\(\int (a+b \sec ^2(e+f x))^{3/2} \sin (e+f x) \, dx\) [82]

Optimal result

Integrand size = 23, antiderivative size = 100 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin (e+f x) \, dx=\frac {3 a \sqrt {b} \text {arctanh}\left (\frac {\sqrt {b} \sec (e+f x)}{\sqrt {a+b \sec ^2(e+f x)}}\right )}{2 f}+\frac {3 b \sec (e+f x) \sqrt {a+b \sec ^2(e+f x)}}{2 f}-\frac {\cos (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2}}{f} \] Output:

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

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 0.48 (sec) , antiderivative size = 73, normalized size of antiderivative = 0.73 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin (e+f x) \, dx=-\frac {a \cos (e+f x) (a+2 b+a \cos (2 (e+f x)))^2 \operatorname {Hypergeometric2F1}\left (2,\frac {5}{2},\frac {7}{2},1+\frac {a \cos ^2(e+f x)}{b}\right ) \sqrt {a+b \sec ^2(e+f x)}}{20 b^2 f} \] Input:

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

Output:

-1/20*(a*Cos[e + f*x]*(a + 2*b + a*Cos[2*(e + f*x)])^2*Hypergeometric2F1[2 
, 5/2, 7/2, 1 + (a*Cos[e + f*x]^2)/b]*Sqrt[a + b*Sec[e + f*x]^2])/(b^2*f)
 

Rubi [A] (verified)

Time = 0.26 (sec) , antiderivative size = 98, normalized size of antiderivative = 0.98, number of steps used = 7, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3042, 4622, 247, 211, 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.

\(\displaystyle \int \sin (e+f x) \left (a+b \sec ^2(e+f x)\right )^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \sin (e+f x) \left (a+b \sec (e+f x)^2\right )^{3/2}dx\)

\(\Big \downarrow \) 4622

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

\(\Big \downarrow \) 247

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

\(\Big \downarrow \) 211

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

\(\Big \downarrow \) 224

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

\(\Big \downarrow \) 219

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 211

Int[((a_) + (b_.)*(x_)^2)^(p_), x_Symbol] :> Simp[x*((a + b*x^2)^p/(2*p + 1 
)), x] + Simp[2*a*(p/(2*p + 1))   Int[(a + b*x^2)^(p - 1), x], x] /; FreeQ[ 
{a, b}, x] && GtQ[p, 0] && (IntegerQ[4*p] || IntegerQ[6*p])
 

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 3042

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

rule 4622

Int[((a_) + (b_.)*((c_.)*sec[(e_.) + (f_.)*(x_)])^(n_))^(p_.)*sin[(e_.) + ( 
f_.)*(x_)]^(m_.), x_Symbol] :> With[{ff = FreeFactors[Cos[e + f*x], x]}, Si 
mp[1/(f*ff^m)   Subst[Int[(-1 + ff^2*x^2)^((m - 1)/2)*((a + b*(c*ff*x)^n)^p 
/x^(m + 1)), x], x, Sec[e + f*x]/ff], x]] /; FreeQ[{a, b, c, e, f, n, p}, x 
] && IntegerQ[(m - 1)/2] && (GtQ[m, 0] || EqQ[n, 2] || EqQ[n, 4])
 

Maple [A] (verified)

Time = 0.22 (sec) , antiderivative size = 121, normalized size of antiderivative = 1.21

method	result	size
derivativedivides	\(-\frac {\left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {5}{2}}}{f a \sec \left (f x +e \right )}+\frac {b \sec \left (f x +e \right ) \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{f a}+\frac {3 b \sec \left (f x +e \right ) \sqrt {a +b \sec \left (f x +e \right )^{2}}}{2 f}+\frac {3 \sqrt {b}\, a \ln \left (\sqrt {b}\, \sec \left (f x +e \right )+\sqrt {a +b \sec \left (f x +e \right )^{2}}\right )}{2 f}\)	\(121\)
default	\(-\frac {\left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {5}{2}}}{f a \sec \left (f x +e \right )}+\frac {b \sec \left (f x +e \right ) \left (a +b \sec \left (f x +e \right )^{2}\right )^{\frac {3}{2}}}{f a}+\frac {3 b \sec \left (f x +e \right ) \sqrt {a +b \sec \left (f x +e \right )^{2}}}{2 f}+\frac {3 \sqrt {b}\, a \ln \left (\sqrt {b}\, \sec \left (f x +e \right )+\sqrt {a +b \sec \left (f x +e \right )^{2}}\right )}{2 f}\)	\(121\)

Input:

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

Output:

-1/f/a/sec(f*x+e)*(a+b*sec(f*x+e)^2)^(5/2)+1/f*b/a*sec(f*x+e)*(a+b*sec(f*x 
+e)^2)^(3/2)+3/2*b*sec(f*x+e)*(a+b*sec(f*x+e)^2)^(1/2)/f+3/2/f*b^(1/2)*a*l 
n(b^(1/2)*sec(f*x+e)+(a+b*sec(f*x+e)^2)^(1/2))
 

