\(\int \cos ^{\frac {7}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [1180]

Optimal result

Integrand size = 31, antiderivative size = 123 \[ \int \cos ^{\frac {7}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {6 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 (5 A+7 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{21 d}+\frac {2 (5 A+7 C) \sqrt {\cos (c+d x)} \sin (c+d x)}{21 d}+\frac {2 B \cos ^{\frac {3}{2}}(c+d x) \sin (c+d x)}{5 d}+\frac {2 A \cos ^{\frac {5}{2}}(c+d x) \sin (c+d x)}{7 d} \] Output:

6/5*B*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2))/d+2/21*(5*A+7*C)*InverseJacobi 
AM(1/2*d*x+1/2*c,2^(1/2))/d+2/21*(5*A+7*C)*cos(d*x+c)^(1/2)*sin(d*x+c)/d+2 
/5*B*cos(d*x+c)^(3/2)*sin(d*x+c)/d+2/7*A*cos(d*x+c)^(5/2)*sin(d*x+c)/d
 

Mathematica [A] (verified)

Time = 1.05 (sec) , antiderivative size = 86, normalized size of antiderivative = 0.70 \[ \int \cos ^{\frac {7}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {126 B E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+10 (5 A+7 C) \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )+\sqrt {\cos (c+d x)} (65 A+70 C+42 B \cos (c+d x)+15 A \cos (2 (c+d x))) \sin (c+d x)}{105 d} \] Input:

Integrate[Cos[c + d*x]^(7/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
 

Output:

(126*B*EllipticE[(c + d*x)/2, 2] + 10*(5*A + 7*C)*EllipticF[(c + d*x)/2, 2 
] + Sqrt[Cos[c + d*x]]*(65*A + 70*C + 42*B*Cos[c + d*x] + 15*A*Cos[2*(c + 
d*x)])*Sin[c + d*x])/(105*d)
 

Rubi [A] (verified)

Time = 0.63 (sec) , antiderivative size = 125, normalized size of antiderivative = 1.02, number of steps used = 12, number of rules used = 12, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.387, Rules used = {3042, 4552, 3042, 3502, 27, 3042, 3227, 3042, 3115, 3042, 3119, 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.

\(\displaystyle \int \cos ^{\frac {7}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \cos (c+d x)^{7/2} \left (A+B \sec (c+d x)+C \sec (c+d x)^2\right )dx\)

\(\Big \downarrow \) 4552

\(\displaystyle \int \cos ^{\frac {3}{2}}(c+d x) \left (A \cos ^2(c+d x)+B \cos (c+d x)+C\right )dx\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 3502

\(\displaystyle \frac {2}{7} \int \frac {1}{2} \cos ^{\frac {3}{2}}(c+d x) (5 A+7 C+7 B \cos (c+d x))dx+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{7} \int \cos ^{\frac {3}{2}}(c+d x) (5 A+7 C+7 B \cos (c+d x))dx+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2} \left (5 A+7 C+7 B \sin \left (c+d x+\frac {\pi }{2}\right )\right )dx+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 3227

\(\displaystyle \frac {1}{7} \left ((5 A+7 C) \int \cos ^{\frac {3}{2}}(c+d x)dx+7 B \int \cos ^{\frac {5}{2}}(c+d x)dx\right )+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left ((5 A+7 C) \int \sin \left (c+d x+\frac {\pi }{2}\right )^{3/2}dx+7 B \int \sin \left (c+d x+\frac {\pi }{2}\right )^{5/2}dx\right )+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 3115

\(\displaystyle \frac {1}{7} \left ((5 A+7 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\cos (c+d x)}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+7 B \left (\frac {3}{5} \int \sqrt {\cos (c+d x)}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\right )+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{7} \left ((5 A+7 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+7 B \left (\frac {3}{5} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\right )+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 3119

\(\displaystyle \frac {1}{7} \left ((5 A+7 C) \left (\frac {1}{3} \int \frac {1}{\sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+7 B \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\right )+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

