Integrand size = 42, antiderivative size = 471 \[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\frac {2 \left (40 a^3 b B-25 a b^3 B-48 a^4 C+24 a^2 b^2 C+9 b^4 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^5 \sqrt {a+b} d}+\frac {2 \left (b^3 (5 B-9 C)+4 a^2 b (10 B-9 C)+6 a b^2 (5 B-2 C)-48 a^3 C\right ) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{15 b^4 \sqrt {a+b} d}+\frac {2 a (b B-a C) \sec ^2(c+d x) \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}}+\frac {2 \left (20 a^2 b B-5 b^3 B-24 a^3 C+9 a b^2 C\right ) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{15 b^3 \left (a^2-b^2\right ) d}-\frac {2 \left (5 a b B-6 a^2 C+b^2 C\right ) \sec (c+d x) \sqrt {a+b \sec (c+d x)} \tan (c+d x)}{5 b^2 \left (a^2-b^2\right ) d} \]
2/15*(40*B*a^3*b-25*B*a*b^3-48*C*a^4+24*C*a^2*b^2+9*C*b^4)*cot(d*x+c)*Elli pticE((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^5/d/(a+b)^(1/2)+2/15* (b^3*(5*B-9*C)+4*a^2*b*(10*B-9*C)+6*a*b^2*(5*B-2*C)-48*a^3*C)*cot(d*x+c)*E llipticF((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/(a+b)^(1/2)+2* a*(B*b-C*a)*sec(d*x+c)^2*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)+2 /15*(20*B*a^2*b-5*B*b^3-24*C*a^3+9*C*a*b^2)*(a+b*sec(d*x+c))^(1/2)*tan(d*x +c)/b^3/(a^2-b^2)/d-2/5*(5*B*a*b-6*C*a^2+C*b^2)*sec(d*x+c)*(a+b*sec(d*x+c) )^(1/2)*tan(d*x+c)/b^2/(a^2-b^2)/d
Leaf count is larger than twice the leaf count of optimal. \(3953\) vs. \(2(471)=942\).
Time = 23.51 (sec) , antiderivative size = 3953, normalized size of antiderivative = 8.39 \[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Result too large to show} \]
((b + a*Cos[c + d*x])^2*Sec[c + d*x]^2*((2*(40*a^3*b*B - 25*a*b^3*B - 48*a ^4*C + 24*a^2*b^2*C + 9*b^4*C)*Sin[c + d*x])/(15*b^4*(-a^2 + b^2)) + (2*Se c[c + d*x]*(5*b*B*Sin[c + d*x] - 9*a*C*Sin[c + d*x]))/(15*b^3) - (2*(a^3*b *B*Sin[c + d*x] - a^4*C*Sin[c + d*x]))/(b^3*(-a^2 + b^2)*(b + a*Cos[c + d* x])) + (2*C*Sec[c + d*x]*Tan[c + d*x])/(5*b^2)))/(d*(a + b*Sec[c + d*x])^( 3/2)) + (2*(b + a*Cos[c + d*x])*((5*a*B)/(3*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (8*a^3*B)/(3*b^2*(-a^2 + b^2)*Sqrt[b + a*Cos [c + d*x]]*Sqrt[Sec[c + d*x]]) + (16*a^4*C)/(5*b^3*(-a^2 + b^2)*Sqrt[b + a *Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (8*a^2*C)/(5*b*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]*Sqrt[Sec[c + d*x]]) - (3*b*C)/(5*(-a^2 + b^2)*Sqrt[b + a*C os[c + d*x]]*Sqrt[Sec[c + d*x]]) - (8*a^4*B*Sqrt[Sec[c + d*x]])/(3*b^3*(-a ^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) + (7*a^2*B*Sqrt[Sec[c + d*x]])/(3*b*(- a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) + (b*B*Sqrt[Sec[c + d*x]])/(3*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) - (4*a*C*Sqrt[Sec[c + d*x]])/(5*(-a^2 + b^ 2)*Sqrt[b + a*Cos[c + d*x]]) + (16*a^5*C*Sqrt[Sec[c + d*x]])/(5*b^4*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) - (12*a^3*C*Sqrt[Sec[c + d*x]])/(5*b^2*(- a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) - (8*a^4*B*Cos[2*(c + d*x)]*Sqrt[Sec[ c + d*x]])/(3*b^3*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) + (5*a^2*B*Cos[2* (c + d*x)]*Sqrt[Sec[c + d*x]])/(3*b*(-a^2 + b^2)*Sqrt[b + a*Cos[c + d*x]]) - (3*a*C*Cos[2*(c + d*x)]*Sqrt[Sec[c + d*x]])/(5*(-a^2 + b^2)*Sqrt[b +...
