3.5.48 \(\int \frac {(a+b \sec (d+e x)+c \tan (d+e x))^{3/2}}{\sec ^{\frac {3}{2}}(d+e x)} \, dx\) [448]

3.5.48.1 Optimal result

Integrand size = 33, antiderivative size = 371 \[ \int \frac {(a+b \sec (d+e x)+c \tan (d+e x))^{3/2}}{\sec ^{\frac {3}{2}}(d+e x)} \, dx=-\frac {2 (c \cos (d+e x)-a \sin (d+e x)) (a+b \sec (d+e x)+c \tan (d+e x))^{3/2}}{3 e \sec ^{\frac {3}{2}}(d+e x) (b+a \cos (d+e x)+c \sin (d+e x))}+\frac {8 b E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right ) (a+b \sec (d+e x)+c \tan (d+e x))^{3/2}}{3 e \sec ^{\frac {3}{2}}(d+e x) (b+a \cos (d+e x)+c \sin (d+e x)) \sqrt {\frac {b+a \cos (d+e x)+c \sin (d+e x)}{b+\sqrt {a^2+c^2}}}}+\frac {2 \left (a^2-b^2+c^2\right ) \operatorname {EllipticF}\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right ),\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right ) \sqrt {\frac {b+a \cos (d+e x)+c \sin (d+e x)}{b+\sqrt {a^2+c^2}}} (a+b \sec (d+e x)+c \tan (d+e x))^{3/2}}{3 e \sec ^{\frac {3}{2}}(d+e x) (b+a \cos (d+e x)+c \sin (d+e x))^2} \]

output

-2/3*(c*cos(e*x+d)-a*sin(e*x+d))*(a+b*sec(e*x+d)+c*tan(e*x+d))^(3/2)/e/sec 
(e*x+d)^(3/2)/(b+a*cos(e*x+d)+c*sin(e*x+d))+8/3*b*(cos(1/2*d+1/2*e*x-1/2*a 
rctan(a,c))^2)^(1/2)/cos(1/2*d+1/2*e*x-1/2*arctan(a,c))*EllipticE(sin(1/2* 
d+1/2*e*x-1/2*arctan(a,c)),2^(1/2)*((a^2+c^2)^(1/2)/(b+(a^2+c^2)^(1/2)))^( 
1/2))*(a+b*sec(e*x+d)+c*tan(e*x+d))^(3/2)/e/sec(e*x+d)^(3/2)/(b+a*cos(e*x+ 
d)+c*sin(e*x+d))/((b+a*cos(e*x+d)+c*sin(e*x+d))/(b+(a^2+c^2)^(1/2)))^(1/2) 
+2/3*(a^2-b^2+c^2)*(cos(1/2*d+1/2*e*x-1/2*arctan(a,c))^2)^(1/2)/cos(1/2*d+ 
1/2*e*x-1/2*arctan(a,c))*EllipticF(sin(1/2*d+1/2*e*x-1/2*arctan(a,c)),2^(1 
/2)*((a^2+c^2)^(1/2)/(b+(a^2+c^2)^(1/2)))^(1/2))*((b+a*cos(e*x+d)+c*sin(e* 
x+d))/(b+(a^2+c^2)^(1/2)))^(1/2)*(a+b*sec(e*x+d)+c*tan(e*x+d))^(3/2)/e/sec 
(e*x+d)^(3/2)/(b+a*cos(e*x+d)+c*sin(e*x+d))^2
 

3.5.48.2 Mathematica [C] (warning: unable to verify)

Result contains higher order function than in optimal. Order 6 vs. order 4 in optimal.

