3.1.89 \(\int (b \cos (c+d x))^{3/2} \sec ^7(c+d x) \, dx\) [89]

3.1.89.1 Optimal result
3.1.89.2 Mathematica [A] (verified)
3.1.89.3 Rubi [A] (verified)
3.1.89.4 Maple [B] (verified)
3.1.89.5 Fricas [C] (verification not implemented)
3.1.89.6 Sympy [F(-1)]
3.1.89.7 Maxima [F]
3.1.89.8 Giac [F]
3.1.89.9 Mupad [F(-1)]

3.1.89.1 Optimal result

Integrand size = 21, antiderivative size = 126 \[ \int (b \cos (c+d x))^{3/2} \sec ^7(c+d x) \, dx=-\frac {14 b \sqrt {b \cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{15 d \sqrt {\cos (c+d x)}}+\frac {2 b^6 \sin (c+d x)}{9 d (b \cos (c+d x))^{9/2}}+\frac {14 b^4 \sin (c+d x)}{45 d (b \cos (c+d x))^{5/2}}+\frac {14 b^2 \sin (c+d x)}{15 d \sqrt {b \cos (c+d x)}} \]

output
2/9*b^6*sin(d*x+c)/d/(b*cos(d*x+c))^(9/2)+14/45*b^4*sin(d*x+c)/d/(b*cos(d* 
x+c))^(5/2)+14/15*b^2*sin(d*x+c)/d/(b*cos(d*x+c))^(1/2)-14/15*b*(cos(1/2*d 
*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticE(sin(1/2*d*x+1/2*c),2^(1/2) 
)*(b*cos(d*x+c))^(1/2)/d/cos(d*x+c)^(1/2)
 
3.1.89.2 Mathematica [A] (verified)

Time = 0.55 (sec) , antiderivative size = 79, normalized size of antiderivative = 0.63 \[ \int (b \cos (c+d x))^{3/2} \sec ^7(c+d x) \, dx=\frac {(b \cos (c+d x))^{3/2} \sec ^6(c+d x) \left (-336 \cos ^{\frac {9}{2}}(c+d x) E\left (\left .\frac {1}{2} (c+d x)\right |2\right )+150 \sin (c+d x)+91 \sin (3 (c+d x))+21 \sin (5 (c+d x))\right )}{360 d} \]

input
Integrate[(b*Cos[c + d*x])^(3/2)*Sec[c + d*x]^7,x]
 
output
((b*Cos[c + d*x])^(3/2)*Sec[c + d*x]^6*(-336*Cos[c + d*x]^(9/2)*EllipticE[ 
(c + d*x)/2, 2] + 150*Sin[c + d*x] + 91*Sin[3*(c + d*x)] + 21*Sin[5*(c + d 
*x)]))/(360*d)
 
3.1.89.3 Rubi [A] (verified)

Time = 0.58 (sec) , antiderivative size = 144, normalized size of antiderivative = 1.14, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.524, Rules used = {3042, 2030, 3116, 3042, 3116, 3042, 3116, 3042, 3121, 3042, 3119}

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.

\(\displaystyle \int \sec ^7(c+d x) (b \cos (c+d x))^{3/2} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}{\sin \left (c+d x+\frac {\pi }{2}\right )^7}dx\)

\(\Big \downarrow \) 2030

\(\displaystyle b^7 \int \frac {1}{\left (b \sin \left (\frac {1}{2} (2 c+\pi )+d x\right )\right )^{11/2}}dx\)

\(\Big \downarrow \) 3116

\(\displaystyle b^7 \left (\frac {7 \int \frac {1}{(b \cos (c+d x))^{7/2}}dx}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^7 \left (\frac {7 \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{7/2}}dx}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\)

\(\Big \downarrow \) 3116

\(\displaystyle b^7 \left (\frac {7 \left (\frac {3 \int \frac {1}{(b \cos (c+d x))^{3/2}}dx}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^7 \left (\frac {7 \left (\frac {3 \int \frac {1}{\left (b \sin \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\)

