\(\int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx\) [568]

Optimal result

Integrand size = 33, antiderivative size = 126 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {2 A \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{a d}-\frac {2 (A b-a B) \sqrt {\cos (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right ) \sqrt {\sec (c+d x)}}{a (a+b) d}+\frac {2 A \sqrt {\sec (c+d x)} \sin (c+d x)}{a d} \] Output:

-2*A*cos(d*x+c)^(1/2)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))*sec(d*x+c)^(1/ 
2)/a/d-2*(A*b-B*a)*cos(d*x+c)^(1/2)*EllipticPi(sin(1/2*d*x+1/2*c),2*b/(a+b 
),2^(1/2))*sec(d*x+c)^(1/2)/a/(a+b)/d+2*A*sec(d*x+c)^(1/2)*sin(d*x+c)/a/d
 

Mathematica [A] (warning: unable to verify)

Time = 3.63 (sec) , antiderivative size = 125, normalized size of antiderivative = 0.99 \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=-\frac {2 \cos (2 (c+d x)) \csc (c+d x) \left (a A E\left (\left .\arcsin \left (\sqrt {\sec (c+d x)}\right )\right |-1\right )-(a A+A b-a B) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )+(A b-a B) \operatorname {EllipticPi}\left (-\frac {a}{b},\arcsin \left (\sqrt {\sec (c+d x)}\right ),-1\right )\right ) \sec (c+d x) \sqrt {-\tan ^2(c+d x)}}{a^2 d \left (-2+\sec ^2(c+d x)\right )} \] Input:

Integrate[((A + B*Cos[c + d*x])*Sec[c + d*x]^(3/2))/(a + b*Cos[c + d*x]),x 
]
 

Output:

(-2*Cos[2*(c + d*x)]*Csc[c + d*x]*(a*A*EllipticE[ArcSin[Sqrt[Sec[c + d*x]] 
], -1] - (a*A + A*b - a*B)*EllipticF[ArcSin[Sqrt[Sec[c + d*x]]], -1] + (A* 
b - a*B)*EllipticPi[-(a/b), ArcSin[Sqrt[Sec[c + d*x]]], -1])*Sec[c + d*x]* 
Sqrt[-Tan[c + d*x]^2])/(a^2*d*(-2 + Sec[c + d*x]^2))
 

Rubi [A] (verified)

Time = 1.10 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 15, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.455, Rules used = {3042, 3439, 3042, 4521, 27, 3042, 4594, 27, 3042, 4258, 3042, 3119, 4336, 3042, 3284}

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.

\(\displaystyle \int \frac {\sec ^{\frac {3}{2}}(c+d x) (A+B \cos (c+d x))}{a+b \cos (c+d x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (A+B \sin \left (c+d x+\frac {\pi }{2}\right )\right )}{a+b \sin \left (c+d x+\frac {\pi }{2}\right )}dx\)

\(\Big \downarrow \) 3439

\(\displaystyle \int \frac {\sec ^{\frac {3}{2}}(c+d x) (A \sec (c+d x)+B)}{a \sec (c+d x)+b}dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (A \csc \left (c+d x+\frac {\pi }{2}\right )+B\right )}{a \csc \left (c+d x+\frac {\pi }{2}\right )+b}dx\)

\(\Big \downarrow \) 4521

\(\displaystyle \frac {2 \int -\frac {(A b-a B) \sec ^2(c+d x)+a A \sec (c+d x)+A b}{2 \sqrt {\sec (c+d x)} (b+a \sec (c+d x))}dx}{a}+\frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)}}{a d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)}}{a d}-\frac {\int \frac {(A b-a B) \sec ^2(c+d x)+a A \sec (c+d x)+A b}{\sqrt {\sec (c+d x)} (b+a \sec (c+d x))}dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)}}{a d}-\frac {\int \frac {(A b-a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+a A \csc \left (c+d x+\frac {\pi }{2}\right )+A b}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )} \left (b+a \csc \left (c+d x+\frac {\pi }{2}\right )\right )}dx}{a}\)