Time = 6.82 (sec) , antiderivative size = 2490, normalized size of antiderivative = 6.71 \[ \int \frac {(a+b \sec (d+e x)+c \tan (d+e x))^{3/2}}{\sec ^{\frac {3}{2}}(d+e x)} \, dx=\text {Result too large to show} \]

input

Integrate[(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(3/2)/Sec[d + e*x]^(3/2),x 
]
 

output

(((8*a*b)/(3*c) - (2*c*Cos[d + e*x])/3 + (2*a*Sin[d + e*x])/3)*(a + b*Sec[ 
d + e*x] + c*Tan[d + e*x])^(3/2))/(e*Sec[d + e*x]^(3/2)*(b + a*Cos[d + e*x 
] + c*Sin[d + e*x])) + (2*a^2*AppellF1[1/2, 1/2, 1/2, 3/2, -((b + Sqrt[1 + 
 a^2/c^2]*c*Sin[d + e*x + ArcTan[a/c]])/(Sqrt[1 + a^2/c^2]*(1 - b/(Sqrt[1 
+ a^2/c^2]*c))*c)), -((b + Sqrt[1 + a^2/c^2]*c*Sin[d + e*x + ArcTan[a/c]]) 
/(Sqrt[1 + a^2/c^2]*(-1 - b/(Sqrt[1 + a^2/c^2]*c))*c))]*Sec[d + e*x + ArcT 
an[a/c]]*Sqrt[(c*Sqrt[(a^2 + c^2)/c^2] - c*Sqrt[(a^2 + c^2)/c^2]*Sin[d + e 
*x + ArcTan[a/c]])/(b + c*Sqrt[(a^2 + c^2)/c^2])]*Sqrt[b + c*Sqrt[(a^2 + c 
^2)/c^2]*Sin[d + e*x + ArcTan[a/c]]]*Sqrt[(c*Sqrt[(a^2 + c^2)/c^2] + c*Sqr 
t[(a^2 + c^2)/c^2]*Sin[d + e*x + ArcTan[a/c]])/(-b + c*Sqrt[(a^2 + c^2)/c^ 
2])]*(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(3/2))/(3*Sqrt[1 + a^2/c^2]*c*e 
*Sec[d + e*x]^(3/2)*(b + a*Cos[d + e*x] + c*Sin[d + e*x])^(3/2)) + (2*b^2* 
AppellF1[1/2, 1/2, 1/2, 3/2, -((b + Sqrt[1 + a^2/c^2]*c*Sin[d + e*x + ArcT 
an[a/c]])/(Sqrt[1 + a^2/c^2]*(1 - b/(Sqrt[1 + a^2/c^2]*c))*c)), -((b + Sqr 
t[1 + a^2/c^2]*c*Sin[d + e*x + ArcTan[a/c]])/(Sqrt[1 + a^2/c^2]*(-1 - b/(S 
qrt[1 + a^2/c^2]*c))*c))]*Sec[d + e*x + ArcTan[a/c]]*Sqrt[(c*Sqrt[(a^2 + c 
^2)/c^2] - c*Sqrt[(a^2 + c^2)/c^2]*Sin[d + e*x + ArcTan[a/c]])/(b + c*Sqrt 
[(a^2 + c^2)/c^2])]*Sqrt[b + c*Sqrt[(a^2 + c^2)/c^2]*Sin[d + e*x + ArcTan[ 
a/c]]]*Sqrt[(c*Sqrt[(a^2 + c^2)/c^2] + c*Sqrt[(a^2 + c^2)/c^2]*Sin[d + e*x 
 + ArcTan[a/c]])/(-b + c*Sqrt[(a^2 + c^2)/c^2])]*(a + b*Sec[d + e*x] + ...
 

3.5.48.3 Rubi [A] (verified)

Time = 1.45 (sec) , antiderivative size = 337, normalized size of antiderivative = 0.91, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.424, Rules used = {3042, 3646, 3042, 3599, 27, 3042, 3628, 3042, 3598, 3042, 3132, 3606, 3042, 3140}

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.