\(\Big \downarrow \) 3116

\(\displaystyle b^7 \left (\frac {7 \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\int \sqrt {b \cos (c+d x)}dx}{b^2}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^7 \left (\frac {7 \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\int \sqrt {b \sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\)

\(\Big \downarrow \) 3121

\(\displaystyle b^7 \left (\frac {7 \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\sqrt {b \cos (c+d x)} \int \sqrt {\cos (c+d x)}dx}{b^2 \sqrt {\cos (c+d x)}}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\)

\(\Big \downarrow \) 3042

\(\displaystyle b^7 \left (\frac {7 \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {\sqrt {b \cos (c+d x)} \int \sqrt {\sin \left (c+d x+\frac {\pi }{2}\right )}dx}{b^2 \sqrt {\cos (c+d x)}}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\)

\(\Big \downarrow \) 3119

\(\displaystyle b^7 \left (\frac {7 \left (\frac {3 \left (\frac {2 \sin (c+d x)}{b d \sqrt {b \cos (c+d x)}}-\frac {2 E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {b \cos (c+d x)}}{b^2 d \sqrt {\cos (c+d x)}}\right )}{5 b^2}+\frac {2 \sin (c+d x)}{5 b d (b \cos (c+d x))^{5/2}}\right )}{9 b^2}+\frac {2 \sin (c+d x)}{9 b d (b \cos (c+d x))^{9/2}}\right )\)

input
Int[(b*Cos[c + d*x])^(3/2)*Sec[c + d*x]^7,x]
 
output
b^7*((2*Sin[c + d*x])/(9*b*d*(b*Cos[c + d*x])^(9/2)) + (7*((2*Sin[c + d*x] 
)/(5*b*d*(b*Cos[c + d*x])^(5/2)) + (3*((-2*Sqrt[b*Cos[c + d*x]]*EllipticE[ 
(c + d*x)/2, 2])/(b^2*d*Sqrt[Cos[c + d*x]]) + (2*Sin[c + d*x])/(b*d*Sqrt[b 
*Cos[c + d*x]])))/(5*b^2)))/(9*b^2))
 

3.1.89.3.1 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 3116
Int[((b_.)*sin[(c_.) + (d_.)*(x_)])^(n_), x_Symbol] :> Simp[Cos[c + d*x]*(( 
b*Sin[c + d*x])^(n + 1)/(b*d*(n + 1))), x] + Simp[(n + 2)/(b^2*(n + 1))   I 
nt[(b*Sin[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[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 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]
 
3.1.89.4 Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(415\) vs. \(2(134)=268\).

Time = 2.99 (sec) , antiderivative size = 416, normalized size of antiderivative = 3.30

method result size
default \(-\frac {2 \sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, b^{2} \left (-\frac {\cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}}{144 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{5}}-\frac {7 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}}{180 b \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )-\frac {1}{2}\right )^{3}}-\frac {14 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )}{15 \sqrt {-\left (-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1\right ) b \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}}+\frac {7 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )}{15 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}}-\frac {7 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {-2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+1}\, \left (F\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-E\left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{15 \sqrt {-b \left (2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-\left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )\right )}}\right )}{\sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) b}\, d}\) \(416\)

input
int((cos(d*x+c)*b)^(3/2)*sec(d*x+c)^7,x,method=_RETURNVERBOSE)
 
output
-2*(-(-2*cos(1/2*d*x+1/2*c)^2+1)*b*sin(1/2*d*x+1/2*c)^2)^(1/2)*b^2*(-1/144 
*cos(1/2*d*x+1/2*c)/b*(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^( 
1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^5-7/180*cos(1/2*d*x+1/2*c)/b*(-b*(2*sin(1/ 
2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2))^(1/2)/(cos(1/2*d*x+1/2*c)^2-1/2)^3-1 
4/15*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c)/(-(-2*cos(1/2*d*x+1/2*c)^2+1) 
*b*sin(1/2*d*x+1/2*c)^2)^(1/2)+7/15*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(-2*cos(1 
/2*d*x+1/2*c)^2+1)^(1/2)/(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(1/2*d*x+1/2*c)^2) 
)^(1/2)*EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-7/15*(sin(1/2*d*x+1/2*c)^2)^ 
(1/2)*(-2*cos(1/2*d*x+1/2*c)^2+1)^(1/2)/(-b*(2*sin(1/2*d*x+1/2*c)^4-sin(1/ 
2*d*x+1/2*c)^2))^(1/2)*(EllipticF(cos(1/2*d*x+1/2*c),2^(1/2))-EllipticE(co 
s(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) 
*b)^(1/2)/d
 
