Integrand size = 27, antiderivative size = 162 \[ \int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {2 A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{a^{5/2} d}-\frac {(43 A-5 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}-\frac {(A+C) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {(11 A-5 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}} \] Output:
2*A*arctan(a^(1/2)*tan(d*x+c)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d-1/32*(43*A -5*C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))*2^(1/2 )/a^(5/2)/d-1/4*(A+C)*tan(d*x+c)/d/(a+a*sec(d*x+c))^(5/2)-1/16*(11*A-5*C)* tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)
Time = 3.47 (sec) , antiderivative size = 153, normalized size of antiderivative = 0.94 \[ \int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {-32 A \arctan \left (\sqrt {-1+\sec (c+d x)}\right ) \sqrt {-1+\sec (c+d x)} \sin (c+d x)+\frac {(43 A-5 C) \arctan \left (\frac {\sqrt {-1+\sec (c+d x)}}{\sqrt {2}}\right ) \sqrt {-1+\sec (c+d x)} \sin (c+d x)}{\sqrt {2}}+(11 A-5 C+(15 A-C) \cos (c+d x)) \tan ^3\left (\frac {1}{2} (c+d x)\right )}{16 a^2 d (-1+\cos (c+d x)) \sqrt {a (1+\sec (c+d x))}} \] Input:
Integrate[(A + C*Sec[c + d*x]^2)/(a + a*Sec[c + d*x])^(5/2),x]
Output:
(-32*A*ArcTan[Sqrt[-1 + Sec[c + d*x]]]*Sqrt[-1 + Sec[c + d*x]]*Sin[c + d*x ] + ((43*A - 5*C)*ArcTan[Sqrt[-1 + Sec[c + d*x]]/Sqrt[2]]*Sqrt[-1 + Sec[c + d*x]]*Sin[c + d*x])/Sqrt[2] + (11*A - 5*C + (15*A - C)*Cos[c + d*x])*Tan [(c + d*x)/2]^3)/(16*a^2*d*(-1 + Cos[c + d*x])*Sqrt[a*(1 + Sec[c + d*x])])
Time = 0.91 (sec) , antiderivative size = 174, normalized size of antiderivative = 1.07, number of steps used = 14, number of rules used = 13, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.481, Rules used = {3042, 4541, 27, 3042, 4410, 27, 3042, 4408, 3042, 4261, 216, 4282, 216}
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.
\(\displaystyle \int \frac {A+C \sec ^2(c+d x)}{(a \sec (c+d x)+a)^{5/2}} \, dx\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \int \frac {A+C \csc \left (c+d x+\frac {\pi }{2}\right )^2}{\left (a \csc \left (c+d x+\frac {\pi }{2}\right )+a\right )^{5/2}}dx\) |
\(\Big \downarrow \) 4541 |
\(\displaystyle -\frac {\int -\frac {8 a A-a (3 A-5 C) \sec (c+d x)}{2 (\sec (c+d x) a+a)^{3/2}}dx}{4 a^2}-\frac {(A+C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\int \frac {8 a A-a (3 A-5 C) \sec (c+d x)}{(\sec (c+d x) a+a)^{3/2}}dx}{8 a^2}-\frac {(A+C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\int \frac {8 a A-a (3 A-5 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\left (\csc \left (c+d x+\frac {\pi }{2}\right ) a+a\right )^{3/2}}dx}{8 a^2}-\frac {(A+C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 4410 |
\(\displaystyle \frac {-\frac {\int -\frac {32 a^2 A-a^2 (11 A-5 C) \sec (c+d x)}{2 \sqrt {\sec (c+d x) a+a}}dx}{2 a^2}-\frac {a (11 A-5 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 27 |
\(\displaystyle \frac {\frac {\int \frac {32 a^2 A-a^2 (11 A-5 C) \sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (11 A-5 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {\int \frac {32 a^2 A-a^2 (11 A-5 C) \csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (11 A-5 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 4408 |
\(\displaystyle \frac {\frac {32 a A \int \sqrt {\sec (c+d x) a+a}dx-a^2 (43 A-5 C) \int \frac {\sec (c+d x)}{\sqrt {\sec (c+d x) a+a}}dx}{4 a^2}-\frac {a (11 A-5 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 3042 |
\(\displaystyle \frac {\frac {32 a A \int \sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}dx-a^2 (43 A-5 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (11 A-5 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 4261 |
\(\displaystyle \frac {\frac {-\left (a^2 (43 A-5 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx\right )-\frac {64 a^2 A \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}}{4 a^2}-\frac {a (11 A-5 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {\frac {64 a^{3/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-a^2 (43 A-5 C) \int \frac {\csc \left (c+d x+\frac {\pi }{2}\right )}{\sqrt {\csc \left (c+d x+\frac {\pi }{2}\right ) a+a}}dx}{4 a^2}-\frac {a (11 A-5 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 4282 |
\(\displaystyle \frac {\frac {\frac {2 a^2 (43 A-5 C) \int \frac {1}{\frac {a^2 \tan ^2(c+d x)}{\sec (c+d x) a+a}+2 a}d\left (-\frac {a \tan (c+d x)}{\sqrt {\sec (c+d x) a+a}}\right )}{d}+\frac {64 a^{3/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {a (11 A-5 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
\(\Big \downarrow \) 216 |
\(\displaystyle \frac {\frac {\frac {64 a^{3/2} A \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {a \sec (c+d x)+a}}\right )}{d}-\frac {\sqrt {2} a^{3/2} (43 A-5 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{d}}{4 a^2}-\frac {a (11 A-5 C) \tan (c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}}{8 a^2}-\frac {(A+C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}}\) |