\(\displaystyle \int \frac {(a+b \sec (d+e x)+c \tan (d+e x))^{3/2}}{\sec ^{\frac {3}{2}}(d+e x)} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {(a+b \sec (d+e x)+c \tan (d+e x))^{3/2}}{\sec (d+e x)^{3/2}}dx\)

\(\Big \downarrow \) 3646

\(\displaystyle \frac {(a+b \sec (d+e x)+c \tan (d+e x))^{3/2} \int (b+a \cos (d+e x)+c \sin (d+e x))^{3/2}dx}{\sec ^{\frac {3}{2}}(d+e x) (a \cos (d+e x)+b+c \sin (d+e x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(a+b \sec (d+e x)+c \tan (d+e x))^{3/2} \int (b+a \cos (d+e x)+c \sin (d+e x))^{3/2}dx}{\sec ^{\frac {3}{2}}(d+e x) (a \cos (d+e x)+b+c \sin (d+e x))^{3/2}}\)

\(\Big \downarrow \) 3599

\(\displaystyle \frac {(a+b \sec (d+e x)+c \tan (d+e x))^{3/2} \left (\frac {2}{3} \int \frac {a^2+4 b \cos (d+e x) a+3 b^2+c^2+4 b c \sin (d+e x)}{2 \sqrt {b+a \cos (d+e x)+c \sin (d+e x)}}dx-\frac {2 (c \cos (d+e x)-a \sin (d+e x)) \sqrt {a \cos (d+e x)+b+c \sin (d+e x)}}{3 e}\right )}{\sec ^{\frac {3}{2}}(d+e x) (a \cos (d+e x)+b+c \sin (d+e x))^{3/2}}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {(a+b \sec (d+e x)+c \tan (d+e x))^{3/2} \left (\frac {1}{3} \int \frac {a^2+4 b \cos (d+e x) a+3 b^2+c^2+4 b c \sin (d+e x)}{\sqrt {b+a \cos (d+e x)+c \sin (d+e x)}}dx-\frac {2 (c \cos (d+e x)-a \sin (d+e x)) \sqrt {a \cos (d+e x)+b+c \sin (d+e x)}}{3 e}\right )}{\sec ^{\frac {3}{2}}(d+e x) (a \cos (d+e x)+b+c \sin (d+e x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(a+b \sec (d+e x)+c \tan (d+e x))^{3/2} \left (\frac {1}{3} \int \frac {a^2+4 b \cos (d+e x) a+3 b^2+c^2+4 b c \sin (d+e x)}{\sqrt {b+a \cos (d+e x)+c \sin (d+e x)}}dx-\frac {2 (c \cos (d+e x)-a \sin (d+e x)) \sqrt {a \cos (d+e x)+b+c \sin (d+e x)}}{3 e}\right )}{\sec ^{\frac {3}{2}}(d+e x) (a \cos (d+e x)+b+c \sin (d+e x))^{3/2}}\)

\(\Big \downarrow \) 3628

\(\displaystyle \frac {(a+b \sec (d+e x)+c \tan (d+e x))^{3/2} \left (\frac {1}{3} \left (\left (a^2-b^2+c^2\right ) \int \frac {1}{\sqrt {b+a \cos (d+e x)+c \sin (d+e x)}}dx+4 b \int \sqrt {b+a \cos (d+e x)+c \sin (d+e x)}dx\right )-\frac {2 (c \cos (d+e x)-a \sin (d+e x)) \sqrt {a \cos (d+e x)+b+c \sin (d+e x)}}{3 e}\right )}{\sec ^{\frac {3}{2}}(d+e x) (a \cos (d+e x)+b+c \sin (d+e x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(a+b \sec (d+e x)+c \tan (d+e x))^{3/2} \left (\frac {1}{3} \left (\left (a^2-b^2+c^2\right ) \int \frac {1}{\sqrt {b+a \cos (d+e x)+c \sin (d+e x)}}dx+4 b \int \sqrt {b+a \cos (d+e x)+c \sin (d+e x)}dx\right )-\frac {2 (c \cos (d+e x)-a \sin (d+e x)) \sqrt {a \cos (d+e x)+b+c \sin (d+e x)}}{3 e}\right )}{\sec ^{\frac {3}{2}}(d+e x) (a \cos (d+e x)+b+c \sin (d+e x))^{3/2}}\)

