\(\int \frac {\sec ^3(c+d x) (A+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\) [193]

Optimal result

Integrand size = 35, antiderivative size = 214 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {(7 A+15 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(15 A+31 C) \tan (c+d x)}{5 a d \sqrt {a+a \sec (c+d x)}}+\frac {(5 A+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {(5 A+13 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{10 a^2 d} \]

[Out]

-1/4*(7*A+15*C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a^(3/2)/d*2^(1/2)-1/2*(A+C)*sec(
d*x+c)^3*tan(d*x+c)/d/(a+a*sec(d*x+c))^(3/2)+1/5*(15*A+31*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)+1/10*(5*A+9
*C)*sec(d*x+c)^2*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(1/2)-1/10*(5*A+13*C)*(a+a*sec(d*x+c))^(1/2)*tan(d*x+c)/a^2/d

Rubi [A] (verified)

Time = 0.79 (sec) , antiderivative size = 214, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 6, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.171, Rules used = {4170, 4106, 4095, 4086, 3880, 209} \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=-\frac {(7 A+15 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(5 A+13 C) \tan (c+d x) \sqrt {a \sec (c+d x)+a}}{10 a^2 d}-\frac {(A+C) \tan (c+d x) \sec ^3(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}+\frac {(5 A+9 C) \tan (c+d x) \sec ^2(c+d x)}{10 a d \sqrt {a \sec (c+d x)+a}}+\frac {(15 A+31 C) \tan (c+d x)}{5 a d \sqrt {a \sec (c+d x)+a}} \]

[In]

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

[Out]

-1/2*((7*A + 15*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(Sqrt[2]*a^(3/2)*d) - ((
A + C)*Sec[c + d*x]^3*Tan[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) + ((15*A + 31*C)*Tan[c + d*x])/(5*a*d*Sqr
t[a + a*Sec[c + d*x]]) + ((5*A + 9*C)*Sec[c + d*x]^2*Tan[c + d*x])/(10*a*d*Sqrt[a + a*Sec[c + d*x]]) - ((5*A +
13*C)*Sqrt[a + a*Sec[c + d*x]]*Tan[c + d*x])/(10*a^2*d)

Rule 209

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1/(Rt[a, 2]*Rt[b, 2]))*ArcTan[Rt[b, 2]*(x/Rt[a, 2])], x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 3880

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[-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
]

Rule 4086

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

Rule 4095

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

Rule 4106

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*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^(n - 1)/(f*(m +
n))), x] + Dist[d/(b*(m + n)), Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^(n - 1)*Simp[b*B*(n - 1) + (A*b*(m
+ n) + a*B*m)*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A*b - a*B, 0] && EqQ[a^2 -
b^2, 0] && GtQ[n, 1]

Rule 4170

Int[((A_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b
_.) + (a_))^(m_), x_Symbol] :> Simp[(-a)*(A + C)*Cot[e + f*x]*(a + b*Csc[e + f*x])^m*((d*Csc[e + f*x])^n/(a*f*
(2*m + 1))), x] + Dist[1/(a*b*(2*m + 1)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n*Simp[b*C*n + A*b
*(2*m + n + 1) - (a*(A*(m + n + 1) - C*(m - n)))*Csc[e + f*x], x], x], x] /; FreeQ[{a, b, d, e, f, A, C, n}, x
] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]

Rubi steps \begin{align*} \text {integral}& = -\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}-\frac {\int \frac {\sec ^3(c+d x) \left (a (A+3 C)-\frac {1}{2} a (5 A+9 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{2 a^2} \\ & = -\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(5 A+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {\int \frac {\sec ^2(c+d x) \left (-a^2 (5 A+9 C)+\frac {3}{4} a^2 (5 A+13 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{5 a^3} \\ & = -\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(5 A+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {(5 A+13 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{10 a^2 d}-\frac {2 \int \frac {\sec (c+d x) \left (\frac {3}{8} a^3 (5 A+13 C)-\frac {3}{4} a^3 (15 A+31 C) \sec (c+d x)\right )}{\sqrt {a+a \sec (c+d x)}} \, dx}{15 a^4} \\ & = -\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(15 A+31 C) \tan (c+d x)}{5 a d \sqrt {a+a \sec (c+d x)}}+\frac {(5 A+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {(5 A+13 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{10 a^2 d}-\frac {(7 A+15 C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{4 a} \\ & = -\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(15 A+31 C) \tan (c+d x)}{5 a d \sqrt {a+a \sec (c+d x)}}+\frac {(5 A+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {(5 A+13 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{10 a^2 d}+\frac {(7 A+15 C) \text {Subst}\left (\int \frac {1}{2 a+x^2} \, dx,x,-\frac {a \tan (c+d x)}{\sqrt {a+a \sec (c+d x)}}\right )}{2 a d} \\ & = -\frac {(7 A+15 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{2 \sqrt {2} a^{3/2} d}-\frac {(A+C) \sec ^3(c+d x) \tan (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac {(15 A+31 C) \tan (c+d x)}{5 a d \sqrt {a+a \sec (c+d x)}}+\frac {(5 A+9 C) \sec ^2(c+d x) \tan (c+d x)}{10 a d \sqrt {a+a \sec (c+d x)}}-\frac {(5 A+13 C) \sqrt {a+a \sec (c+d x)} \tan (c+d x)}{10 a^2 d} \\ \end{align*}