\(\Big \downarrow \) 4594

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)}}{a d}-\frac {(A b-a B) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{b+a \sec (c+d x)}dx+\frac {\int \frac {A b^2}{\sqrt {\sec (c+d x)}}dx}{b^2}}{a}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)}}{a d}-\frac {(A b-a B) \int \frac {\sec ^{\frac {3}{2}}(c+d x)}{b+a \sec (c+d x)}dx+A \int \frac {1}{\sqrt {\sec (c+d x)}}dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)}}{a d}-\frac {(A b-a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx+A \int \frac {1}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{a}\)

\(\Big \downarrow \) 4258

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)}}{a d}-\frac {(A b-a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx+A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\cos (c+d x)}dx}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)}}{a d}-\frac {(A b-a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx+A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{a}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)}}{a d}-\frac {(A b-a B) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^{3/2}}{b+a \csc \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}\)

\(\Big \downarrow \) 4336

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)}}{a d}-\frac {(A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)} (a+b \cos (c+d x))}dx+\frac {2 A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)}}{a d}-\frac {(A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )} \left (a+b \sin \left (c+d x+\frac {\pi }{2}\right )\right )}dx+\frac {2 A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}\)

\(\Big \downarrow \) 3284

\(\displaystyle \frac {2 A \sin (c+d x) \sqrt {\sec (c+d x)}}{a d}-\frac {\frac {2 (A b-a B) \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{d (a+b)}+\frac {2 A \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}}{a}\)

Input:

Int[((A + B*Cos[c + d*x])*Sec[c + d*x]^(3/2))/(a + b*Cos[c + d*x]),x]
 

Output:

-(((2*A*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d 
 + (2*(A*b - a*B)*Sqrt[Cos[c + d*x]]*EllipticPi[(2*b)/(a + b), (c + d*x)/2 
, 2]*Sqrt[Sec[c + d*x]])/((a + b)*d))/a) + (2*A*Sqrt[Sec[c + d*x]]*Sin[c + 
 d*x])/(a*d)
 

Defintions of rubi rules used

rule 27

Int[(a_)*(Fx_), x_Symbol] :> Simp[a   Int[Fx, x], x] /; FreeQ[a, x] &&  !Ma 
tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
 

rule 3042

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

rule 3119

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2/d)*EllipticE[(1/2)* 
(c - Pi/2 + d*x), 2], x] /; FreeQ[{c, d}, x]
 

rule 3284

Int[1/(((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])*Sqrt[(c_.) + (d_.)*sin[(e_.) 
 + (f_.)*(x_)]]), x_Symbol] :> Simp[(2/(f*(a + b)*Sqrt[c + d]))*EllipticPi[ 
2*(b/(a + b)), (1/2)*(e - Pi/2 + f*x), 2*(d/(c + d))], x] /; FreeQ[{a, b, c 
, d, e, f}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 
0] && GtQ[c + d, 0]
 

rule 3439

Int[(csc[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*((a_.) + (b_.)*sin[(e_.) + (f_.)* 
(x_)])^(m_.)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Sim 
p[g^(m + n)   Int[(g*Csc[e + f*x])^(p - m - n)*(b + a*Csc[e + f*x])^m*(d + 
c*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, c, d, e, f, g, p}, x] && NeQ[b*c 
- a*d, 0] &&  !IntegerQ[p] && IntegerQ[m] && IntegerQ[n]
 

rule 4258

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(b*Csc[c + d*x] 
)^n*Sin[c + d*x]^n   Int[1/Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && 
 EqQ[n^2, 1/4]
 

rule 4336

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(3/2)/(csc[(e_.) + (f_.)*(x_)]*(b_.) + 
(a_)), x_Symbol] :> Simp[d*Sqrt[d*Sin[e + f*x]]*Sqrt[d*Csc[e + f*x]]   Int[ 
1/(Sqrt[d*Sin[e + f*x]]*(b + a*Sin[e + f*x])), x], x] /; FreeQ[{a, b, d, e, 
 f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4521