\(\Big \downarrow \) 3598

\(\displaystyle \frac {(a+b \sec (d+e x)+c \tan (d+e x))^{3/2} \left (\frac {1}{3} \left (\left (a^2-b^2+c^2\right ) \int \frac {1}{\sqrt {b+a \cos (d+e x)+c \sin (d+e x)}}dx+\frac {4 b \sqrt {a \cos (d+e x)+b+c \sin (d+e x)} \int \sqrt {\frac {b}{b+\sqrt {a^2+c^2}}+\frac {\sqrt {a^2+c^2} \cos \left (d+e x-\tan ^{-1}(a,c)\right )}{b+\sqrt {a^2+c^2}}}dx}{\sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}}}\right )-\frac {2 (c \cos (d+e x)-a \sin (d+e x)) \sqrt {a \cos (d+e x)+b+c \sin (d+e x)}}{3 e}\right )}{\sec ^{\frac {3}{2}}(d+e x) (a \cos (d+e x)+b+c \sin (d+e x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(a+b \sec (d+e x)+c \tan (d+e x))^{3/2} \left (\frac {1}{3} \left (\left (a^2-b^2+c^2\right ) \int \frac {1}{\sqrt {b+a \cos (d+e x)+c \sin (d+e x)}}dx+\frac {4 b \sqrt {a \cos (d+e x)+b+c \sin (d+e x)} \int \sqrt {\frac {b}{b+\sqrt {a^2+c^2}}+\frac {\sqrt {a^2+c^2} \sin \left (d+e x-\tan ^{-1}(a,c)+\frac {\pi }{2}\right )}{b+\sqrt {a^2+c^2}}}dx}{\sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}}}\right )-\frac {2 (c \cos (d+e x)-a \sin (d+e x)) \sqrt {a \cos (d+e x)+b+c \sin (d+e x)}}{3 e}\right )}{\sec ^{\frac {3}{2}}(d+e x) (a \cos (d+e x)+b+c \sin (d+e x))^{3/2}}\)

\(\Big \downarrow \) 3132

\(\displaystyle \frac {(a+b \sec (d+e x)+c \tan (d+e x))^{3/2} \left (\frac {1}{3} \left (\left (a^2-b^2+c^2\right ) \int \frac {1}{\sqrt {b+a \cos (d+e x)+c \sin (d+e x)}}dx+\frac {8 b \sqrt {a \cos (d+e x)+b+c \sin (d+e x)} E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{e \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}}}\right )-\frac {2 (c \cos (d+e x)-a \sin (d+e x)) \sqrt {a \cos (d+e x)+b+c \sin (d+e x)}}{3 e}\right )}{\sec ^{\frac {3}{2}}(d+e x) (a \cos (d+e x)+b+c \sin (d+e x))^{3/2}}\)

\(\Big \downarrow \) 3606

\(\displaystyle \frac {(a+b \sec (d+e x)+c \tan (d+e x))^{3/2} \left (\frac {1}{3} \left (\frac {\left (a^2-b^2+c^2\right ) \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}} \int \frac {1}{\sqrt {\frac {b}{b+\sqrt {a^2+c^2}}+\frac {\sqrt {a^2+c^2} \cos \left (d+e x-\tan ^{-1}(a,c)\right )}{b+\sqrt {a^2+c^2}}}}dx}{\sqrt {a \cos (d+e x)+b+c \sin (d+e x)}}+\frac {8 b \sqrt {a \cos (d+e x)+b+c \sin (d+e x)} E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{e \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}}}\right )-\frac {2 (c \cos (d+e x)-a \sin (d+e x)) \sqrt {a \cos (d+e x)+b+c \sin (d+e x)}}{3 e}\right )}{\sec ^{\frac {3}{2}}(d+e x) (a \cos (d+e x)+b+c \sin (d+e x))^{3/2}}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {(a+b \sec (d+e x)+c \tan (d+e x))^{3/2} \left (\frac {1}{3} \left (\frac {\left (a^2-b^2+c^2\right ) \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}} \int \frac {1}{\sqrt {\frac {b}{b+\sqrt {a^2+c^2}}+\frac {\sqrt {a^2+c^2} \sin \left (d+e x-\tan ^{-1}(a,c)+\frac {\pi }{2}\right )}{b+\sqrt {a^2+c^2}}}}dx}{\sqrt {a \cos (d+e x)+b+c \sin (d+e x)}}+\frac {8 b \sqrt {a \cos (d+e x)+b+c \sin (d+e x)} E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{e \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}}}\right )-\frac {2 (c \cos (d+e x)-a \sin (d+e x)) \sqrt {a \cos (d+e x)+b+c \sin (d+e x)}}{3 e}\right )}{\sec ^{\frac {3}{2}}(d+e x) (a \cos (d+e x)+b+c \sin (d+e x))^{3/2}}\)

