\(\int (b \cos (c+d x))^{3/2} (A+C \cos ^2(c+d x)) \sec ^4(c+d x) \, dx\) [50]

Optimal result

Integrand size = 33, antiderivative size = 78 \[ \int (b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {2 b^2 (A+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d \sqrt {b \cos (c+d x)}}+\frac {2 A b^3 \sin (c+d x)}{3 d (b \cos (c+d x))^{3/2}} \] Output:

2/3*b^2*(A+3*C)*cos(d*x+c)^(1/2)*InverseJacobiAM(1/2*d*x+1/2*c,2^(1/2))/d/ 
(b*cos(d*x+c))^(1/2)+2/3*A*b^3*sin(d*x+c)/d/(b*cos(d*x+c))^(3/2)
 

Mathematica [A] (verified)

Time = 0.66 (sec) , antiderivative size = 58, normalized size of antiderivative = 0.74 \[ \int (b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\frac {2 b^2 \left ((A+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+A \tan (c+d x)\right )}{3 d \sqrt {b \cos (c+d x)}} \] Input:

Integrate[(b*Cos[c + d*x])^(3/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]
 

Output:

(2*b^2*((A + 3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2] + A*Tan[c + 
 d*x]))/(3*d*Sqrt[b*Cos[c + d*x]])
 

Rubi [A] (verified)

Time = 0.43 (sec) , antiderivative size = 82, normalized size of antiderivative = 1.05, number of steps used = 7, number of rules used = 7, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.212, Rules used = {3042, 2030, 3491, 3042, 3121, 3042, 3120}

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.

\(\displaystyle \int \sec ^4(c+d x) (b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2} \left (A+C \sin \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\sin \left (c+d x+\frac {\pi }{2}\right )^4}dx\)

\(\Big \downarrow \) 2030

\(\displaystyle b^4 \int \frac {C \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )^2+A}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{5/2}}dx\)

\(\Big \downarrow \) 3491

\(\displaystyle b^4 \left (\frac {(A+3 C) \int \frac {1}{\sqrt {b \cos (c+d x)}}dx}{3 b^2}+\frac {2 A \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^4 \left (\frac {(A+3 C) \int \frac {1}{\sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2}+\frac {2 A \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )\)

\(\Big \downarrow \) 3121

\(\displaystyle b^4 \left (\frac {(A+3 C) \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\cos (c+d x)}}dx}{3 b^2 \sqrt {b \cos (c+d x)}}+\frac {2 A \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^4 \left (\frac {(A+3 C) \sqrt {\cos (c+d x)} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b^2 \sqrt {b \cos (c+d x)}}+\frac {2 A \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )\)

\(\Big \downarrow \) 3120

\(\displaystyle b^4 \left (\frac {2 (A+3 C) \sqrt {\cos (c+d x)} \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 b^2 d \sqrt {b \cos (c+d x)}}+\frac {2 A \sin (c+d x)}{3 b d (b \cos (c+d x))^{3/2}}\right )\)

Input:

Int[(b*Cos[c + d*x])^(3/2)*(A + C*Cos[c + d*x]^2)*Sec[c + d*x]^4,x]
 

Output:

b^4*((2*(A + 3*C)*Sqrt[Cos[c + d*x]]*EllipticF[(c + d*x)/2, 2])/(3*b^2*d*S 
qrt[b*Cos[c + d*x]]) + (2*A*Sin[c + d*x])/(3*b*d*(b*Cos[c + d*x])^(3/2)))
 

Defintions of rubi rules used

rule 2030

Int[(Fx_.)*(v_)^(m_.)*((b_)*(v_))^(n_), x_Symbol] :> Simp[1/b^m   Int[(b*v) 
^(m + n)*Fx, x], x] /; FreeQ[{b, n}, x] && IntegerQ[m]
 

rule 3042

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

rule 3120

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

rule 3121

Int[((b_)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[(b*Sin[c + d*x]) 
^n/Sin[c + d*x]^n   Int[Sin[c + d*x]^n, x], x] /; FreeQ[{b, c, d}, x] && Lt 
Q[-1, n, 1] && IntegerQ[2*n]
 

rule 3491

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((A_) + (C_.)*sin[(e_.) + (f_.)*(x 
_)]^2), x_Symbol] :> Simp[A*Cos[e + f*x]*((b*Sin[e + f*x])^(m + 1)/(b*f*(m 
+ 1))), x] + Simp[(A*(m + 2) + C*(m + 1))/(b^2*(m + 1))   Int[(b*Sin[e + f* 
x])^(m + 2), x], x] /; FreeQ[{b, e, f, A, C}, x] && LtQ[m, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(293\) vs. \(2(69)=138\).

