\(\int (a+b \sec (c+d x))^{3/2} (A+B \sec (c+d x)+C \sec ^2(c+d x)) \, dx\) [945]

Optimal result

Integrand size = 35, antiderivative size = 443 \[ \int (a+b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=-\frac {2 (a-b) \sqrt {a+b} \left (15 A b^2+20 a b B+3 a^2 C+9 b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^2 d}+\frac {2 \sqrt {a+b} \left (3 a^2 (5 B-C)+2 a b (15 A-10 B+6 C)-b^2 (15 A-5 B+9 C)\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b d}-\frac {2 a A \sqrt {a+b} \cot (c+d x) \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{d}+\frac {2 (5 b B+3 a C) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 d}+\frac {2 C (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 d} \] Output:

-2/15*(a-b)*(a+b)^(1/2)*(15*A*b^2+20*B*a*b+3*C*a^2+9*C*b^2)*cot(d*x+c)*Ell 
ipticE((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^2/d+2/15*(a+b)^(1/2) 
*(3*a^2*(5*B-C)+2*a*b*(15*A-10*B+6*C)-b^2*(15*A-5*B+9*C))*cot(d*x+c)*Ellip 
ticF((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/d-2*a*A*(a+b)^(1/2)*co 
t(d*x+c)*EllipticPi((a+b*sec(d*x+c))^(1/2)/(a+b)^(1/2),(a+b)/a,((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)/d 
+2/15*(5*B*b+3*C*a)*(a+b*sec(d*x+c))^(1/2)*tan(d*x+c)/d+2/5*C*(a+b*sec(d*x 
+c))^(3/2)*tan(d*x+c)/d
 

Mathematica [B] (warning: unable to verify)

Leaf count is larger than twice the leaf count of optimal. \(6946\) vs. \(2(443)=886\).

Time = 25.39 (sec) , antiderivative size = 6946, normalized size of antiderivative = 15.68 \[ \int (a+b \sec (c+d x))^{3/2} \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \, dx=\text {Result too large to show} \] Input:

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

Output:

Result too large to show
 

Rubi [A] (verified)

Time = 1.77 (sec) , antiderivative size = 449, normalized size of antiderivative = 1.01, number of steps used = 14, number of rules used = 14, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.400, Rules used = {3042, 4544, 27, 3042, 4544, 27, 3042, 4546, 3042, 4409, 3042, 4271, 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.

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

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4544

\(\displaystyle \frac {2}{5} \int \frac {1}{2} \sqrt {a+b \sec (c+d x)} \left ((5 b B+3 a C) \sec ^2(c+d x)+(5 A b+3 C b+5 a B) \sec (c+d x)+5 a A\right )dx+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \int \sqrt {a+b \sec (c+d x)} \left ((5 b B+3 a C) \sec ^2(c+d x)+(5 A b+3 C b+5 a B) \sec (c+d x)+5 a A\right )dx+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

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

\(\Big \downarrow \) 4544

\(\displaystyle \frac {1}{5} \left (\frac {2}{3} \int \frac {15 A a^2+\left (3 C a^2+20 b B a+15 A b^2+9 b^2 C\right ) \sec ^2(c+d x)+\left (15 B a^2+30 A b a+12 b C a+5 b^2 B\right ) \sec (c+d x)}{2 \sqrt {a+b \sec (c+d x)}}dx+\frac {2 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 27

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {15 A a^2+\left (3 C a^2+20 b B a+15 A b^2+9 b^2 C\right ) \sec ^2(c+d x)+\left (15 B a^2+30 A b a+12 b C a+5 b^2 B\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+\frac {2 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \int \frac {15 A a^2+\left (3 C a^2+20 b B a+15 A b^2+9 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2+\left (15 B a^2+30 A b a+12 b C a+5 b^2 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 4546

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\left (3 a^2 C+20 a b B+15 A b^2+9 b^2 C\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx+\int \frac {15 A a^2+\left (15 B a^2-3 C a^2+30 A b a-20 b B a+12 b C a-15 A b^2+5 b^2 B-9 b^2 C\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {2 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\left (3 a^2 C+20 a b B+15 A b^2+9 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\int \frac {15 A a^2+\left (15 B a^2-3 C a^2+30 A b a-20 b B a+12 b C a-15 A b^2+5 b^2 B-9 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 4409

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\left (3 a^2 C+20 a b B+15 A b^2+9 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (3 a^2 (5 B-C)+2 a b (15 A-10 B+6 C)-b^2 (15 A-5 B+9 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx+15 a^2 A \int \frac {1}{\sqrt {a+b \sec (c+d x)}}dx\right )+\frac {2 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 3042

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\left (3 a^2 (5 B-C)+2 a b (15 A-10 B+6 C)-b^2 (15 A-5 B+9 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (3 a^2 C+20 a b B+15 A b^2+9 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+15 a^2 A \int \frac {1}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx\right )+\frac {2 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 4271

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\left (3 a^2 (5 B-C)+2 a b (15 A-10 B+6 C)-b^2 (15 A-5 B+9 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\left (3 a^2 C+20 a b B+15 A b^2+9 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {30 a A \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )+\frac {2 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 4319

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\left (3 a^2 C+20 a b B+15 A b^2+9 b^2 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx+\frac {2 \sqrt {a+b} \cot (c+d x) \left (3 a^2 (5 B-C)+2 a b (15 A-10 B+6 C)-b^2 (15 A-5 B+9 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {30 a A \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )+\frac {2 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