\(\Big \downarrow \) 3120

\(\displaystyle \frac {1}{7} \left ((5 A+7 C) \left (\frac {2 \operatorname {EllipticF}\left (\frac {1}{2} (c+d x),2\right )}{3 d}+\frac {2 \sin (c+d x) \sqrt {\cos (c+d x)}}{3 d}\right )+7 B \left (\frac {6 E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d}+\frac {2 \sin (c+d x) \cos ^{\frac {3}{2}}(c+d x)}{5 d}\right )\right )+\frac {2 A \sin (c+d x) \cos ^{\frac {5}{2}}(c+d x)}{7 d}\)

Input:

Int[Cos[c + d*x]^(7/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2),x]
 

Output:

(2*A*Cos[c + d*x]^(5/2)*Sin[c + d*x])/(7*d) + ((5*A + 7*C)*((2*EllipticF[( 
c + d*x)/2, 2])/(3*d) + (2*Sqrt[Cos[c + d*x]]*Sin[c + d*x])/(3*d)) + 7*B*( 
(6*EllipticE[(c + d*x)/2, 2])/(5*d) + (2*Cos[c + d*x]^(3/2)*Sin[c + d*x])/ 
(5*d)))/7
 

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 3115

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

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 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 3227

Int[((b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_) + (d_.)*sin[(e_.) + (f_.)*(x 
_)]), x_Symbol] :> Simp[c   Int[(b*Sin[e + f*x])^m, x], x] + Simp[d/b   Int 
[(b*Sin[e + f*x])^(m + 1), x], x] /; FreeQ[{b, c, d, e, f, m}, x]
 

rule 3502

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

rule 4552

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(m_)*((A_.) + (B_.)*sec[(e_.) + (f_.)*( 
x_)] + (C_.)*sec[(e_.) + (f_.)*(x_)]^2), x_Symbol] :> Simp[b^2   Int[(b*Cos 
[e + f*x])^(m - 2)*(C + B*Cos[e + f*x] + A*Cos[e + f*x]^2), x], x] /; FreeQ 
[{b, e, f, A, B, C, m}, x] &&  !IntegerQ[m]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(341\) vs. \(2(110)=220\).

Time = 17.72 (sec) , antiderivative size = 342, normalized size of antiderivative = 2.78

method	result	size
default	\(-\frac {2 \sqrt {\left (2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \left (240 A \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{8}+\left (-360 A -168 B \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{6} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (280 A +168 B +140 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+\left (-80 A -42 B -70 C \right ) \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2} \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+25 A \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-63 B \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticE}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+35 C \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, \operatorname {EllipticF}\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{105 \sqrt {-2 \sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{4}+\sin \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \cos \left (\frac {d x}{2}+\frac {c}{2}\right )^{2}-1}\, d}\)	\(342\)

Input:

int(cos(d*x+c)^(7/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETURNVERBO 
SE)
 

Output:

-2/105*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*(240*A*cos( 
1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^8+(-360*A-168*B)*sin(1/2*d*x+1/2*c)^6*co 
s(1/2*d*x+1/2*c)+(280*A+168*B+140*C)*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2* 
c)+(-80*A-42*B-70*C)*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)+25*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))-63*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)*EllipticE(cos(1/2*d*x+1/2*c),2^(1/2))+35*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)))/(-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*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d
 

Fricas [C] (verification not implemented)

Result contains complex when optimal does not.

