\(\int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx\) [405]

Optimal result

Integrand size = 34, antiderivative size = 126 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {(3 B+5 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a+a \sec (c+d x)}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(B-C) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(3 B+5 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}} \]

[Out]

1/32*(3*B+5*C)*arctan(1/2*a^(1/2)*tan(d*x+c)*2^(1/2)/(a+a*sec(d*x+c))^(1/2))/a^(5/2)/d*2^(1/2)+1/4*(B-C)*tan(d
*x+c)/d/(a+a*sec(d*x+c))^(5/2)+1/16*(3*B+5*C)*tan(d*x+c)/a/d/(a+a*sec(d*x+c))^(3/2)

Rubi [A] (verified)

Time = 0.20 (sec) , antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {4137, 12, 3881, 3880, 209} \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {(3 B+5 C) \arctan \left (\frac {\sqrt {a} \tan (c+d x)}{\sqrt {2} \sqrt {a \sec (c+d x)+a}}\right )}{16 \sqrt {2} a^{5/2} d}+\frac {(3 B+5 C) \tan (c+d x)}{16 a d (a \sec (c+d x)+a)^{3/2}}+\frac {(B-C) \tan (c+d x)}{4 d (a \sec (c+d x)+a)^{5/2}} \]

[In]

Int[(B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

((3*B + 5*C)*ArcTan[(Sqrt[a]*Tan[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(16*Sqrt[2]*a^(5/2)*d) + ((B -
C)*Tan[c + d*x])/(4*d*(a + a*Sec[c + d*x])^(5/2)) + ((3*B + 5*C)*Tan[c + d*x])/(16*a*d*(a + a*Sec[c + d*x])^(
3/2))

Rule 12

Int[(a_)*(u_), x_Symbol] :> Dist[a, Int[u, x], x] /; FreeQ[a, x] &&  !MatchQ[u, (b_)*(v_) /; FreeQ[b, x]]

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 3881

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

Rule 4137

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

Rubi steps \begin{align*} \text {integral}& = \frac {(B-C) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}-\frac {\int -\frac {a (3 B+5 C) \sec (c+d x)}{2 (a+a \sec (c+d x))^{3/2}} \, dx}{4 a^2} \\ & = \frac {(B-C) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(3 B+5 C) \int \frac {\sec (c+d x)}{(a+a \sec (c+d x))^{3/2}} \, dx}{8 a} \\ & = \frac {(B-C) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(3 B+5 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}+\frac {(3 B+5 C) \int \frac {\sec (c+d x)}{\sqrt {a+a \sec (c+d x)}} \, dx}{32 a^2} \\ & = \frac {(B-C) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(3 B+5 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}}-\frac {(3 B+5 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 )}{16 a^2 d} \\ & = \frac {(3 B+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 {(B-C) \tan (c+d x)}{4 d (a+a \sec (c+d x))^{5/2}}+\frac {(3 B+5 C) \tan (c+d x)}{16 a d (a+a \sec (c+d x))^{3/2}} \\ \end{align*}

Mathematica [C] (verified)

Result contains higher order function than in optimal. Order 5 vs. order 3 in optimal.

Time = 1.06 (sec) , antiderivative size = 206, normalized size of antiderivative = 1.63 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\frac {40 \sqrt {2} C \text {arctanh}\left (\frac {\sqrt {1-\sec (c+d x)}}{\sqrt {2}}\right ) \cos ^5\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x) \sin \left (\frac {1}{2} (c+d x)\right )+64 B \cos ^5\left (\frac {1}{2} (c+d x)\right ) \operatorname {Hypergeometric2F1}\left (\frac {1}{2},3,\frac {3}{2},\frac {1}{2} (1-\sec (c+d x))\right ) \sqrt {1-\sec (c+d x)} \sec (c+d x) \sin \left (\frac {1}{2} (c+d x)\right )+C \sqrt {1-\sec (c+d x)} (10 \sin (c+d x)+\sin (2 (c+d x)))}{32 a^2 d (1+\cos (c+d x))^2 \sqrt {1-\sec (c+d x)} \sqrt {a (1+\sec (c+d x))}} \]

[In]

Integrate[(B*Sec[c + d*x] + C*Sec[c + d*x]^2)/(a + a*Sec[c + d*x])^(5/2),x]

[Out]

(40*Sqrt[2]*C*ArcTanh[Sqrt[1 - Sec[c + d*x]]/Sqrt[2]]*Cos[(c + d*x)/2]^5*Sec[c + d*x]*Sin[(c + d*x)/2] + 64*B*
Cos[(c + d*x)/2]^5*Hypergeometric2F1[1/2, 3, 3/2, (1 - Sec[c + d*x])/2]*Sqrt[1 - Sec[c + d*x]]*Sec[c + d*x]*Si
n[(c + d*x)/2] + C*Sqrt[1 - Sec[c + d*x]]*(10*Sin[c + d*x] + Sin[2*(c + d*x)]))/(32*a^2*d*(1 + Cos[c + d*x])^2
*Sqrt[1 - Sec[c + d*x]]*Sqrt[a*(1 + Sec[c + d*x])])

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(324\) vs. \(2(107)=214\).