Time = 0.39 (sec) , antiderivative size = 294, normalized size of antiderivative = 3.77

method	result	size
default	\(-\frac {2 \left (-2 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \left (A +3 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+3 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) b^{2} \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 \sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, d}\)	\(294\)
parts	\(-\frac {2 A \left (-2 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right ) b^{2} \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}}{3 \sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, d}-\frac {2 C \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, b^{2} \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}+1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{\sqrt {-b \left (2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}-\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {b \left (-1+2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}\right )}\, d}\)	\(387\)

Input:

int((b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x,method=_RETURNV 
ERBOSE)
 

Output:

-2/3*(-2*A*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^2-2*EllipticF(cos(1/2*d*x 
+1/2*c),2^(1/2))*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^( 
1/2)*(A+3*C)*sin(1/2*d*x+1/2*c)^2+A*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/ 
2*d*x+1/2*c)^2-1)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))+3*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)*EllipticF(cos(1/2*d* 
x+1/2*c),2^(1/2)))*b^2*(b*(-1+2*cos(1/2*d*x+1/2*c)^2)*sin(1/2*d*x+1/2*c)^2 
)^(1/2)/(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/(-1+2*cos 
(1/2*d*x+1/2*c)^2)/sin(1/2*d*x+1/2*c)/(b*(-1+2*cos(1/2*d*x+1/2*c)^2))^(1/2 
)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 112, normalized size of antiderivative = 1.44 \[ \int (b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=-\frac {2 \, {\left (i \, \sqrt {\frac {1}{2}} {\left (A + 3 \, C\right )} b^{\frac {3}{2}} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - i \, \sqrt {\frac {1}{2}} {\left (A + 3 \, C\right )} b^{\frac {3}{2}} \cos \left (d x + c\right )^{2} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - \sqrt {b \cos \left (d x + c\right )} A b \sin \left (d x + c\right )\right )}}{3 \, d \cos \left (d x + c\right )^{2}} \] Input:

integrate((b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorith 
m="fricas")
 

Output:

-2/3*(I*sqrt(1/2)*(A + 3*C)*b^(3/2)*cos(d*x + c)^2*weierstrassPInverse(-4, 
 0, cos(d*x + c) + I*sin(d*x + c)) - I*sqrt(1/2)*(A + 3*C)*b^(3/2)*cos(d*x 
 + c)^2*weierstrassPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - sqrt(b 
*cos(d*x + c))*A*b*sin(d*x + c))/(d*cos(d*x + c)^2)
 

Sympy [F(-1)]

Timed out. \[ \int (b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\text {Timed out} \] Input:

integrate((b*cos(d*x+c))**(3/2)*(A+C*cos(d*x+c)**2)*sec(d*x+c)**4,x)
 

Output:

Timed out
 

Maxima [F]

\[ \int (b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}} \sec \left (d x + c\right )^{4} \,d x } \] Input:

integrate((b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorith 
m="maxima")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(3/2)*sec(d*x + c)^4, x)
 

Giac [F]

\[ \int (b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int { {\left (C \cos \left (d x + c\right )^{2} + A\right )} \left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}} \sec \left (d x + c\right )^{4} \,d x } \] Input:

integrate((b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x, algorith 
m="giac")
 

Output:

integrate((C*cos(d*x + c)^2 + A)*(b*cos(d*x + c))^(3/2)*sec(d*x + c)^4, x)
 

Mupad [F(-1)]

Timed out. \[ \int (b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\int \frac {\left (C\,{\cos \left (c+d\,x\right )}^2+A\right )\,{\left (b\,\cos \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^4} \,d x \] Input:

int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(3/2))/cos(c + d*x)^4,x)
 

Output:

int(((A + C*cos(c + d*x)^2)*(b*cos(c + d*x))^(3/2))/cos(c + d*x)^4, x)
 

Reduce [F]

\[ \int (b \cos (c+d x))^{3/2} \left (A+C \cos ^2(c+d x)\right ) \sec ^4(c+d x) \, dx=\sqrt {b}\, b \left (\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right ) \sec \left (d x +c \right )^{4}d x \right ) a +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{4}d x \right ) c \right ) \] Input:

int((b*cos(d*x+c))^(3/2)*(A+C*cos(d*x+c)^2)*sec(d*x+c)^4,x)
 

Output:

sqrt(b)*b*(int(sqrt(cos(c + d*x))*cos(c + d*x)*sec(c + d*x)**4,x)*a + int( 
sqrt(cos(c + d*x))*cos(c + d*x)**3*sec(c + d*x)**4,x)*c)