\(\int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx\) [852]

Optimal result

Integrand size = 34, antiderivative size = 353 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {2 \left (4 a b B-a^2 C-3 b^2 C\right ) \cot (c+d x) E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b^2 (a+b)^{3/2} d}+\frac {2 (3 a B-b B+a C-3 b C) \cot (c+d x) \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{3 (a-b) b (a+b)^{3/2} d}-\frac {2 (b B-a C) \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (4 a b B-a^2 C-3 b^2 C\right ) \tan (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \]

[Out]

-2/3*(4*B*a*b-C*a^2-3*C*b^2)*cot(d*x+c)*EllipticE((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)/(a-b)/b^2/(a+b)^(3/2)/d+2/3*(3*B*a-B*b+C*a-3*C*b)*c
ot(d*x+c)*EllipticF((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)/(a-b)/b/(a+b)^(3/2)/d-2/3*(B*b-C*a)*tan(d*x+c)/(a^2-b^2)/d/(a+b*sec(d*x+c))^(3/2)
-2/3*(4*B*a*b-C*a^2-3*C*b^2)*tan(d*x+c)/(a^2-b^2)^2/d/(a+b*sec(d*x+c))^(1/2)

Rubi [A] (verified)

Time = 0.65 (sec) , antiderivative size = 353, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.147, Rules used = {4145, 4143, 12, 3917, 4089} \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=-\frac {2 \left (a^2 (-C)+4 a b B-3 b^2 C\right ) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} E\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{3 b^2 d (a-b) (a+b)^{3/2}}-\frac {2 \left (a^2 (-C)+4 a b B-3 b^2 C\right ) \tan (c+d x)}{3 d \left (a^2-b^2\right )^2 \sqrt {a+b \sec (c+d x)}}-\frac {2 (b B-a C) \tan (c+d x)}{3 d \left (a^2-b^2\right ) (a+b \sec (c+d x))^{3/2}}+\frac {2 (3 a B+a C-b B-3 b C) \cot (c+d x) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (\sec (c+d x)+1)}{a-b}} \operatorname {EllipticF}\left (\arcsin \left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right ),\frac {a+b}{a-b}\right )}{3 b d (a-b) (a+b)^{3/2}} \]

[In]

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

[Out]

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

Rule 12

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

Rule 3917

Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> 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 4089

Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(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 + C
sc[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 4143

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] + Dist[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
]

Rule 4145

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

Rubi steps \begin{align*} \text {integral}& = -\frac {2 (b B-a C) \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \int \frac {-\frac {3}{2} a (a B-b C) \sec (c+d x)+\frac {1}{2} a (b B-a C) \sec ^2(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx}{3 a \left (a^2-b^2\right )} \\ & = -\frac {2 (b B-a C) \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (4 a b B-a^2 C-3 b^2 C\right ) \tan (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {4 \int \frac {\frac {1}{4} a^2 \left (3 a^2 B+b^2 B-4 a b C\right ) \sec (c+d x)+\frac {1}{4} a^2 \left (4 a b B-a^2 C-3 b^2 C\right ) \sec ^2(c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2} \\ & = -\frac {2 (b B-a C) \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (4 a b B-a^2 C-3 b^2 C\right ) \tan (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {4 \int \frac {\left (\frac {1}{4} a^2 \left (3 a^2 B+b^2 B-4 a b C\right )-\frac {1}{4} a^2 \left (4 a b B-a^2 C-3 b^2 C\right )\right ) \sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 a^2 \left (a^2-b^2\right )^2}+\frac {\left (4 a b B-a^2 C-3 b^2 C\right ) \int \frac {\sec (c+d x) (1+\sec (c+d x))}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 \left (a^2-b^2\right )^2} \\ & = -\frac {2 \left (4 a b B-a^2 C-3 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}}}{3 (a-b) b^2 (a+b)^{3/2} d}-\frac {2 (b B-a C) \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (4 a b B-a^2 C-3 b^2 C\right ) \tan (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}}+\frac {(a (3 B+C)-b (B+3 C)) \int \frac {\sec (c+d x)}{\sqrt {a+b \sec (c+d x)}} \, dx}{3 (a-b) (a+b)^2} \\ & = -\frac {2 \left (4 a b B-a^2 C-3 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}}}{3 (a-b) b^2 (a+b)^{3/2} d}+\frac {2 (a (3 B+C)-b (B+3 C)) \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}}}{3 (a-b) b (a+b)^{3/2} d}-\frac {2 (b B-a C) \tan (c+d x)}{3 \left (a^2-b^2\right ) d (a+b \sec (c+d x))^{3/2}}-\frac {2 \left (4 a b B-a^2 C-3 b^2 C\right ) \tan (c+d x)}{3 \left (a^2-b^2\right )^2 d \sqrt {a+b \sec (c+d x)}} \\ \end{align*}

