\(\int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx\) [371]

Optimal result

Integrand size = 33, antiderivative size = 329 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {2 (a-b) \sqrt {a+b} \left (10 a A b-8 a^2 B-9 b^2 B\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^4 d}+\frac {2 \sqrt {a+b} \left (b^2 (5 A-9 B)-8 a^2 B+2 a b (5 A+B)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^3 d}+\frac {2 (5 A b-4 a B) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 b^2 d}+\frac {2 B \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b d} \] Output:

2/15*(a-b)*(a+b)^(1/2)*(10*A*a*b-8*B*a^2-9*B*b^2)*cot(d*x+c)*EllipticE((a+ 
b*sec(d*x+c))^(1/2)/(a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+ 
b))^(1/2)*(-b*(1+sec(d*x+c))/(a-b))^(1/2)/b^4/d+2/15*(a+b)^(1/2)*(b^2*(5*A 
-9*B)-8*B*a^2+2*a*b*(5*A+B))*cot(d*x+c)*EllipticF((a+b*sec(d*x+c))^(1/2)/( 
a+b)^(1/2),((a+b)/(a-b))^(1/2))*(b*(1-sec(d*x+c))/(a+b))^(1/2)*(-b*(1+sec( 
d*x+c))/(a-b))^(1/2)/b^3/d+2/15*(5*A*b-4*B*a)*(a+b*sec(d*x+c))^(1/2)*tan(d 
*x+c)/b^2/d+2/5*B*sec(d*x+c)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/b/d
 

Mathematica [A] (warning: unable to verify)

Time = 14.16 (sec) , antiderivative size = 395, normalized size of antiderivative = 1.20 \[ \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx=\frac {2 \sqrt {\sec (c+d x)} \left (-\frac {\sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 (a+b) \left (-10 a A b+8 a^2 B+9 b^2 B\right ) E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {a+b \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}-2 b \left (8 a^2 B+2 a b (-5 A+B)+b^2 (5 A+9 B)\right ) \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right ) \sqrt {\frac {1}{1+\sec (c+d x)}} \sqrt {\frac {a+b \sec (c+d x)}{(a+b) (1+\sec (c+d x))}}+\left (-10 a A b+8 a^2 B+9 b^2 B\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{\sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )}}+(b+a \cos (c+d x)) \sqrt {\sec (c+d x)} \left (\left (-10 a A b+8 a^2 B+9 b^2 B\right ) \sin (c+d x)+b (5 A b-4 a B+3 b B \sec (c+d x)) \tan (c+d x)\right )\right )}{15 b^3 d \sqrt {a+b \sec (c+d x)}} \] Input:

Integrate[(Sec[c + d*x]^3*(A + B*Sec[c + d*x]))/Sqrt[a + b*Sec[c + d*x]],x 
]
 

Output:

(2*Sqrt[Sec[c + d*x]]*(-((Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(2*(a + b) 
*(-10*a*A*b + 8*a^2*B + 9*b^2*B)*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - 
b)/(a + b)]*Sqrt[(1 + Sec[c + d*x])^(-1)]*Sqrt[(a + b*Sec[c + d*x])/((a + 
b)*(1 + Sec[c + d*x]))] - 2*b*(8*a^2*B + 2*a*b*(-5*A + B) + b^2*(5*A + 9*B 
))*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)]*Sqrt[(1 + Sec[c + 
d*x])^(-1)]*Sqrt[(a + b*Sec[c + d*x])/((a + b)*(1 + Sec[c + d*x]))] + (-10 
*a*A*b + 8*a^2*B + 9*b^2*B)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x 
)/2]^2*Tan[(c + d*x)/2]))/Sqrt[Sec[(c + d*x)/2]^2]) + (b + a*Cos[c + d*x]) 
*Sqrt[Sec[c + d*x]]*((-10*a*A*b + 8*a^2*B + 9*b^2*B)*Sin[c + d*x] + b*(5*A 
*b - 4*a*B + 3*b*B*Sec[c + d*x])*Tan[c + d*x])))/(15*b^3*d*Sqrt[a + b*Sec[ 
c + d*x]])
 

Rubi [A] (verified)

Time = 1.27 (sec) , antiderivative size = 341, normalized size of antiderivative = 1.04, number of steps used = 11, number of rules used = 11, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.333, Rules used = {3042, 4521, 27, 3042, 4570, 27, 3042, 4493, 3042, 4319, 4492}

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.