Time = 0.69 (sec) , antiderivative size = 325, normalized size of antiderivative = 2.58

method	result	size
default	\(\frac {\sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (2 B \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}} \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-2 C \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}} \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-3 B \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-5 C \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+3 B \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )+5 C \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )\right )}{32 a^{3} d}\)	\(325\)
parts	\(\frac {B \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (2 \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}} \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-3 \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+3 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )\right )}{32 d \,a^{3}}-\frac {C \sqrt {-\frac {2 a}{\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}}\, \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (2 \left (\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1\right )^{\frac {3}{2}} \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )+5 \sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\, \left (-\cot \left (d x +c \right )+\csc \left (d x +c \right )\right )-5 \ln \left (\csc \left (d x +c \right )-\cot \left (d x +c \right )+\sqrt {\left (1-\cos \left (d x +c \right )\right )^{2} \csc \left (d x +c \right )^{2}-1}\right )\right )}{32 d \,a^{3}}\)	\(386\)

[In]

int((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x,method=_RETURNVERBOSE)

[Out]

1/32/a^3/d*(-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*B*((1-cos
(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)*(-cot(d*x+c)+csc(d*x+c))-2*C*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(3/2)*(-cot(d*
x+c)+csc(d*x+c))-3*B*((1-cos(d*x+c))^2*csc(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c))-5*C*((1-cos(d*x+c))^2*cs
c(d*x+c)^2-1)^(1/2)*(-cot(d*x+c)+csc(d*x+c))+3*B*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)))

Fricas [A] (verification not implemented)

none

Time = 0.29 (sec) , antiderivative size = 475, normalized size of antiderivative = 3.77 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\left [-\frac {\sqrt {2} {\left ({\left (3 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, B + 5 \, C\right )} \cos \left (d x + c\right ) + 3 \, B + 5 \, C\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 (7 \, B + C\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, B + 5 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{64 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}, -\frac {\sqrt {2} {\left ({\left (3 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{3} + 3 \, {\left (3 \, B + 5 \, C\right )} \cos \left (d x + c\right )^{2} + 3 \, {\left (3 \, B + 5 \, C\right )} \cos \left (d x + c\right ) + 3 \, B + 5 \, C\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 (7 \, B + C\right )} \cos \left (d x + c\right )^{2} + {\left (3 \, B + 5 \, C\right )} \cos \left (d x + c\right )\right )} \sqrt {\frac {a \cos \left (d x + c\right ) + a}{\cos \left (d x + c\right )}} \sin \left (d x + c\right )}{32 \, {\left (a^{3} d \cos \left (d x + c\right )^{3} + 3 \, a^{3} d \cos \left (d x + c\right )^{2} + 3 \, a^{3} d \cos \left (d x + c\right ) + a^{3} d\right )}}\right ] \]

[In]

integrate((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="fricas")

[Out]

[-1/64*(sqrt(2)*((3*B + 5*C)*cos(d*x + c)^3 + 3*(3*B + 5*C)*cos(d*x + c)^2 + 3*(3*B + 5*C)*cos(d*x + c) + 3*B
+ 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)) - 4*((7*B + C)*cos(d*x + c)^2
+ (3*B + 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)*((3*B + 5*C)*cos(d*x + c)^3 + 3*(3*B + 5*C)
*cos(d*x + c)^2 + 3*(3*B + 5*C)*cos(d*x + c) + 3*B + 5*C)*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*((7*B + C)*cos(d*x + c)^2 + (3*B + 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)]

Sympy [F]

\[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\left (B + C \sec {\left (c + d x \right )}\right ) \sec {\left (c + d x \right )}}{\left (a \left (\sec {\left (c + d x \right )} + 1\right )\right )^{\frac {5}{2}}}\, dx \]

[In]

integrate((B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(5/2),x)

[Out]

Integral((B + C*sec(c + d*x))*sec(c + d*x)/(a*(sec(c + d*x) + 1))**(5/2), x)

Maxima [F]

\[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int { \frac {C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right )}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac {5}{2}}} \,d x } \]

[In]

integrate((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="maxima")

[Out]

integrate((C*sec(d*x + c)^2 + B*sec(d*x + c))/(a*sec(d*x + c) + a)^(5/2), x)

Giac [A] (verification not implemented)

none

Time = 1.96 (sec) , antiderivative size = 170, normalized size of antiderivative = 1.35 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=-\frac {\sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} {\left (\frac {2 \, \sqrt {2} {\left (B a^{5} - C a^{5}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2}}{a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )} - \frac {\sqrt {2} {\left (5 \, B a^{5} + 3 \, C a^{5}\right )}}{a^{8} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}\right )} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \frac {\sqrt {2} {\left (3 \, B + 5 \, C\right )} \log \left ({\left | -\sqrt {-a} \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right ) + \sqrt {-a \tan \left (\frac {1}{2} \, d x + \frac {1}{2} \, c\right )^{2} + a} \right |}\right )}{\sqrt {-a} a^{2} \mathrm {sgn}\left (\cos \left (d x + c\right )\right )}}{32 \, d} \]

[In]

integrate((B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(5/2),x, algorithm="giac")

[Out]

-1/32*(sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)*(2*sqrt(2)*(B*a^5 - C*a^5)*tan(1/2*d*x + 1/2*c)^2/(a^8*sgn(cos(d*x
+ c))) - sqrt(2)*(5*B*a^5 + 3*C*a^5)/(a^8*sgn(cos(d*x + c))))*tan(1/2*d*x + 1/2*c) + sqrt(2)*(3*B + 5*C)*log(a
bs(-sqrt(-a)*tan(1/2*d*x + 1/2*c) + sqrt(-a*tan(1/2*d*x + 1/2*c)^2 + a)))/(sqrt(-a)*a^2*sgn(cos(d*x + c))))/d

Mupad [F(-1)]

Timed out. \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+a \sec (c+d x))^{5/2}} \, dx=\int \frac {\frac {B}{\cos \left (c+d\,x\right )}+\frac {C}{{\cos \left (c+d\,x\right )}^2}}{{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

[In]

int((B/cos(c + d*x) + C/cos(c + d*x)^2)/(a + a/cos(c + d*x))^(5/2),x)

[Out]

int((B/cos(c + d*x) + C/cos(c + d*x)^2)/(a + a/cos(c + d*x))^(5/2), x)