Mathematica [A] (verified)

Time = 3.44 (sec) , antiderivative size = 154, normalized size of antiderivative = 0.72 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\left (-5 \sqrt {2} (7 A+15 C) \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)+\sqrt {1-\sec (c+d x)} \left (25 A+49 C+4 (5 A+9 C) \sec (c+d x)-4 C \sec ^2(c+d x)+4 C \sec ^3(c+d x)\right )\right ) \tan (c+d x)}{10 d \sqrt {1-\sec (c+d x)} (a (1+\sec (c+d x)))^{3/2}} \]

[In]

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

[Out]

((-5*Sqrt[2]*(7*A + 15*C)*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2]^2*Sec[c + d*x] + Sqrt[1 - S
ec[c + d*x]]*(25*A + 49*C + 4*(5*A + 9*C)*Sec[c + d*x] - 4*C*Sec[c + d*x]^2 + 4*C*Sec[c + d*x]^3))*Tan[c + d*x
])/(10*d*Sqrt[1 - Sec[c + d*x]]*(a*(1 + Sec[c + d*x]))^(3/2))

Maple [A] (warning: unable to verify)

Time = 0.83 (sec) , antiderivative size = 362, normalized size of antiderivative = 1.69

method	result	size
parts	\(-\frac {A \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (\left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+7 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}+9 \cot \left (d x +c \right )-9 \csc \left (d x +c \right )\right )}{4 d \,a^{2}}-\frac {C \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (5 \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}+75 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right ) \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {5}{2}}-127 \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+175 \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+85 \cot \left (d x +c \right )-85 \csc \left (d x +c \right )\right )}{20 d \,a^{2} \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{2}}\)	\(362\)
default	\(-\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \left (5 A \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}+5 C \left (1-\cos \left (d x +c \right )\right )^{7} \csc \left (d x +c \right )^{7}+35 A \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {5}{2}} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )+75 C \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {5}{2}} \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )-55 A \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}-127 C \left (1-\cos \left (d x +c \right )\right )^{5} \csc \left (d x +c \right )^{5}+95 A \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}+175 C \left (1-\cos \left (d x +c \right )\right )^{3} \csc \left (d x +c \right )^{3}-45 A \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-85 C \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )\right )}{20 a^{2} d \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{2}}\)	\(377\)

[In]

int(sec(d*x+c)^3*(A+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/4*A/d/a^2*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*((1-cos(d*x+c))^3*csc(d*x+c)^3+7*ln(csc(d*x+c)-cot
(d*x+c)+((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)+9*cot(d*x+c)-9*csc(d*
x+c))-1/20*C/d/a^2*(-2*a/((1-cos(d*x+c))^2*csc(d*x+c)^2-1))^(1/2)*(5*(1-cos(d*x+c))^7*csc(d*x+c)^7+75*ln(csc(d
*x+c)-cot(d*x+c)+((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)^(5/2)-127*(1-cos(d
*x+c))^5*csc(d*x+c)^5+175*(1-cos(d*x+c))^3*csc(d*x+c)^3+85*cot(d*x+c)-85*csc(d*x+c))/((1-cos(d*x+c))^2*csc(d*x
+c)^2-1)^2

