)*sec(f*x+e)-4*b*sec(f*x+e))*b^(1/2)*a-3*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*a+
4*((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)*b)*(a+b*tan(f*x+e)^2)^(1/2)*cos(f*x+e)/(cos(f*x+e
)+1)/((a*cos(f*x+e)^2-b*cos(f*x+e)^2+b)/(cos(f*x+e)+1)^2)^(1/2)

Fricas [A] (verification not implemented)

none

Time = 0.40 (sec) , antiderivative size = 277, normalized size of antiderivative = 2.45 \[ \int \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\left [\frac {3 \, {\left (a - b\right )} \sqrt {b} \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} - {\left (3 \, a - 4 \, b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{6 \, {\left (a - b\right )} f}, -\frac {3 \, {\left (a - b\right )} \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{b}\right ) - {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} - {\left (3 \, a - 4 \, b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, {\left (a - b\right )} f}\right ] \]

[In]

integrate(sin(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="fricas")

[Out]

[1/6*(3*(a - b)*sqrt(b)*log(-((a - b)*cos(f*x + e)^2 + 2*sqrt(b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e
)^2)*cos(f*x + e) + 2*b)/cos(f*x + e)^2) + 2*((a - b)*cos(f*x + e)^3 - (3*a - 4*b)*cos(f*x + e))*sqrt(((a - b)
*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a - b)*f), -1/3*(3*(a - b)*sqrt(-b)*arctan(sqrt(-b)*sqrt(((a - b)*cos(
f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x + e)/b) - ((a - b)*cos(f*x + e)^3 - (3*a - 4*b)*cos(f*x + e))*sqrt(((a
- b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2))/((a - b)*f)]

Sympy [F(-1)]

Timed out. \[ \int \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\text {Timed out} \]

[In]

integrate(sin(f*x+e)**3*(a+b*tan(f*x+e)**2)**(1/2),x)

[Out]

Timed out

Maxima [A] (verification not implemented)

none

Time = 0.33 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.17 \[ \int \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {\frac {2 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3}}{a - b} - 6 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - 3 \, \sqrt {b} \log \left (\frac {\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right )}{6 \, f} \]

[In]

integrate(sin(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="maxima")

[Out]

1/6*(2*(a - b + b/cos(f*x + e)^2)^(3/2)*cos(f*x + e)^3/(a - b) - 6*sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e)
- 3*sqrt(b)*log((sqrt(a - b + b/cos(f*x + e)^2)*cos(f*x + e) - sqrt(b))/(sqrt(a - b + b/cos(f*x + e)^2)*cos(f
*x + e) + sqrt(b))))/f

Giac [B] (verification not implemented)

Leaf count of result is larger than twice the leaf count of optimal. 1182 vs. \(2 (101) = 202\).

Time = 0.79 (sec) , antiderivative size = 1182, normalized size of antiderivative = 10.46 \[ \int \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\text {Too large to display} \]

[In]

integrate(sin(f*x+e)^3*(a+b*tan(f*x+e)^2)^(1/2),x, algorithm="giac")

[Out]

2/3*(3*b*arctan(-1/2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e
)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a) - sqrt(a))/sqrt(-b))/sqrt(-b) - 2*(3*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - s
qrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^5*b - 12*(sqrt(a)
*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e
)^2 + a))^4*a^(3/2) + 21*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1
/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^4*sqrt(a)*b + 16*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f
*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^3*a^2 - 50*(sqrt(a)*tan(1/2*f*x
+ 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^3*a
*b + 40*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan
(1/2*f*x + 1/2*e)^2 + a))^3*b^2 + 24*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan
(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^2*a^(5/2) - 54*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a
*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))^2*a^(3/2)*b + 24*(sqrt
(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/
2*e)^2 + a))^2*sqrt(a)*b^2 - 48*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*
f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a^3 + 159*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f
*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a^2*b - 168*(sqrt(a)*tan(1/2*f*x
+ 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 + a))*a*
b^2 + 48*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*ta
n(1/2*f*x + 1/2*e)^2 + a))*b^3 + 20*a^(7/2) - 79*a^(5/2)*b + 104*a^(3/2)*b^2 - 48*sqrt(a)*b^3)/((sqrt(a)*tan(1
/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*tan(1/2*f*x + 1/2*e)^2 +
a))^2 + 2*(sqrt(a)*tan(1/2*f*x + 1/2*e)^2 - sqrt(a*tan(1/2*f*x + 1/2*e)^4 - 2*a*tan(1/2*f*x + 1/2*e)^2 + 4*b*t
an(1/2*f*x + 1/2*e)^2 + a))*sqrt(a) - 3*a + 4*b)^3)*sgn(tan(1/2*f*x + 1/2*e)^2 - 1)/f

Mupad [F(-1)]

Timed out. \[ \int \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\int {\sin \left (e+f\,x\right )}^3\,\sqrt {b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a} \,d x \]

[In]

int(sin(e + f*x)^3*(a + b*tan(e + f*x)^2)^(1/2),x)

[Out]

int(sin(e + f*x)^3*(a + b*tan(e + f*x)^2)^(1/2), x)