\(\Big \downarrow \) 3140

\(\displaystyle \frac {(a+b \sec (d+e x)+c \tan (d+e x))^{3/2} \left (\frac {1}{3} \left (\frac {2 \left (a^2-b^2+c^2\right ) \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}} \operatorname {EllipticF}\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right ),\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{e \sqrt {a \cos (d+e x)+b+c \sin (d+e x)}}+\frac {8 b \sqrt {a \cos (d+e x)+b+c \sin (d+e x)} E\left (\frac {1}{2} \left (d+e x-\tan ^{-1}(a,c)\right )|\frac {2 \sqrt {a^2+c^2}}{b+\sqrt {a^2+c^2}}\right )}{e \sqrt {\frac {a \cos (d+e x)+b+c \sin (d+e x)}{\sqrt {a^2+c^2}+b}}}\right )-\frac {2 (c \cos (d+e x)-a \sin (d+e x)) \sqrt {a \cos (d+e x)+b+c \sin (d+e x)}}{3 e}\right )}{\sec ^{\frac {3}{2}}(d+e x) (a \cos (d+e x)+b+c \sin (d+e x))^{3/2}}\)

input

Int[(a + b*Sec[d + e*x] + c*Tan[d + e*x])^(3/2)/Sec[d + e*x]^(3/2),x]
 

output

(((-2*(c*Cos[d + e*x] - a*Sin[d + e*x])*Sqrt[b + a*Cos[d + e*x] + c*Sin[d 
+ e*x]])/(3*e) + ((8*b*EllipticE[(d + e*x - ArcTan[a, c])/2, (2*Sqrt[a^2 + 
 c^2])/(b + Sqrt[a^2 + c^2])]*Sqrt[b + a*Cos[d + e*x] + c*Sin[d + e*x]])/( 
e*Sqrt[(b + a*Cos[d + e*x] + c*Sin[d + e*x])/(b + Sqrt[a^2 + c^2])]) + (2* 
(a^2 - b^2 + c^2)*EllipticF[(d + e*x - ArcTan[a, c])/2, (2*Sqrt[a^2 + c^2] 
)/(b + Sqrt[a^2 + c^2])]*Sqrt[(b + a*Cos[d + e*x] + c*Sin[d + e*x])/(b + S 
qrt[a^2 + c^2])])/(e*Sqrt[b + a*Cos[d + e*x] + c*Sin[d + e*x]]))/3)*(a + b 
*Sec[d + e*x] + c*Tan[d + e*x])^(3/2))/(Sec[d + e*x]^(3/2)*(b + a*Cos[d + 
e*x] + c*Sin[d + e*x])^(3/2))
 

3.5.48.3.1 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 3132

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

rule 3140

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

rule 3598

Int[Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_ 
)]], x_Symbol] :> Simp[Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]]/Sqrt[(a + 
b*Cos[d + e*x] + c*Sin[d + e*x])/(a + Sqrt[b^2 + c^2])]   Int[Sqrt[a/(a + S 
qrt[b^2 + c^2]) + (Sqrt[b^2 + c^2]/(a + Sqrt[b^2 + c^2]))*Cos[d + e*x - Arc 
Tan[b, c]]], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] 
 && NeQ[b^2 + c^2, 0] &&  !GtQ[a + Sqrt[b^2 + c^2], 0]
 