\(\displaystyle \int \frac {\sec ^3(c+d x) (A+B \sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx\)

\(\Big \downarrow \) 3042

\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (A+B \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\)

\(\Big \downarrow \) 4521

\(\displaystyle \frac {2 \int \frac {\sec (c+d x) \left ((5 A b-4 a B) \sec ^2(c+d x)+3 b B \sec (c+d x)+2 a B\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{5 b}+\frac {2 B \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\int \frac {\sec (c+d x) \left ((5 A b-4 a B) \sec ^2(c+d x)+3 b B \sec (c+d x)+2 a B\right )}{\sqrt {a+b \sec (c+d x)}}dx}{5 b}+\frac {2 B \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left ((5 A b-4 a B) \csc \left (c+d x+\frac {\pi }{2}\right )^2+3 b B \csc \left (c+d x+\frac {\pi }{2}\right )+2 a B\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b}+\frac {2 B \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 4570

\(\displaystyle \frac {\frac {2 \int \frac {\sec (c+d x) \left (b (5 A b+2 a B)-\left (-8 B a^2+10 A b a-9 b^2 B\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 B \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {\frac {\int \frac {\sec (c+d x) \left (b (5 A b+2 a B)-\left (-8 B a^2+10 A b a-9 b^2 B\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 B \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b (5 A b+2 a B)+\left (8 B a^2-10 A b a+9 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 B \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 4493

\(\displaystyle \frac {\frac {\left (-8 a^2 B+2 a b (5 A+B)+b^2 (5 A-9 B)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx-\left (-8 a^2 B+10 a A b-9 b^2 B\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}+\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 B \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {\frac {\left (-8 a^2 B+2 a b (5 A+B)+b^2 (5 A-9 B)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\left (-8 a^2 B+10 a A b-9 b^2 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 B \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {\frac {\frac {2 \sqrt {a+b} \left (-8 a^2 B+2 a b (5 A+B)+b^2 (5 A-9 B)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\left (-8 a^2 B+10 a A b-9 b^2 B\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}+\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 B \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {\frac {\frac {2 \sqrt {a+b} \left (-8 a^2 B+2 a b (5 A+B)+b^2 (5 A-9 B)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}+\frac {2 (a-b) \sqrt {a+b} \left (-8 a^2 B+10 a A b-9 b^2 B\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}}{3 b}+\frac {2 (5 A b-4 a B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}+\frac {2 B \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}\)

Input:

Int[(Sec[c + d*x]^3*(A + B*Sec[c + d*x]))/Sqrt[a + b*Sec[c + d*x]],x]
 

Output:

(2*B*Sec[c + d*x]*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(5*b*d) + (((2*(a 
 - b)*Sqrt[a + b]*(10*a*A*b - 8*a^2*B - 9*b^2*B)*Cot[c + d*x]*EllipticE[Ar 
cSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - 
Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b^2*d) + 
 (2*Sqrt[a + b]*(b^2*(5*A - 9*B) - 8*a^2*B + 2*a*b*(5*A + B))*Cot[c + d*x] 
*EllipticF[ArcSin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]* 
Sqrt[(b*(1 - Sec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b) 
)])/(b*d))/(3*b) + (2*(5*A*b - 4*a*B)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x 
])/(3*b*d))/(5*b)
 

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 4319

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S 
ymbol] :> Simp[-2*(Rt[a + b, 2]/(b*f*Cot[e + f*x]))*Sqrt[(b*(1 - Csc[e + f* 
x]))/(a + b)]*Sqrt[(-b)*((1 + Csc[e + f*x])/(a - b))]*EllipticF[ArcSin[Sqrt 
[a + b*Csc[e + f*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, e, 
 f}, x] && NeQ[a^2 - b^2, 0]
 

rule 4492

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c 
sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(A*b - a*B)*Rt[a 
 + b*(B/A), 2]*Sqrt[b*((1 - Csc[e + f*x])/(a + b))]*(Sqrt[(-b)*((1 + Csc[e 
+ f*x])/(a - b))]/(b^2*f*Cot[e + f*x]))*EllipticE[ArcSin[Sqrt[a + b*Csc[e + 
 f*x]]/Rt[a + b*(B/A), 2]], (a*A + b*B)/(a*A - b*B)], x] /; FreeQ[{a, b, e, 
 f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
 