Time = 2.20 (sec) , antiderivative size = 486, normalized size of antiderivative = 1.03, number of steps used = 16, number of rules used = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.381, Rules used = {3042, 4560, 3042, 4517, 27, 3042, 4580, 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.
\(\displaystyle \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^3 \left (B \csc \left (c+d x+\frac {\pi }{2}\right )+C \csc \left (c+d x+\frac {\pi }{2}\right )^2\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4560 |
\(\displaystyle \int \frac {\sec ^4(c+d x) (B+C \sec (c+d x))}{(a+b \sec (c+d x))^{3/2}}dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^4 \left (B+C \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\left (a+b \csc \left (c+d x+\frac {\pi }{2}\right )\right )^{3/2}}dx\) |
\(\Big \downarrow \) 4517 |
\(\displaystyle \frac {2 \int \frac {\sec ^2(c+d x) \left (-\left (\left (-6 C a^2+5 b B a+b^2 C\right ) \sec ^2(c+d x)\right )-b (b B-a C) \sec (c+d x)+4 a (b B-a C)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}+\frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {\sec ^2(c+d x) \left (-\left (\left (-6 C a^2+5 b B a+b^2 C\right ) \sec ^2(c+d x)\right )-b (b B-a C) \sec (c+d x)+4 a (b B-a C)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{b \left (a^2-b^2\right )}+\frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )^2 \left (\left (6 C a^2-5 b B a-b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-b (b B-a C) \csc \left (c+d x+\frac {\pi }{2}\right )+4 a (b B-a C)\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{b \left (a^2-b^2\right )}+\frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
\(\Big \downarrow \) 4580 |
\(\displaystyle \frac {\frac {2 \int -\frac {\sec (c+d x) \left (-\left (\left (-24 C a^3+20 b B a^2+9 b^2 C a-5 b^3 B\right ) \sec ^2(c+d x)\right )-b \left (-2 C a^2+5 b B a-3 b^2 C\right ) \sec (c+d x)+2 a \left (-6 C a^2+5 b B a+b^2 C\right )\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{5 b}-\frac {2 \left (-6 a^2 C+5 a b B+b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}+\frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {-\frac {\int \frac {\sec (c+d x) \left (-\left (\left (-24 C a^3+20 b B a^2+9 b^2 C a-5 b^3 B\right ) \sec ^2(c+d x)\right )-b \left (-2 C a^2+5 b B a-3 b^2 C\right ) \sec (c+d x)+2 a \left (-6 C a^2+5 b B a+b^2 C\right )\right )}{\sqrt {a+b \sec (c+d x)}}dx}{5 b}-\frac {2 \left (-6 a^2 C+5 a b B+b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}+\frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\left (24 C a^3-20 b B a^2-9 b^2 C a+5 b^3 B\right ) \csc \left (c+d x+\frac {\pi }{2}\right )^2-b \left (-2 C a^2+5 b B a-3 b^2 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )+2 a \left (-6 C a^2+5 b B a+b^2 C\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{5 b}-\frac {2 \left (-6 a^2 C+5 a b B+b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}+\frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
\(\Big \downarrow \) 4570 |
\(\displaystyle \frac {-\frac {\frac {2 \int \frac {\sec (c+d x) \left (b \left (-12 C a^3+10 b B a^2-3 b^2 C a+5 b^3 B\right )+\left (-48 C a^4+40 b B a^3+24 b^2 C a^2-25 b^3 B a+9 b^4 C\right ) \sec (c+d x)\right )}{2 \sqrt {a+b \sec (c+d x)}}dx}{3 b}-\frac {2 \left (-24 a^3 C+20 a^2 b B+9 a b^2 C-5 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (-6 a^2 C+5 a b B+b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}+\frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {-\frac {\frac {\int \frac {\sec (c+d x) \left (b \left (-12 C a^3+10 b B a^2-3 b^2 C a+5 b^3 B\right )+\left (-48 C a^4+40 b B a^3+24 b^2 C a^2-25 b^3 B a+9 b^4 C\right ) \sec (c+d x)\right )}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}-\frac {2 \left (-24 a^3 C+20 a^2 b B+9 a b^2 C-5 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (-6 a^2 C+5 a b B+b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}+\frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-\frac {\frac {\int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (b \left (-12 C a^3+10 b B a^2-3 b^2 C a+5 b^3 B\right )+\left (-48 C a^4+40 b B a^3+24 b^2 C a^2-25 b^3 B a+9 b^4 C\right ) \csc \left (c+d x+\frac {\pi }{2}\right )\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}-\frac {2 \left (-24 a^3 C+20 a^2 b B+9 a b^2 C-5 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (-6 a^2 C+5 a b B+b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}+\frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
\(\Big \downarrow \) 4493 |
\(\displaystyle \frac {-\frac {\frac {\left (-48 a^4 C+40 a^3 b B+24 a^2 b^2 C-25 a b^3 B+9 b^4 C\right ) \int \frac {\sec (c+d x) (\sec (c+d x)+1)}{\sqrt {a+b \sec (c+d x)}}dx-(a-b) \left (-48 a^3 C+a^2 b (40 B-36 C)+6 a b^2 (5 B-2 C)+b^3 (5 B-9 C)\right ) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}}dx}{3 b}-\frac {2 \left (-24 a^3 C+20 a^2 b B+9 a b^2 C-5 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (-6 a^2 C+5 a b B+b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}+\frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {-\frac {\frac {\left (-48 a^4 C+40 a^3 b B+24 a^2 b^2 C-25 a b^3 B+9 b^4 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-(a-b) \left (-48 a^3 C+a^2 b (40 B-36 C)+6 a b^2 (5 B-2 C)+b^3 (5 B-9 C)\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx}{3 b}-\frac {2 \left (-24 a^3 C+20 a^2 b B+9 a b^2 C-5 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (-6 a^2 C+5 a b B+b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}+\frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
\(\Big \downarrow \) 4319 |
\(\displaystyle \frac {-\frac {\frac {\left (-48 a^4 C+40 a^3 b B+24 a^2 b^2 C-25 a b^3 B+9 b^4 C\right ) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right ) \left (\csc \left (c+d x+\frac {\pi }{2}\right )+1\right )}{\sqrt {a+b \csc \left (c+d x+\frac {\pi }{2}\right )}}dx-\frac {2 (a-b) \sqrt {a+b} \left (-48 a^3 C+a^2 b (40 B-36 C)+6 a b^2 (5 B-2 C)+b^3 (5 B-9 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}}{3 b}-\frac {2 \left (-24 a^3 C+20 a^2 b B+9 a b^2 C-5 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}-\frac {2 \left (-6 a^2 C+5 a b B+b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}}{b \left (a^2-b^2\right )}+\frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}\) |
\(\Big \downarrow \) 4492 |
\(\displaystyle \frac {2 a (b B-a C) \tan (c+d x) \sec ^2(c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}+\frac {-\frac {2 \left (-6 a^2 C+5 a b B+b^2 C\right ) \tan (c+d x) \sec (c+d x) \sqrt {a+b \sec (c+d x)}}{5 b d}-\frac {\frac {-\frac {2 (a-b) \sqrt {a+b} \left (-48 a^3 C+a^2 b (40 B-36 C)+6 a b^2 (5 B-2 C)+b^3 (5 B-9 C)\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{b d}-\frac {2 (a-b) \sqrt {a+b} \left (-48 a^4 C+40 a^3 b B+24 a^2 b^2 C-25 a b^3 B+9 b^4 C\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d}}{3 b}-\frac {2 \left (-24 a^3 C+20 a^2 b B+9 a b^2 C-5 b^3 B\right ) \tan (c+d x) \sqrt {a+b \sec (c+d x)}}{3 b d}}{5 b}}{b \left (a^2-b^2\right )}\) |
(2*a*(b*B - a*C)*Sec[c + d*x]^2*Tan[c + d*x])/(b*(a^2 - b^2)*d*Sqrt[a + b* Sec[c + d*x]]) + ((-2*(5*a*b*B - 6*a^2*C + b^2*C)*Sec[c + d*x]*Sqrt[a + b* Sec[c + d*x]]*Tan[c + d*x])/(5*b*d) - (((-2*(a - b)*Sqrt[a + b]*(40*a^3*b* B - 25*a*b^3*B - 48*a^4*C + 24*a^2*b^2*C + 9*b^4*C)*Cot[c + d*x]*EllipticE [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^2*d ) - (2*(a - b)*Sqrt[a + b]*(a^2*b*(40*B - 36*C) + b^3*(5*B - 9*C) + 6*a*b^ 2*(5*B - 2*C) - 48*a^3*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))/(3*b) - (2*(20*a^2*b*B - 5*b^3*B - 24*a^3*C + 9*a*b^2*C)*Sqrt[a + b*Sec[c + d*x]]*Tan[c + d*x])/(3* b*d))/(5*b))/(b*(a^2 - b^2))
