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

Optimal result

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

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

Mathematica [A] (warning: unable to verify)

Time = 0.97 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.50 \[ \int \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {\cos (e+f x) \left (6 \sqrt {2} (a-b) \sqrt {b} \text {arctanh}\left (\frac {\sqrt {a+b+(a-b) \cos (2 (e+f x))}}{\sqrt {2} \sqrt {b}}\right )+\sqrt {a+b+(a-b) \cos (2 (e+f x))} (-5 a+7 b+(a-b) \cos (2 (e+f x)))\right ) \sqrt {(a+b+(a-b) \cos (2 (e+f x))) \sec ^2(e+f x)}}{6 \sqrt {2} (a-b) f \sqrt {a+b+(a-b) \cos (2 (e+f x))}} \] Input:

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

Output:

(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])] + Sqrt[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)]])
 

Rubi [A] (verified)

Time = 0.48 (sec) , antiderivative size = 108, normalized size of antiderivative = 0.96, number of steps used = 8, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.280, Rules used = {3042, 4147, 25, 358, 247, 224, 219}

Below are the steps used by Rubi to obtain the solution. The rule number used for the transformation is given above next to the arrow. The rules definitions used are listed below.

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

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (e+f x)^3 \sqrt {a+b \tan (e+f x)^2}dx\)

\(\Big \downarrow \) 4147

\(\displaystyle \frac {\int -\cos ^4(e+f x) \left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}d\sec (e+f x)}{f}\)

\(\Big \downarrow \) 25

\(\displaystyle -\frac {\int \cos ^4(e+f x) \left (1-\sec ^2(e+f x)\right ) \sqrt {b \sec ^2(e+f x)+a-b}d\sec (e+f x)}{f}\)

\(\Big \downarrow \) 358

\(\displaystyle \frac {\int \cos ^2(e+f x) \sqrt {b \sec ^2(e+f x)+a-b}d\sec (e+f x)+\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{3 (a-b)}}{f}\)

\(\Big \downarrow \) 247

\(\displaystyle \frac {b \int \frac {1}{\sqrt {b \sec ^2(e+f x)+a-b}}d\sec (e+f x)+\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{3 (a-b)}-\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{f}\)

\(\Big \downarrow \) 224

\(\displaystyle \frac {b \int \frac {1}{1-\frac {b \sec ^2(e+f x)}{b \sec ^2(e+f x)+a-b}}d\frac {\sec (e+f x)}{\sqrt {b \sec ^2(e+f x)+a-b}}+\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{3 (a-b)}-\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{f}\)

\(\Big \downarrow \) 219

\(\displaystyle \frac {\sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)-b}}\right )+\frac {\cos ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^{3/2}}{3 (a-b)}-\cos (e+f x) \sqrt {a+b \sec ^2(e+f x)-b}}{f}\)

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

Int[-(Fx_), x_Symbol] :> Simp[Identity[-1]   Int[Fx, x], x]
 

rule 219

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] && (Gt 
Q[a, 0] || LtQ[b, 0])
 

rule 224

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 247

Int[((c_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[(c*x)^ 
(m + 1)*((a + b*x^2)^p/(c*(m + 1))), x] - Simp[2*b*(p/(c^2*(m + 1)))   Int[ 
(c*x)^(m + 2)*(a + b*x^2)^(p - 1), x], x] /; FreeQ[{a, b, c}, x] && GtQ[p, 
0] && LtQ[m, -1] &&  !ILtQ[(m + 2*p + 3)/2, 0] && IntBinomialQ[a, b, c, 2, 
m, p, x]
 

rule 358

Int[((e_.)*(x_))^(m_)*((a_) + (b_.)*(x_)^2)^(p_.)*((c_) + (d_.)*(x_)^2), x_ 
Symbol] :> Simp[c*(e*x)^(m + 1)*((a + b*x^2)^(p + 1)/(a*e*(m + 1))), x] + S 
imp[d/e^2   Int[(e*x)^(m + 2)*(a + b*x^2)^p, x], x] /; FreeQ[{a, b, c, d, e 
, m, p}, x] && NeQ[b*c - a*d, 0] && EqQ[Simplify[m + 2*p + 3], 0] && NeQ[m, 
 -1]
 

rule 3042

Int[u_, x_Symbol] :> Int[DeactivateTrig[u, x], x] /; FunctionOfTrigOfLinear 
Q[u, x]
 

rule 4147

Int[sin[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^ 
(p_.), x_Symbol] :> With[{ff = FreeFactors[Sec[e + f*x], x]}, Simp[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]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(442\) vs. \(2(101)=202\).