Input:
Int[(A + C*Sec[c + d*x]^2)/(a + a*Sec[c + d*x])^(5/2),x]
Output:
-1/4*((A + C)*Tan[c + d*x])/(d*(a + a*Sec[c + d*x])^(5/2)) + (((64*a^(3/2) *A*ArcTan[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/d - (Sqrt[2]*a ^(3/2)*(43*A - 5*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[ c + d*x]])])/d)/(4*a^2) - (a*(11*A - 5*C)*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)))/(8*a^2)
Int[(a_)*(Fx_), x_Symbol] :> Simp[a Int[Fx, x], x] /; FreeQ[a, x] && !Ma tchQ[Fx, (b_)*(Gx_) /; FreeQ[b, x]]
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*A rcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /; FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a , 0] || GtQ[b, 0])
Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[-2*(b/d) Subst[Int[1/(a + x^2), x], x, b*(Cot[c + d*x]/Sqrt[a + b*Csc[c + d*x]])], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_S ymbol] :> Simp[-2/f Subst[Int[1/(2*a + x^2), x], x, b*(Cot[e + f*x]/Sqrt[ a + b*Csc[e + f*x]])], x] /; FreeQ[{a, b, e, f}, x] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_ .) + (a_)], x_Symbol] :> Simp[c/a Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Simp[(b*c - a*d)/a Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /; F reeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]
Int[(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*(d _.) + (c_)), x_Symbol] :> Simp[(-(b*c - a*d))*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(b*f*(2*m + 1))), x] + Simp[1/(a^2*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[a*c*(2*m + 1) - (b*c - a*d)*(m + 1)*Csc[e + f*x], x], x] , x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && LtQ[m, -1] && EqQ[a^2 - b^2, 0] && IntegerQ[2*m]
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(b_. ) + (a_))^(m_), x_Symbol] :> Simp[(-a)*(A + C)*Cot[e + f*x]*((a + b*Csc[e + f*x])^m/(a*f*(2*m + 1))), x] + Simp[1/(a*b*(2*m + 1)) Int[(a + b*Csc[e + f*x])^(m + 1)*Simp[A*b*(2*m + 1) - a*(A*(m + 1) - C*m)*Csc[e + f*x], x], x ], x] /; FreeQ[{a, b, e, f, A, C}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1 )]
Leaf count of result is larger than twice the leaf count of optimal. \(375\) vs. \(2(137)=274\).
Time = 1.02 (sec) , antiderivative size = 376, normalized size of antiderivative = 2.32
method | result | size |
default | \(\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-2 A \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}} \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-2 C \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}} \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )+32 A \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )+11 A \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-5 C \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-43 A \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )+5 C \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )\right )}{32 d \,a^{3}}\) | \(376\) |
parts | \(\frac {A \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (-2 \left (1-\cos \left (d x +c \right )\right )^{3} \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \csc \left (d x +c \right )^{3}+32 \sqrt {2}\, \operatorname {arctanh}\left (\frac {\sqrt {2}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )}{\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\right )+13 \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-43 \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{32 d \,a^{3}}-\frac {C \sqrt {a \left (1+\sec \left (d x +c \right )\right )}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\frac {4 \sqrt {2}\, \left (\cos \left (d x +c \right )-1\right ) \sqrt {-\frac {\cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \cot \left (d x +c \right )}{\cos \left (d x +c \right )+1}+5 \sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}\, \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )\right )-5 \ln \left (\sqrt {-\frac {2 \cos \left (d x +c \right )}{\cos \left (d x +c \right )+1}}-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{32 d \,a^{3}}\) | \(377\) |
Input:
int((A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)
Output:
1/32/d/a^3*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*((1-cos(d*x+c))^ 2*csc(d*x+c)^2-1)^(1/2)*(-2*A*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)*(csc (d*x+c)-cot(d*x+c))-2*C*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)*(csc(d*x+c )-cot(d*x+c))+32*A*2^(1/2)*arctanh(2^(1/2)/((1-cos(d*x+c))^2*csc(d*x+c)^2- 1)^(1/2)*(csc(d*x+c)-cot(d*x+c)))+11*A*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^( 1/2)*(csc(d*x+c)-cot(d*x+c))-5*C*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*( csc(d*x+c)-cot(d*x+c))-43*A*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc (d*x+c)^2-1)^(1/2))+5*C*ln(csc(d*x+c)-cot(d*x+c)+((1-cos(d*x+c))^2*csc(d*x +c)^2-1)^(1/2)))
Leaf count of result is larger than twice the leaf count of optimal. 292 vs. \(2 (137) = 274\).