3.9.43.3.1 Defintions of rubi rules used
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
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]
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]
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]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + ( a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)), x_Symbol] :> Simp[a*d^2*( A*b - a*B)*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*((d*Csc[e + f*x])^(n - 2)/(b*f*(m + 1)*(a^2 - b^2))), x] - Simp[d/(b*(m + 1)*(a^2 - b^2)) Int[( a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2)*Simp[a*d*(A*b - a*B)*( n - 2) + b*d*(A*b - a*B)*(m + 1)*Csc[e + f*x] - (a*A*b*d*(m + n) - d*B*(a^2 *(n - 1) + b^2*(m + 1)))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, d, e, f , A, B}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && GtQ[ n, 1]
Int[((a_.) + csc[(e_.) + (f_.)*(x_)]*(b_.))^(m_.)*((A_.) + csc[(e_.) + (f_. )*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*((c_.) + csc[(e_.) + (f_.) *(x_)]*(d_.))^(n_.), x_Symbol] :> Simp[1/b^2 Int[(a + b*Csc[e + f*x])^(m + 1)*(c + d*Csc[e + f*x])^n*(b*B - a*C + b*C*Csc[e + f*x]), x], x] /; FreeQ [{a, b, c, d, e, f, A, B, C, m, n}, x] && EqQ[A*b^2 - a*b*B + a^2*C, 0]
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]
Int[csc[(e_.) + (f_.)*(x_)]^2*((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[ (e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x _Symbol] :> Simp[(-C)*Csc[e + f*x]*Cot[e + f*x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 3))), x] + Simp[1/(b*(m + 3)) Int[Csc[e + f*x]*(a + b*Csc[e + f*x])^m*Simp[a*C + b*(C*(m + 2) + A*(m + 3))*Csc[e + f*x] - (2*a*C - b*B* (m + 3))*Csc[e + f*x]^2, x], x], x] /; FreeQ[{a, b, e, f, A, B, C, m}, x] & & NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Leaf count of result is larger than twice the leaf count of optimal. \(5460\) vs. \(2(439)=878\).
Time = 22.52 (sec) , antiderivative size = 5461, normalized size of antiderivative = 11.59
method | result | size |
parts | \(\text {Expression too large to display}\) | \(5461\) |
default | \(\text {Expression too large to display}\) | \(5511\) |
int(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2),x,me thod=_RETURNVERBOSE)
\[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sec \left (d x + c\right )^{3}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2 ),x, algorithm="fricas")
integral((C*sec(d*x + c)^5 + B*sec(d*x + c)^4)*sqrt(b*sec(d*x + c) + a)/(b ^2*sec(d*x + c)^2 + 2*a*b*sec(d*x + c) + a^2), x)
\[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \sec ^{4}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \]
Timed out. \[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\text {Timed out} \]
integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2 ),x, algorithm="maxima")
\[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int { \frac {{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )\right )} \sec \left (d x + c\right )^{3}}{{\left (b \sec \left (d x + c\right ) + a\right )}^{\frac {3}{2}}} \,d x } \]
integrate(sec(d*x+c)^3*(B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+b*sec(d*x+c))^(3/2 ),x, algorithm="giac")
Timed out. \[ \int \frac {\sec ^3(c+d x) \left (B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+b \sec (c+d x))^{3/2}} \, dx=\int \frac {\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\cos \left (c+d\,x\right )}^3\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]