Time = 6.44 (sec) , antiderivative size = 443, normalized size of antiderivative = 3.92

method	result	size
default	\(-\frac {\left (3 b^{\frac {3}{2}} \ln \left (\frac {4 b \cot \left (f x +e \right )^{2}-8 b \cot \left (f x +e \right ) \csc \left (f x +e \right )+4 b \csc \left (f x +e \right )^{2}+8 \sqrt {b}\, \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 b}{\cot \left (f x +e \right )^{2}-2 \csc \left (f x +e \right ) \cot \left (f x +e \right )+\csc \left (f x +e \right )^{2}-1}\right )-3 \sqrt {b}\, \ln \left (\frac {4 b \cot \left (f x +e \right )^{2}-8 b \cot \left (f x +e \right ) \csc \left (f x +e \right )+4 b \csc \left (f x +e \right )^{2}+8 \sqrt {b}\, \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}+4 b}{\cot \left (f x +e \right )^{2}-2 \csc \left (f x +e \right ) \cot \left (f x +e \right )+\csc \left (f x +e \right )^{2}-1}\right ) a +\left (-\cos \left (f x +e \right )^{3}-\cos \left (f x +e \right )^{2}+3 \cos \left (f x +e \right )+3\right ) \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, a +\left (\cos \left (f x +e \right )^{3}+\cos \left (f x +e \right )^{2}-4 \cos \left (f x +e \right )-4\right ) \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}\, b \right ) \cos \left (f x +e \right ) \sqrt {a +b \tan \left (f x +e \right )^{2}}}{3 f \left (a -b \right ) \left (\cos \left (f x +e \right )+1\right ) \sqrt {\frac {a \cos \left (f x +e \right )^{2}+b \sin \left (f x +e \right )^{2}}{\left (\cos \left (f x +e \right )+1\right )^{2}}}}\)	\(443\)

Input:

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

Output:

-1/3/f/(a-b)*(3*b^(3/2)*ln(4*(b*cot(f*x+e)^2-2*b*cot(f*x+e)*csc(f*x+e)+b*c 
sc(f*x+e)^2+2*b^(1/2)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^( 
1/2)+b)/(cot(f*x+e)^2-2*csc(f*x+e)*cot(f*x+e)+csc(f*x+e)^2-1))-3*b^(1/2)*l 
n(4*(b*cot(f*x+e)^2-2*b*cot(f*x+e)*csc(f*x+e)+b*csc(f*x+e)^2+2*b^(1/2)*((a 
*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)+b)/(cot(f*x+e)^2-2*c 
sc(f*x+e)*cot(f*x+e)+csc(f*x+e)^2-1))*a+(-cos(f*x+e)^3-cos(f*x+e)^2+3*cos( 
f*x+e)+3)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)*a+(cos( 
f*x+e)^3+cos(f*x+e)^2-4*cos(f*x+e)-4)*((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(co 
s(f*x+e)+1)^2)^(1/2)*b)*cos(f*x+e)*(a+b*tan(f*x+e)^2)^(1/2)/(cos(f*x+e)+1) 
/((a*cos(f*x+e)^2+b*sin(f*x+e)^2)/(cos(f*x+e)+1)^2)^(1/2)
 

Fricas [A] (verification not implemented)

Time = 0.22 (sec) , antiderivative size = 292, normalized size of antiderivative = 2.58 \[ \int \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\left [\frac {3 \, {\left (a - b\right )} \sqrt {b} \log \left (-\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + 2 \, \sqrt {b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) + 2 \, {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} - {\left (3 \, a - 4 \, b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{6 \, {\left (a - b\right )} f}, \frac {3 \, {\left (a - b\right )} \sqrt {-b} \arctan \left (-\frac {\sqrt {-b} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}\right ) + {\left ({\left (a - b\right )} \cos \left (f x + e\right )^{3} - {\left (3 \, a - 4 \, b\right )} \cos \left (f x + e\right )\right )} \sqrt {\frac {{\left (a - b\right )} \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{3 \, {\left (a - b\right )} f}\right ] \] Input:

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

Output:

[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)*c 
os(f*x + e)^2 + b)/cos(f*x + e)^2))/((a - b)*f), 1/3*(3*(a - b)*sqrt(-b)*a 
rctan(-sqrt(-b)*sqrt(((a - b)*cos(f*x + e)^2 + b)/cos(f*x + e)^2)*cos(f*x 
+ e)/((a - b)*cos(f*x + e)^2 + 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. \[ \int \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Maxima [A] (verification not implemented)

Time = 0.12 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.17 \[ \int \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\frac {\frac {2 \, {\left (a - b + \frac {b}{\cos \left (f x + e\right )^{2}}\right )}^{\frac {3}{2}} \cos \left (f x + e\right )^{3}}{a - b} - 6 \, \sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - 3 \, \sqrt {b} \log \left (\frac {\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a - b + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right )}{6 \, f} \] Input:

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

Output:

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 = 0.92 (sec) , antiderivative size = 1182, normalized size of antiderivative = 10.46 \[ \int \sin ^3(e+f x) \sqrt {a+b \tan ^2(e+f x)} \, dx=\text {Too large to display} \] Input:

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

Output:

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 - sqr 
t(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/...
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(tan(e + f*x)**2*b + a)*sin(e + f*x)**3,x)