\(\Big \downarrow \) 4492

\(\displaystyle \frac {1}{5} \left (\frac {1}{3} \left (\frac {2 \sqrt {a+b} \cot (c+d x) \left (3 a^2 (5 B-C)+2 a b (15 A-10 B+6 C)-b^2 (15 A-5 B+9 C)\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \cot (c+d x) \left (3 a^2 C+20 a b B+15 A b^2+9 b^2 C\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}-\frac {30 a A \sqrt {a+b} \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticPi}\left (\frac {a+b}{a},\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{d}\right )+\frac {2 (3 a C+5 b B) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 d}\right )+\frac {2 C \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 d}\)

Input:

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

Output:

(2*C*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*d) + (((-2*(a - b)*Sqrt[a 
 + b]*(15*A*b^2 + 20*a*b*B + 3*a^2*C + 9*b^2*C)*Cot[c + d*x]*EllipticE[Arc 
Sin[Sqrt[a + b*Sec[c + d*x]]/Sqrt[a + b]], (a + b)/(a - b)]*Sqrt[(b*(1 - S 
ec[c + d*x]))/(a + b)]*Sqrt[-((b*(1 + Sec[c + d*x]))/(a - b))])/(b^2*d) + 
(2*Sqrt[a + b]*(3*a^2*(5*B - C) + 2*a*b*(15*A - 10*B + 6*C) - b^2*(15*A - 
5*B + 9*C))*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) - (30*a*A*Sqrt[a + b]*Cot[c + d*x]*Ellip 
ticPi[(a + b)/a, 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))])/d)/3 + (2*(5*b*B + 3*a*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x]) 
/(3*d))/5
 

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 4271

Int[1/Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[2*(Rt[a 
 + b, 2]/(a*d*Cot[c + d*x]))*Sqrt[b*((1 - Csc[c + d*x])/(a + b))]*Sqrt[(-b) 
*((1 + Csc[c + d*x])/(a - b))]*EllipticPi[(a + b)/a, ArcSin[Sqrt[a + b*Csc[ 
c + d*x]]/Rt[a + b, 2]], (a + b)/(a - b)], x] /; FreeQ[{a, b, c, d}, x] && 
NeQ[a^2 - b^2, 0]
 

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 4409

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ 
.) + (a_)], x_Symbol] :> Simp[c   Int[1/Sqrt[a + b*Csc[e + f*x]], x], x] + 
Simp[d   Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; FreeQ[{a, b, 
c, d, e, f}, x] && NeQ[b*c - a*d, 0] && 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 4544

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

rule 4546

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

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(2076\) vs. \(2(404)=808\).

Time = 70.24 (sec) , antiderivative size = 2077, normalized size of antiderivative = 4.69

method	result	size
default	\(\text {Expression too large to display}\)	\(2077\)
parts	\(\text {Expression too large to display}\)	\(2097\)

Input:

int((a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x,method=_RETUR 
NVERBOSE)
 

Output:

2/15/d/b*(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)*(3*(cos(d*x+c)^2+2*cos(d*x+c)+1)*C*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1 
/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^3*EllipticE(-csc(d*x+c)+co 
t(d*x+c),((a-b)/(a+b))^(1/2))+15*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*B*(1/(a+b) 
*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)* 
a^2*b*EllipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+3*C*a^3*cos(d* 
x+c)*sin(d*x+c)+15*A*a*b^2*cos(d*x+c)*sin(d*x+c)+30*(-cos(d*x+c)^2-2*cos(d 
*x+c)-1)*A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(co 
s(d*x+c)+1))^(1/2)*a^2*b*EllipticPi(-csc(d*x+c)+cot(d*x+c),-1,((a-b)/(a+b) 
)^(1/2))+15*(cos(d*x+c)^2+2*cos(d*x+c)+1)*A*(1/(a+b)*(b+a*cos(d*x+c))/(cos 
(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(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)*C*(1 
/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^ 
(1/2)*a*b^2*EllipticE(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+15*(cos( 
d*x+c)^2+2*cos(d*x+c)+1)*A*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2) 
*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a^2*b*EllipticF(-csc(d*x+c)+cot(d*x+c), 
((a-b)/(a+b))^(1/2))+30*(-cos(d*x+c)^2-2*cos(d*x+c)-1)*A*(1/(a+b)*(b+a*cos 
(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*a*b^2*Ell 
ipticF(-csc(d*x+c)+cot(d*x+c),((a-b)/(a+b))^(1/2))+12*(-cos(d*x+c)^2-2*cos 
(d*x+c)-1)*C*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c...
 

Fricas [F]

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

integrate((a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algori 
thm="fricas")
 

Output:

integral((C*b*sec(d*x + c)^3 + (C*a + B*b)*sec(d*x + c)^2 + A*a + (B*a + A 
*b)*sec(d*x + c))*sqrt(b*sec(d*x + c) + a), x)
 

Sympy [F]

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

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

Output:

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

Maxima [F]

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

integrate((a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algori 
thm="maxima")
 

Output:

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

Giac [F]

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

integrate((a+b*sec(d*x+c))^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2),x, algori 
thm="giac")
 

Output:

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

Mupad [F(-1)]

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

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

Output:

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

Reduce [F]

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

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

Output:

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