rule 3599

Int[(cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*(x_)])^ 
(n_), x_Symbol] :> Simp[(-(c*Cos[d + e*x] - b*Sin[d + e*x]))*((a + b*Cos[d 
+ e*x] + c*Sin[d + e*x])^(n - 1)/(e*n)), x] + Simp[1/n   Int[Simp[n*a^2 + ( 
n - 1)*(b^2 + c^2) + a*b*(2*n - 1)*Cos[d + e*x] + a*c*(2*n - 1)*Sin[d + e*x 
], x]*(a + b*Cos[d + e*x] + c*Sin[d + e*x])^(n - 2), x], x] /; FreeQ[{a, b, 
 c, d, e}, x] && NeQ[a^2 - b^2 - c^2, 0] && GtQ[n, 1]
 

rule 3606

Int[1/Sqrt[cos[(d_.) + (e_.)*(x_)]*(b_.) + (a_) + (c_.)*sin[(d_.) + (e_.)*( 
x_)]], x_Symbol] :> Simp[Sqrt[(a + b*Cos[d + e*x] + c*Sin[d + e*x])/(a + Sq 
rt[b^2 + c^2])]/Sqrt[a + b*Cos[d + e*x] + c*Sin[d + e*x]]   Int[1/Sqrt[a/(a 
 + Sqrt[b^2 + c^2]) + (Sqrt[b^2 + c^2]/(a + Sqrt[b^2 + c^2]))*Cos[d + e*x - 
 ArcTan[b, c]]], x], x] /; FreeQ[{a, b, c, d, e}, x] && NeQ[a^2 - b^2 - c^2 
, 0] && NeQ[b^2 + c^2, 0] &&  !GtQ[a + Sqrt[b^2 + c^2], 0]
 

rule 3628

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

rule 3646

Int[sec[(d_.) + (e_.)*(x_)]^(n_.)*((a_.) + (b_.)*sec[(d_.) + (e_.)*(x_)] + 
(c_.)*tan[(d_.) + (e_.)*(x_)])^(m_), x_Symbol] :> Simp[Sec[d + e*x]^n*((b + 
 a*Cos[d + e*x] + c*Sin[d + e*x])^n/(a + b*Sec[d + e*x] + c*Tan[d + e*x])^n 
)   Int[1/(b + a*Cos[d + e*x] + c*Sin[d + e*x])^n, x], x] /; FreeQ[{a, b, c 
, d, e}, x] && EqQ[m + n, 0] &&  !IntegerQ[n]
 

3.5.48.4 Maple [C] (warning: unable to verify)

Result contains complex when optimal does not.

Time = 29.33 (sec) , antiderivative size = 64069, normalized size of antiderivative = 172.69

method	result	size
default	\(\text {Expression too large to display}\)	\(64069\)

input

int((a+b*sec(e*x+d)+c*tan(e*x+d))^(3/2)/sec(e*x+d)^(3/2),x,method=_RETURNV 
ERBOSE)
 

output

result too large to display
 

3.5.48.5 Fricas [C] (verification not implemented)

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

Time = 0.16 (sec) , antiderivative size = 1504, normalized size of antiderivative = 4.05 \[ \int \frac {(a+b \sec (d+e x)+c \tan (d+e x))^{3/2}}{\sec ^{\frac {3}{2}}(d+e x)} \, dx=\text {Too large to display} \]

input

integrate((a+b*sec(e*x+d)+c*tan(e*x+d))^(3/2)/sec(e*x+d)^(3/2),x, algorith 
m="fricas")
 

