\(\int \frac {a B+b B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx\) [393]

Optimal result

Integrand size = 36, antiderivative size = 80 \[ \int \frac {a B+b B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=-\frac {2 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{a d}-\frac {2 b B \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a (a+b) d}+\frac {2 B \sin (c+d x)}{a d \sqrt {\cos (c+d x)}} \] Output:

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

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(196\) vs. \(2(80)=160\).

Time = 3.32 (sec) , antiderivative size = 196, normalized size of antiderivative = 2.45 \[ \int \frac {a B+b B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=-\frac {B \left (\frac {6 b \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}+\frac {2 a \left (2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )-\frac {2 a \operatorname {EllipticPi}\left (\frac {2 b}{a+b},\frac {1}{2} (c+d x),2\right )}{a+b}\right )}{b}-\frac {4 \sin (c+d x)}{\sqrt {\cos (c+d x)}}+\frac {2 \left (-2 a b E\left (\left .\arcsin \left (\sqrt {\cos (c+d x)}\right )\right |-1\right )+2 a (a+b) \operatorname {EllipticF}\left (\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )+\left (-2 a^2+b^2\right ) \operatorname {EllipticPi}\left (-\frac {b}{a},\arcsin \left (\sqrt {\cos (c+d x)}\right ),-1\right )\right ) \sin (c+d x)}{a b \sqrt {\sin ^2(c+d x)}}\right )}{2 a d} \] Input:

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

Output:

-1/2*(B*((6*b*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2])/(a + b) + (2*a*(2 
*EllipticF[(c + d*x)/2, 2] - (2*a*EllipticPi[(2*b)/(a + b), (c + d*x)/2, 2 
])/(a + b)))/b - (4*Sin[c + d*x])/Sqrt[Cos[c + d*x]] + (2*(-2*a*b*Elliptic 
E[ArcSin[Sqrt[Cos[c + d*x]]], -1] + 2*a*(a + b)*EllipticF[ArcSin[Sqrt[Cos[ 
c + d*x]]], -1] + (-2*a^2 + b^2)*EllipticPi[-(b/a), ArcSin[Sqrt[Cos[c + d* 
x]]], -1])*Sin[c + d*x])/(a*b*Sqrt[Sin[c + d*x]^2])))/(a*d)
 

Rubi [A] (verified)

Time = 0.60 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.99, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.306, Rules used = {2011, 3042, 3281, 27, 3042, 3538, 25, 27, 3042, 3119, 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.

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

\(\Big \downarrow \) 2011

\(\displaystyle B \int \frac {1}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))}dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3281

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3538

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

\(\Big \downarrow \) 25

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

\(\Big \downarrow \) 27

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3119

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

\(\Big \downarrow \) 3284

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

Input:

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

Output:

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

Defintions of rubi rules used

rule 25

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

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 2011

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> 
 Simp[(b/d)^m   Int[u*(c + d*v)^(m + n), x], x] /; FreeQ[{a, b, c, d, n}, x 
] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c + d*x 
, a + 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 3281

Int[((a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + 
 (f_.)*(x_)])^(n_), x_Symbol] :> Simp[(-b^2)*Cos[e + f*x]*(a + b*Sin[e + f* 
x])^(m + 1)*((c + d*Sin[e + f*x])^(n + 1)/(f*(m + 1)*(b*c - a*d)*(a^2 - b^2 
))), x] + Simp[1/((m + 1)*(b*c - a*d)*(a^2 - b^2))   Int[(a + b*Sin[e + f*x 
])^(m + 1)*(c + d*Sin[e + f*x])^n*Simp[a*(b*c - a*d)*(m + 1) + b^2*d*(m + n 
 + 2) - (b^2*c + b*(b*c - a*d)*(m + 1))*Sin[e + f*x] - b^2*d*(m + n + 3)*Si 
n[e + f*x]^2, x], x], x] /; FreeQ[{a, b, c, d, e, f, n}, x] && NeQ[b*c - a* 
d, 0] && NeQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0] && LtQ[m, -1] && IntegersQ[ 
2*m, 2*n] && ((EqQ[a, 0] && IntegerQ[m] &&  !IntegerQ[n]) ||  !(IntegerQ[2* 
n] && LtQ[n, -1] && ((IntegerQ[n] &&  !IntegerQ[m]) || EqQ[a, 0])))
 

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 3538

Int[((A_.) + (B_.)*sin[(e_.) + (f_.)*(x_)] + (C_.)*sin[(e_.) + (f_.)*(x_)]^ 
2)/(Sqrt[(a_.) + (b_.)*sin[(e_.) + (f_.)*(x_)]]*((c_.) + (d_.)*sin[(e_.) + 
(f_.)*(x_)])), x_Symbol] :> Simp[C/(b*d)   Int[Sqrt[a + b*Sin[e + f*x]], x] 
, x] - Simp[1/(b*d)   Int[Simp[a*c*C - A*b*d + (b*c*C - b*B*d + a*C*d)*Sin[ 
e + f*x], x]/(Sqrt[a + b*Sin[e + f*x]]*(c + d*Sin[e + f*x])), x], x] /; Fre 
eQ[{a, b, c, d, e, f, A, B, C}, x] && NeQ[b*c - a*d, 0] && NeQ[a^2 - b^2, 0 
] && NeQ[c^2 - d^2, 0]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(354\) vs. \(2(82)=164\).

Time = 4.12 (sec) , antiderivative size = 355, normalized size of antiderivative = 4.44

method	result	size
default	\(-\frac {2 B \left (-2 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (a -b \right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) a -\sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) b -b \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticPi}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), -\frac {2 b}{a -b}, \sqrt {2}\right ) \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\right )}{a \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (a -b \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(355\)

Input:

int((B*a+b*B*cos(d*x+c))/cos(d*x+c)^(3/2)/(a+cos(d*x+c)*b)^2,x,method=_RET 
URNVERBOSE)
 

Output:

-2*B*(-2*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)*(a-b)*cos(1/ 
2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2+(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2 
*c)^2)^(1/2)*(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))*a-(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2* 
d*x+1/2*c)^2)^(1/2)*(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))*b-b*(sin(1/2*d*x+1/2*c)^2)^( 
1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticPi(cos(1/2*d*x+1/2*c),-2*b/( 
a-b),2^(1/2))*(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2))/a/(-2* 
sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*c)^2)^(1/2)/(a-b)/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. \[ \int \frac {a B+b B \cos (c+d x)}{\cos ^{\frac {3}{2}}(c+d x) (a+b \cos (c+d x))^2} \, dx=\text {Timed out} \] Input:

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

Output:

Timed out
 

Sympy [F(-1)]

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

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

Output:

Timed out
 

Maxima [F]

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

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

Output:

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

Giac [F]

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

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

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

int(sqrt(cos(c + d*x))/(cos(c + d*x)**3*b + cos(c + d*x)**2*a),x)*b