3.1.89.5 Fricas [C] (verification not implemented)

Result contains higher order function than in optimal. Order 9 vs. order 4.

Time = 0.11 (sec) , antiderivative size = 132, normalized size of antiderivative = 1.05 \[ \int (b \cos (c+d x))^{3/2} \sec ^7(c+d x) \, dx=\frac {-21 i \, \sqrt {2} b^{\frac {3}{2}} \cos \left (d x + c\right )^{5} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 21 i \, \sqrt {2} b^{\frac {3}{2}} \cos \left (d x + c\right )^{5} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) + 2 \, {\left (21 \, b \cos \left (d x + c\right )^{4} + 7 \, b \cos \left (d x + c\right )^{2} + 5 \, b\right )} \sqrt {b \cos \left (d x + c\right )} \sin \left (d x + c\right )}{45 \, d \cos \left (d x + c\right )^{5}} \]

input
integrate((b*cos(d*x+c))^(3/2)*sec(d*x+c)^7,x, algorithm="fricas")
 
output
1/45*(-21*I*sqrt(2)*b^(3/2)*cos(d*x + c)^5*weierstrassZeta(-4, 0, weierstr 
assPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 21*I*sqrt(2)*b^(3/2)* 
cos(d*x + c)^5*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, cos(d*x + 
 c) - I*sin(d*x + c))) + 2*(21*b*cos(d*x + c)^4 + 7*b*cos(d*x + c)^2 + 5*b 
)*sqrt(b*cos(d*x + c))*sin(d*x + c))/(d*cos(d*x + c)^5)
 
3.1.89.6 Sympy [F(-1)]

Timed out. \[ \int (b \cos (c+d x))^{3/2} \sec ^7(c+d x) \, dx=\text {Timed out} \]

input
integrate((b*cos(d*x+c))**(3/2)*sec(d*x+c)**7,x)
 
output
Timed out
 
3.1.89.7 Maxima [F]

\[ \int (b \cos (c+d x))^{3/2} \sec ^7(c+d x) \, dx=\int { \left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}} \sec \left (d x + c\right )^{7} \,d x } \]

input
integrate((b*cos(d*x+c))^(3/2)*sec(d*x+c)^7,x, algorithm="maxima")
 
output
integrate((b*cos(d*x + c))^(3/2)*sec(d*x + c)^7, x)
 
3.1.89.8 Giac [F]

\[ \int (b \cos (c+d x))^{3/2} \sec ^7(c+d x) \, dx=\int { \left (b \cos \left (d x + c\right )\right )^{\frac {3}{2}} \sec \left (d x + c\right )^{7} \,d x } \]

input
integrate((b*cos(d*x+c))^(3/2)*sec(d*x+c)^7,x, algorithm="giac")
 
output
integrate((b*cos(d*x + c))^(3/2)*sec(d*x + c)^7, x)
 
3.1.89.9 Mupad [F(-1)]

Timed out. \[ \int (b \cos (c+d x))^{3/2} \sec ^7(c+d x) \, dx=\int \frac {{\left (b\,\cos \left (c+d\,x\right )\right )}^{3/2}}{{\cos \left (c+d\,x\right )}^7} \,d x \]

input
int((b*cos(c + d*x))^(3/2)/cos(c + d*x)^7,x)
 
output
int((b*cos(c + d*x))^(3/2)/cos(c + d*x)^7, x)