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*d^ 
2*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 2)/(b*f* 
(m + n))), x] + Simp[d^2/(b*(m + n))   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e 
+ f*x])^(n - 2)*Simp[a*B*(n - 2) + B*b*(m + n - 1)*Csc[e + f*x] + (A*b*(m + 
 n) - a*B*(n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B 
, m}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && NeQ[m + 
n, 0] &&  !IGtQ[m, 1]
 

rule 4594

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_. 
))/(Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a 
_))), x_Symbol] :> Simp[(A*b^2 - a*b*B + a^2*C)/(a^2*d^2)   Int[(d*Csc[e + 
f*x])^(3/2)/(a + b*Csc[e + f*x]), x], x] + Simp[1/a^2   Int[(a*A - (A*b - a 
*B)*Csc[e + f*x])/Sqrt[d*Csc[e + f*x]], x], x] /; FreeQ[{a, b, d, e, f, A, 
B, C}, x] && NeQ[a^2 - b^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(299\) vs. \(2(120)=240\).

Time = 5.15 (sec) , antiderivative size = 300, normalized size of antiderivative = 2.38

method	result	size
default	\(-\frac {\sqrt {-\left (-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (\frac {2 A \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{a \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right )}+\frac {4 \left (A b -B a \right ) b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right )}{a \left (-2 a b +2 b^{2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(300\)

Input:

int((A+B*cos(d*x+c))*sec(d*x+c)^(3/2)/(a+cos(d*x+c)*b),x,method=_RETURNVER 
BOSE)
 

Output:

-(-(-2*cos(1/2*d*x+1/2*c)^2+1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(2/a*A/sin(1/2* 
d*x+1/2*c)^2/(2*sin(1/2*d*x+1/2*c)^2-1)*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d 
*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)-(sin(1/2*d*x 
+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(cos(1/2*d*x+1/ 
2*c),2^(1/2)))+4*(A*b-B*a)/a/(-2*a*b+2*b^2)*b*(sin(1/2*d*x+1/2*c)^2)^(1/2) 
*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/ 
2*c)^2)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/(a-b),2^(1/2)))/sin(1/2*d 
*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
 

Fricas [F(-1)]

Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\text {Timed out} \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(3/2)/(a+b*cos(d*x+c)),x, algorithm= 
"fricas")
 

Output:

Timed out
 

Sympy [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\int \frac {\left (A + B \cos {\left (c + d x \right )}\right ) \sec ^{\frac {3}{2}}{\left (c + d x \right )}}{a + b \cos {\left (c + d x \right )}}\, dx \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)**(3/2)/(a+b*cos(d*x+c)),x)
 

Output:

Integral((A + B*cos(c + d*x))*sec(c + d*x)**(3/2)/(a + b*cos(c + d*x)), x)
 

Maxima [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}{b \cos \left (d x + c\right ) + a} \,d x } \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(3/2)/(a+b*cos(d*x+c)),x, algorithm= 
"maxima")
 

Output:

integrate((B*cos(d*x + c) + A)*sec(d*x + c)^(3/2)/(b*cos(d*x + c) + a), x)
 

Giac [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\int { \frac {{\left (B \cos \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac {3}{2}}}{b \cos \left (d x + c\right ) + a} \,d x } \] Input:

integrate((A+B*cos(d*x+c))*sec(d*x+c)^(3/2)/(a+b*cos(d*x+c)),x, algorithm= 
"giac")
 

Output:

integrate((B*cos(d*x + c) + A)*sec(d*x + c)^(3/2)/(b*cos(d*x + c) + a), x)
 

Mupad [F(-1)]

Timed out. \[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\int \frac {\left (A+B\,\cos \left (c+d\,x\right )\right )\,{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{a+b\,\cos \left (c+d\,x\right )} \,d x \] Input:

int(((A + B*cos(c + d*x))*(1/cos(c + d*x))^(3/2))/(a + b*cos(c + d*x)),x)
 

Output:

int(((A + B*cos(c + d*x))*(1/cos(c + d*x))^(3/2))/(a + b*cos(c + d*x)), x)
 

Reduce [F]

\[ \int \frac {(A+B \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x)}{a+b \cos (c+d x)} \, dx=\int \sqrt {\sec \left (d x +c \right )}\, \sec \left (d x +c \right )d x \] Input:

int((A+B*cos(d*x+c))*sec(d*x+c)^(3/2)/(a+b*cos(d*x+c)),x)
 

Output:

int(sqrt(sec(c + d*x))*sec(c + d*x),x)