Mathematica [A] (warning: unable to verify)

Time = 18.43 (sec) , antiderivative size = 559, normalized size of antiderivative = 1.58 \[ \int \frac {B \sec (c+d x)+C \sec ^2(c+d x)}{(a+b \sec (c+d x))^{5/2}} \, dx=\frac {(b+a \cos (c+d x))^3 \sec ^3(c+d x) \left (-\frac {2 \left (-4 a b B+a^2 C+3 b^2 C\right ) \sin (c+d x)}{3 b \left (-a^2+b^2\right )^2}+\frac {2 \left (b^2 B \sin (c+d x)-a b C \sin (c+d x)\right )}{3 a \left (a^2-b^2\right ) (b+a \cos (c+d x))^2}+\frac {2 \left (-5 a^2 b B \sin (c+d x)+b^3 B \sin (c+d x)+2 a^3 C \sin (c+d x)+2 a b^2 C \sin (c+d x)\right )}{3 a \left (a^2-b^2\right )^2 (b+a \cos (c+d x))}\right )}{d (a+b \sec (c+d x))^{5/2}}+\frac {2 (b+a \cos (c+d x))^2 \sec ^{\frac {5}{2}}(c+d x) \sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 (a+b) \left (-4 a b B+a^2 C+3 b^2 C\right ) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} E\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+2 b (a+b) (3 a B+b B-a C-3 b C) \sqrt {\frac {\cos (c+d x)}{1+\cos (c+d x)}} \sqrt {\frac {b+a \cos (c+d x)}{(a+b) (1+\cos (c+d x))}} \operatorname {EllipticF}\left (\arcsin \left (\tan \left (\frac {1}{2} (c+d x)\right )\right ),\frac {a-b}{a+b}\right )+\left (-4 a b B+a^2 C+3 b^2 C\right ) \cos (c+d x) (b+a \cos (c+d x)) \sec ^2\left (\frac {1}{2} (c+d x)\right ) \tan \left (\frac {1}{2} (c+d x)\right )\right )}{3 b \left (a^2-b^2\right )^2 d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} (a+b \sec (c+d x))^{5/2}} \]

[In]

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

[Out]

((b + a*Cos[c + d*x])^3*Sec[c + d*x]^3*((-2*(-4*a*b*B + a^2*C + 3*b^2*C)*Sin[c + d*x])/(3*b*(-a^2 + b^2)^2) +
(2*(b^2*B*Sin[c + d*x] - a*b*C*Sin[c + d*x]))/(3*a*(a^2 - b^2)*(b + a*Cos[c + d*x])^2) + (2*(-5*a^2*b*B*Sin[c
+ d*x] + b^3*B*Sin[c + d*x] + 2*a^3*C*Sin[c + d*x] + 2*a*b^2*C*Sin[c + d*x]))/(3*a*(a^2 - b^2)^2*(b + a*Cos[c
+ d*x]))))/(d*(a + b*Sec[c + d*x])^(5/2)) + (2*(b + a*Cos[c + d*x])^2*Sec[c + d*x]^(5/2)*Sqrt[Cos[(c + d*x)/2]
^2*Sec[c + d*x]]*(2*(a + b)*(-4*a*b*B + a^2*C + 3*b^2*C)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos
[c + d*x])/((a + b)*(1 + Cos[c + d*x]))]*EllipticE[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + 2*b*(a + b)*(3
*a*B + b*B - a*C - 3*b*C)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sqrt[(b + a*Cos[c + d*x])/((a + b)*(1 + Cos[c
+ d*x]))]*EllipticF[ArcSin[Tan[(c + d*x)/2]], (a - b)/(a + b)] + (-4*a*b*B + a^2*C + 3*b^2*C)*Cos[c + d*x]*(b
+ a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(3*b*(a^2 - b^2)^2*d*Sqrt[Sec[(c + d*x)/2]^2]*(a + b*S
ec[c + d*x])^(5/2))

Maple [B] (verified)

Leaf count of result is larger than twice the leaf count of optimal. \(5261\) vs. \(2(323)=646\).

Time = 11.62 (sec) , antiderivative size = 5262, normalized size of antiderivative = 14.91

method	result	size
parts	\(\text {Expression too large to display}\)	\(5262\)
default	\(\text {Expression too large to display}\)	\(5303\)

[In]

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

[Out]

result too large to display

Fricas [F]

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

[In]

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

[Out]

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

Sympy [F]

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

[In]

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

[Out]

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

Maxima [F(-1)]

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

[In]

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

[Out]

Timed out

Giac [F]

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

[In]

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

[Out]

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

Mupad [F(-1)]

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

[In]

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

[Out]

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