rule 4493

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[c 
sc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(A - B)   Int[Csc[e 
 + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Simp[B   Int[Csc[e + f*x]*((1 + 
Csc[e + f*x])/Sqrt[a + b*Csc[e + f*x]]), x], x] /; FreeQ[{a, b, e, f, A, B} 
, x] && NeQ[a^2 - b^2, 0] && NeQ[A^2 - B^2, 0]
 

rule 4521

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( 
a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[(-B)*d^ 
2*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 2)/(b*f* 
(m + n))), x] + Simp[d^2/(b*(m + n))   Int[(a + b*Csc[e + f*x])^m*(d*Csc[e 
+ f*x])^(n - 2)*Simp[a*B*(n - 2) + B*b*(m + n - 1)*Csc[e + f*x] + (A*b*(m + 
 n) - a*B*(n - 1))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f, A, B 
, m}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[n, 1] && NeQ[m + 
n, 0] &&  !IGtQ[m, 1]
 

rule 4570

Int[csc[(e_.) + (f_.)*(x_)]*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e 
_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_S 
ymbol] :> Simp[(-C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 2) 
)), x] + Simp[1/(b*(m + 2))   Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[ 
b*A*(m + 2) + b*C*(m + 1) + (b*B*(m + 2) - a*C)*Csc[e + f*x], x], x], x] /; 
 FreeQ[{a, b, e, f, A, B, C, m}, x] &&  !LtQ[m, -1]
 

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(1336\) vs. \(2(299)=598\).

Time = 59.12 (sec) , antiderivative size = 1337, normalized size of antiderivative = 4.06

method	result	size
default	\(\text {Expression too large to display}\)	\(1337\)
parts	\(\text {Expression too large to display}\)	\(1361\)

Input:

int(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x,method=_RETURNV 
ERBOSE)
 

Output:

2/15/d/b^3*(a+b*sec(d*x+c))^(1/2)/(cos(d*x+c)^2*a+a*cos(d*x+c)+b*cos(d*x+c 
)+b)*(10*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*A*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos 
(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a^2*b*EllipticE(-csc(d*x 
+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+10*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*A*(1 
/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^ 
(1/2)*a*b^2*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+8*(cos(d 
*x+c)^2+2*cos(d*x+c)+1)*B*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)* 
(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a^3*EllipticE(-csc(d*x+c)+cot(d*x+c),((a 
-b)/(a+b))^(1/2))+8*(cos(d*x+c)^2+2*cos(d*x+c)+1)*B*(1/(a+b)*(b+a*cos(d*x+ 
c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a^2*b*Elliptic 
E(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+9*(cos(d*x+c)^2+2*cos(d*x+c) 
+1)*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(1+cos(d 
*x+c)))^(1/2)*a*b^2*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+ 
9*(cos(d*x+c)^2+2*cos(d*x+c)+1)*B*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+ 
b)*(b+a*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*b^3*EllipticE(-csc(d*x+c)+cot(d* 
x+c),((a-b)/(a+b))^(1/2))+10*(cos(d*x+c)^2+2*cos(d*x+c)+1)*A*(1/(a+b)*(b+a 
*cos(d*x+c))/(1+cos(d*x+c)))^(1/2)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*a*b^2 
*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+5*(-cos(d*x+c)^2-2* 
cos(d*x+c)-1)*A*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c) 
)/(1+cos(d*x+c)))^(1/2)*b^3*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+...
 

Fricas [F]

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

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x, algorith 
m="fricas")
 

Output:

integral((B*sec(d*x + c)^4 + A*sec(d*x + c)^3)/sqrt(b*sec(d*x + c) + a), x 
)
 

Sympy [F]

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

integrate(sec(d*x+c)**3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))**(1/2),x)
 

Output:

Integral((A + B*sec(c + d*x))*sec(c + d*x)**3/sqrt(a + b*sec(c + d*x)), x)
 

Maxima [F]

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

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x, algorith 
m="maxima")
 

Output:

integrate((B*sec(d*x + c) + A)*sec(d*x + c)^3/sqrt(b*sec(d*x + c) + a), x)
 

Giac [F]

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

integrate(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x, algorith 
m="giac")
 

Output:

integrate((B*sec(d*x + c) + A)*sec(d*x + c)^3/sqrt(b*sec(d*x + c) + a), x)
 

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

int(sec(d*x+c)^3*(A+B*sec(d*x+c))/(a+b*sec(d*x+c))^(1/2),x)
 

Output:

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