Fricas [A] (verification not implemented)

none

Time = 0.31 (sec) , antiderivative size = 492, normalized size of antiderivative = 2.30 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\left [-\frac {5 \, \sqrt {2} {\left ({\left (7 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (7 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (7 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {-a} \log \left (-\frac {2 \, \sqrt {2} \sqrt {-a} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right ) \sin \left (d x + c\right ) - 3 \, a \cos \left (d x + c\right )^{2} - 2 \, a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )^{2} + 2 \, \cos \left (d x + c\right ) + 1}\right ) - 4 \, {\left ({\left (25 \, A + 49 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (5 \, A + 9 \, C\right )} \cos \left (d x + c\right )^{2} - 4 \, C \cos \left (d x + c\right ) + 4 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{40 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}}, \frac {5 \, \sqrt {2} {\left ({\left (7 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{4} + 2 \, {\left (7 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{3} + {\left (7 \, A + 15 \, C\right )} \cos \left (d x + c\right )^{2}\right )} \sqrt {a} \arctan \left (\frac {\sqrt {2} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \cos \left (d x + c\right )}{\sqrt {a} \sin \left (d x + c\right )}\right ) + 2 \, {\left ({\left (25 \, A + 49 \, C\right )} \cos \left (d x + c\right )^{3} + 4 \, {\left (5 \, A + 9 \, C\right )} \cos \left (d x + c\right )^{2} - 4 \, C \cos \left (d x + c\right ) + 4 \, C\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{20 \, {\left (a^{2} d \cos \left (d x + c\right )^{4} + 2 \, a^{2} d \cos \left (d x + c\right )^{3} + a^{2} d \cos \left (d x + c\right )^{2}\right )}}\right ] \]

[In]

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

[Out]

[-1/40*(5*sqrt(2)*((7*A + 15*C)*cos(d*x + c)^4 + 2*(7*A + 15*C)*cos(d*x + c)^3 + (7*A + 15*C)*cos(d*x + c)^2)*
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)) - 4*((25*A + 49*C)*cos(d*x + c)^3 +
4*(5*A + 9*C)*cos(d*x + c)^2 - 4*C*cos(d*x + c) + 4*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(
a^2*d*cos(d*x + c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2), 1/20*(5*sqrt(2)*((7*A + 15*C)*cos(d*x +
c)^4 + 2*(7*A + 15*C)*cos(d*x + c)^3 + (7*A + 15*C)*cos(d*x + c)^2)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(d*x +
c) + a)/cos(d*x + c))*cos(d*x + c)/(sqrt(a)*sin(d*x + c))) + 2*((25*A + 49*C)*cos(d*x + c)^3 + 4*(5*A + 9*C)*c
os(d*x + c)^2 - 4*C*cos(d*x + c) + 4*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c))/(a^2*d*cos(d*x +
c)^4 + 2*a^2*d*cos(d*x + c)^3 + a^2*d*cos(d*x + c)^2)]

Sympy [F]

\[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\int \frac {\left (A + C \sec ^{2}{\left (c + d x \right )}\right ) \sec ^{3}{\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {3}{2}}}\, dx \]

[In]

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

[Out]

Integral((A + C*sec(c + d*x)**2)*sec(c + d*x)**3/(a*(sec(c + d*x) + 1))**(3/2), x)

Maxima [F]

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

[In]

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

[Out]

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

Giac [A] (verification not implemented)

none

Time = 1.76 (sec) , antiderivative size = 276, normalized size of antiderivative = 1.29 \[ \int \frac {\sec ^3(c+d x) \left (A+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx=\frac {\frac {5 \, \sqrt {2} {\left (7 \, A + 15 \, C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {{\left ({\left ({\left (\frac {5 \, \sqrt {2} {\left (A a^{3} + C a^{3}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {\sqrt {2} {\left (55 \, A a^{3} + 127 \, C a^{3}\right )}}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + \frac {5 \, \sqrt {2} {\left (19 \, A a^{3} + 35 \, C a^{3}\right )}}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - \frac {5 \, \sqrt {2} {\left (9 \, A a^{3} + 17 \, C a^{3}\right )}}{a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )}{{\left (a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} - a\right )}^{2} \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a}}}{20 \, d} \]

[In]

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

[Out]

1/20*(5*sqrt(2)*(7*A + 15*C)*log(abs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/(s
qrt(-a)*a*sgn(cos(d*x + c))) - (((5*sqrt(2)*(A*a^3 + C*a^3)*tan(1/2*d*x + 1/2*c)^2/(a^2*sgn(cos(d*x + c))) - s
qrt(2)*(55*A*a^3 + 127*C*a^3)/(a^2*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 + 5*sqrt(2)*(19*A*a^3 + 35*C*a^3
)/(a^2*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c)^2 - 5*sqrt(2)*(9*A*a^3 + 17*C*a^3)/(a^2*sgn(cos(d*x + c))))*ta
n(1/2*d*x + 1/2*c)/((a*tan(1/2*d*x + 1/2*c)^2 - a)^2*sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/d

Mupad [F(-1)]

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

[In]

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

[Out]

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