Time = 3.27 (sec) , antiderivative size = 670, normalized size of antiderivative = 4.14 \[ \int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx =\text {Too large to display} \] Input:
integrate((A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")
Output:
[1/64*(sqrt(2)*((43*A - 5*C)*cos(d*x + c)^3 + 3*(43*A - 5*C)*cos(d*x + c)^ 2 + 3*(43*A - 5*C)*cos(d*x + c) + 43*A - 5*C)*sqrt(-a)*log((2*sqrt(2)*sqrt (-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)*sin(d*x + c) + 3 *a*cos(d*x + c)^2 + 2*a*cos(d*x + c) - a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 64*(A*cos(d*x + c)^3 + 3*A*cos(d*x + c)^2 + 3*A*cos(d*x + c) + A) *sqrt(-a)*log((2*a*cos(d*x + c)^2 + 2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/c os(d*x + c))*cos(d*x + c)*sin(d*x + c) + a*cos(d*x + c) - a)/(cos(d*x + c) + 1)) - 4*((15*A - C)*cos(d*x + c)^2 + (11*A - 5*C)*cos(d*x + c))*sqrt((a *cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a ^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d), 1/32*(sqrt(2)*((43*A - 5*C)*cos(d*x + c)^3 + 3*(43*A - 5*C)*cos(d*x + c)^2 + 3*(43*A - 5*C)*cos (d*x + c) + 43*A - 5*C)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x + c) + a)/c os(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - 64*(A*cos(d*x + c)^3 + 3*A*cos(d*x + c)^2 + 3*A*cos(d*x + c) + A)*sqrt(a)*arctan(sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) - 2*((15*A - C)*cos(d*x + c)^2 + (11*A - 5*C)*cos(d*x + c))*sqrt((a*cos(d*x + c) + a)/ cos(d*x + c))*sin(d*x + c))/(a^3*d*cos(d*x + c)^3 + 3*a^3*d*cos(d*x + c)^2 + 3*a^3*d*cos(d*x + c) + a^3*d)]
\[ \int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {A + C \sec ^{2}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \] Input:
integrate((A+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(5/2),x)
Output:
Integral((A + C*sec(c + d*x)**2)/(a*(sec(c + d*x) + 1))**(5/2), x)
\[ \int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + A}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \] Input:
integrate((A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")
Output:
integrate((C*sec(d*x + c)^2 + A)/(a*sec(d*x + c) + a)^(5/2), x)
Exception generated. \[ \int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\text {Exception raised: TypeError} \] Input:
integrate((A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")
Output:
Exception raised: TypeError >> an error occurred running a Giac command:IN PUT:sage2:=int(sage0,sageVARx):;OUTPUT:index.cc index_m i_lex_is_greater E rror: Bad Argument Value
Timed out. \[ \int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {A+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \] Input:
int((A + C/cos(c + d*x)^2)/(a + a/cos(c + d*x))^(5/2),x)
Output:
int((A + C/cos(c + d*x)^2)/(a + a/cos(c + d*x))^(5/2), x)
\[ \int \frac {A+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {\sqrt {a}\, \left (\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) a +\left (\int \frac {\sqrt {\sec \left (d x +c \right )+1}\, \sec \left (d x +c \right )^{2}}{\sec \left (d x +c \right )^{3}+3 \sec \left (d x +c \right )^{2}+3 \sec \left (d x +c \right )+1}d x \right ) c \right )}{a^{3}} \] Input:
int((A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x)
Output:
(sqrt(a)*(int(sqrt(sec(c + d*x) + 1)/(sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x)*a + int((sqrt(sec(c + d*x) + 1)*sec(c + d*x)**2)/ (sec(c + d*x)**3 + 3*sec(c + d*x)**2 + 3*sec(c + d*x) + 1),x)*c))/a**3