Fricas [A] (verification not implemented)

Time = 0.24 (sec) , antiderivative size = 242, normalized size of antiderivative = 2.42 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin (e+f x) \, dx=\left [\frac {3 \, a \sqrt {b} \cos \left (f x + e\right ) \log \left (\frac {a \cos \left (f x + e\right )^{2} + 2 \, \sqrt {b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + 2 \, b}{\cos \left (f x + e\right )^{2}}\right ) - 2 \, {\left (2 \, a \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{4 \, f \cos \left (f x + e\right )}, -\frac {3 \, a \sqrt {-b} \arctan \left (\frac {\sqrt {-b} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right )}{a \cos \left (f x + e\right )^{2} + b}\right ) \cos \left (f x + e\right ) + {\left (2 \, a \cos \left (f x + e\right )^{2} - b\right )} \sqrt {\frac {a \cos \left (f x + e\right )^{2} + b}{\cos \left (f x + e\right )^{2}}}}{2 \, f \cos \left (f x + e\right )}\right ] \] Input:

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

Output:

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

Sympy [F]

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

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

Output:

Integral((a + b*sec(e + f*x)**2)**(3/2)*sin(e + f*x), x)
 

Maxima [A] (verification not implemented)

Time = 0.11 (sec) , antiderivative size = 142, normalized size of antiderivative = 1.42 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin (e+f x) \, dx=-\frac {4 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} a \cos \left (f x + e\right ) - \frac {2 \, \sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} a b \cos \left (f x + e\right )}{{\left (a + \frac {b}{\cos \left (f x + e\right )^{2}}\right )} \cos \left (f x + e\right )^{2} - b} + 3 \, a \sqrt {b} \log \left (\frac {\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) - \sqrt {b}}{\sqrt {a + \frac {b}{\cos \left (f x + e\right )^{2}}} \cos \left (f x + e\right ) + \sqrt {b}}\right )}{4 \, f} \] Input:

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

Output:

-1/4*(4*sqrt(a + b/cos(f*x + e)^2)*a*cos(f*x + e) - 2*sqrt(a + b/cos(f*x + 
 e)^2)*a*b*cos(f*x + e)/((a + b/cos(f*x + e)^2)*cos(f*x + e)^2 - b) + 3*a* 
sqrt(b)*log((sqrt(a + b/cos(f*x + e)^2)*cos(f*x + e) - sqrt(b))/(sqrt(a + 
b/cos(f*x + e)^2)*cos(f*x + e) + sqrt(b))))/f
 

Giac [A] (verification not implemented)

Time = 0.20 (sec) , antiderivative size = 87, normalized size of antiderivative = 0.87 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin (e+f x) \, dx=-\frac {{\left (\frac {3 \, b \arctan \left (\frac {\sqrt {a \cos \left (f x + e\right )^{2} + b}}{\sqrt {-b}}\right )}{\sqrt {-b}} + 2 \, \sqrt {a \cos \left (f x + e\right )^{2} + b} - \frac {\sqrt {a \cos \left (f x + e\right )^{2} + b} b}{a \cos \left (f x + e\right )^{2}}\right )} a \mathrm {sgn}\left (\cos \left (f x + e\right )\right )}{2 \, f} \] Input:

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

Output:

-1/2*(3*b*arctan(sqrt(a*cos(f*x + e)^2 + b)/sqrt(-b))/sqrt(-b) + 2*sqrt(a* 
cos(f*x + e)^2 + b) - sqrt(a*cos(f*x + e)^2 + b)*b/(a*cos(f*x + e)^2))*a*s 
gn(cos(f*x + e))/f
 

Mupad [B] (verification not implemented)

Time = 14.11 (sec) , antiderivative size = 61, normalized size of antiderivative = 0.61 \[ \int \left (a+b \sec ^2(e+f x)\right )^{3/2} \sin (e+f x) \, dx=-\frac {\cos \left (e+f\,x\right )\,{\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},-\frac {1}{2};\ \frac {1}{2};\ -\frac {b}{a\,{\cos \left (e+f\,x\right )}^2}\right )}{f\,{\left (\frac {b}{a\,{\cos \left (e+f\,x\right )}^2}+1\right )}^{3/2}} \] Input:

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

Output:

-(cos(e + f*x)*(a + b/cos(e + f*x)^2)^(3/2)*hypergeom([-3/2, -1/2], 1/2, - 
b/(a*cos(e + f*x)^2)))/(f*(b/(a*cos(e + f*x)^2) + 1)^(3/2))
 

Reduce [F]

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

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

Output:

int(sqrt(sec(e + f*x)**2*b + a)*sec(e + f*x)**2*sin(e + f*x),x)*b + int(sq 
rt(sec(e + f*x)**2*b + a)*sin(e + f*x),x)*a