output

1/9*((-3*I*a^3 - I*a*b^2 - 3*I*a*c^2 + 3*c^3 + (3*a^2 + b^2)*c)*sqrt(2*a - 
 2*I*c)*weierstrassPInverse(-4/3*(3*a^4 - 4*a^2*b^2 + 4*b^2*c^2 + 6*I*a*c^ 
3 - 3*c^4 + 2*I*(3*a^3 - 4*a*b^2)*c)/(a^4 + 2*a^2*c^2 + c^4), 8/27*(9*a^5* 
b - 8*a^3*b^3 - 27*a*b*c^4 - 9*I*b*c^5 + 2*I*(9*a^2*b + 4*b^3)*c^3 - 6*(3* 
a^3*b - 4*a*b^3)*c^2 + 3*I*(9*a^4*b - 8*a^2*b^3)*c)/(a^6 + 3*a^4*c^2 + 3*a 
^2*c^4 + c^6), 1/3*(2*a*b + 2*I*b*c + 3*(a^2 + c^2)*cos(e*x + d) - 3*(-I*a 
^2 - I*c^2)*sin(e*x + d))/(a^2 + c^2)) + (3*I*a^3 + I*a*b^2 + 3*I*a*c^2 + 
3*c^3 + (3*a^2 + b^2)*c)*sqrt(2*a + 2*I*c)*weierstrassPInverse(-4/3*(3*a^4 
 - 4*a^2*b^2 + 4*b^2*c^2 - 6*I*a*c^3 - 3*c^4 - 2*I*(3*a^3 - 4*a*b^2)*c)/(a 
^4 + 2*a^2*c^2 + c^4), 8/27*(9*a^5*b - 8*a^3*b^3 - 27*a*b*c^4 + 9*I*b*c^5 
- 2*I*(9*a^2*b + 4*b^3)*c^3 - 6*(3*a^3*b - 4*a*b^3)*c^2 - 3*I*(9*a^4*b - 8 
*a^2*b^3)*c)/(a^6 + 3*a^4*c^2 + 3*a^2*c^4 + c^6), 1/3*(2*a*b - 2*I*b*c + 3 
*(a^2 + c^2)*cos(e*x + d) - 3*(I*a^2 + I*c^2)*sin(e*x + d))/(a^2 + c^2)) - 
 12*(-I*a^2*b - I*b*c^2)*sqrt(2*a - 2*I*c)*weierstrassZeta(-4/3*(3*a^4 - 4 
*a^2*b^2 + 4*b^2*c^2 + 6*I*a*c^3 - 3*c^4 + 2*I*(3*a^3 - 4*a*b^2)*c)/(a^4 + 
 2*a^2*c^2 + c^4), 8/27*(9*a^5*b - 8*a^3*b^3 - 27*a*b*c^4 - 9*I*b*c^5 + 2* 
I*(9*a^2*b + 4*b^3)*c^3 - 6*(3*a^3*b - 4*a*b^3)*c^2 + 3*I*(9*a^4*b - 8*a^2 
*b^3)*c)/(a^6 + 3*a^4*c^2 + 3*a^2*c^4 + c^6), weierstrassPInverse(-4/3*(3* 
a^4 - 4*a^2*b^2 + 4*b^2*c^2 + 6*I*a*c^3 - 3*c^4 + 2*I*(3*a^3 - 4*a*b^2)*c) 
/(a^4 + 2*a^2*c^2 + c^4), 8/27*(9*a^5*b - 8*a^3*b^3 - 27*a*b*c^4 - 9*I*...
 

3.5.48.6 Sympy [F]

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

input

integrate((a+b*sec(e*x+d)+c*tan(e*x+d))**(3/2)/sec(e*x+d)**(3/2),x)
 

output

Integral((a + b*sec(d + e*x) + c*tan(d + e*x))**(3/2)/sec(d + e*x)**(3/2), 
 x)
 

3.5.48.7 Maxima [F]

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

input

integrate((a+b*sec(e*x+d)+c*tan(e*x+d))^(3/2)/sec(e*x+d)^(3/2),x, algorith 
m="maxima")
 

output

integrate((b*sec(e*x + d) + c*tan(e*x + d) + a)^(3/2)/sec(e*x + d)^(3/2), 
x)
 

3.5.48.8 Giac [F]

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

input

integrate((a+b*sec(e*x+d)+c*tan(e*x+d))^(3/2)/sec(e*x+d)^(3/2),x, algorith 
m="giac")
 

output

integrate((b*sec(e*x + d) + c*tan(e*x + d) + a)^(3/2)/sec(e*x + d)^(3/2), 
x)
 

3.5.48.9 Mupad [F(-1)]

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

input

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

output

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