Time = 0.09 (sec) , antiderivative size = 163, normalized size of antiderivative = 1.33 \[ \int \cos ^{\frac {7}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2 \, {\left (15 \, A \cos \left (d x + c\right )^{2} + 21 \, B \cos \left (d x + c\right ) + 25 \, A + 35 \, C\right )} \sqrt {\cos \left (d x + c\right )} \sin \left (d x + c\right ) - 5 \, \sqrt {2} {\left (5 i \, A + 7 i \, C\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 5 \, \sqrt {2} {\left (-5 i \, A - 7 i \, C\right )} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) + 63 i \, \sqrt {2} B {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) - 63 i \, \sqrt {2} B {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right )}{105 \, d} \] Input:

integrate(cos(d*x+c)^(7/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="f 
ricas")
 

Output:

1/105*(2*(15*A*cos(d*x + c)^2 + 21*B*cos(d*x + c) + 25*A + 35*C)*sqrt(cos( 
d*x + c))*sin(d*x + c) - 5*sqrt(2)*(5*I*A + 7*I*C)*weierstrassPInverse(-4, 
 0, cos(d*x + c) + I*sin(d*x + c)) - 5*sqrt(2)*(-5*I*A - 7*I*C)*weierstras 
sPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) + 63*I*sqrt(2)*B*weierstra 
ssZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) - 
 63*I*sqrt(2)*B*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x 
+ c) - I*sin(d*x + c))))/d
 

Sympy [F(-1)]

Timed out. \[ \int \cos ^{\frac {7}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Timed out} \] Input:

integrate(cos(d*x+c)**(7/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2),x)
 

Output:

Timed out
 

Maxima [F]

\[ \int \cos ^{\frac {7}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {7}{2}} \,d x } \] Input:

integrate(cos(d*x+c)^(7/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="m 
axima")
 

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^(7/2), x)
 

Giac [F]

\[ \int \cos ^{\frac {7}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\int { {\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \cos \left (d x + c\right )^{\frac {7}{2}} \,d x } \] Input:

integrate(cos(d*x+c)^(7/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algorithm="g 
iac")
 

Output:

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*cos(d*x + c)^(7/2), x)
 

Mupad [B] (verification not implemented)

Time = 13.33 (sec) , antiderivative size = 123, normalized size of antiderivative = 1.00 \[ \int \cos ^{\frac {7}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\frac {2\,C\,\mathrm {F}\left (\frac {c}{2}+\frac {d\,x}{2}\middle |2\right )}{3\,d}+\frac {2\,C\,\sqrt {\cos \left (c+d\,x\right )}\,\sin \left (c+d\,x\right )}{3\,d}-\frac {2\,A\,{\cos \left (c+d\,x\right )}^{9/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {9}{4};\ \frac {13}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{9\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}}-\frac {2\,B\,{\cos \left (c+d\,x\right )}^{7/2}\,\sin \left (c+d\,x\right )\,{{}}_2{\mathrm {F}}_1\left (\frac {1}{2},\frac {7}{4};\ \frac {11}{4};\ {\cos \left (c+d\,x\right )}^2\right )}{7\,d\,\sqrt {{\sin \left (c+d\,x\right )}^2}} \] Input:

int(cos(c + d*x)^(7/2)*(A + B/cos(c + d*x) + C/cos(c + d*x)^2),x)
 

Output:

(2*C*ellipticF(c/2 + (d*x)/2, 2))/(3*d) + (2*C*cos(c + d*x)^(1/2)*sin(c + 
d*x))/(3*d) - (2*A*cos(c + d*x)^(9/2)*sin(c + d*x)*hypergeom([1/2, 9/4], 1 
3/4, cos(c + d*x)^2))/(9*d*(sin(c + d*x)^2)^(1/2)) - (2*B*cos(c + d*x)^(7/ 
2)*sin(c + d*x)*hypergeom([1/2, 7/4], 11/4, cos(c + d*x)^2))/(7*d*(sin(c + 
 d*x)^2)^(1/2))
 

Reduce [F]

\[ \int \cos ^{\frac {7}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )^{2}d x \right ) c +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{3} \sec \left (d x +c \right )d x \right ) b +\left (\int \sqrt {\cos \left (d x +c \right )}\, \cos \left (d x +c \right )^{3}d x \right ) a \] Input:

int(cos(d*x+c)^(7/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x)
 

Output:

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