3.180 \(\int \frac{c+d \sec (e+f x)}{(a+a \sec (e+f x))^{5/2}} \, dx\)
Optimal. Leaf size=164 \[ -\frac{(43 c-3 d) \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a}}\right )}{16 \sqrt{2} a^{5/2} f}+\frac{2 c \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{a^{5/2} f}-\frac{(11 c-3 d) \tan (e+f x)}{16 a f (a \sec (e+f x)+a)^{3/2}}-\frac{(c-d) \tan (e+f x)}{4 f (a \sec (e+f x)+a)^{5/2}} \]
[Out]
(2*c*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(a^(5/2)*f) - ((43*c - 3*d)*ArcTan[(Sqrt[a]*Tan[
e + f*x])/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]])])/(16*Sqrt[2]*a^(5/2)*f) - ((c - d)*Tan[e + f*x])/(4*f*(a + a*Sec
[e + f*x])^(5/2)) - ((11*c - 3*d)*Tan[e + f*x])/(16*a*f*(a + a*Sec[e + f*x])^(3/2))
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Rubi [A] time = 0.26493, antiderivative size = 164, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used =
{3922, 3920, 3774, 203, 3795} \[ -\frac{(43 c-3 d) \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a \sec (e+f x)+a}}\right )}{16 \sqrt{2} a^{5/2} f}+\frac{2 c \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a \sec (e+f x)+a}}\right )}{a^{5/2} f}-\frac{(11 c-3 d) \tan (e+f x)}{16 a f (a \sec (e+f x)+a)^{3/2}}-\frac{(c-d) \tan (e+f x)}{4 f (a \sec (e+f x)+a)^{5/2}} \]
Antiderivative was successfully verified.
[In]
Int[(c + d*Sec[e + f*x])/(a + a*Sec[e + f*x])^(5/2),x]
[Out]
(2*c*ArcTan[(Sqrt[a]*Tan[e + f*x])/Sqrt[a + a*Sec[e + f*x]]])/(a^(5/2)*f) - ((43*c - 3*d)*ArcTan[(Sqrt[a]*Tan[
e + f*x])/(Sqrt[2]*Sqrt[a + a*Sec[e + f*x]])])/(16*Sqrt[2]*a^(5/2)*f) - ((c - d)*Tan[e + f*x])/(4*f*(a + a*Sec
[e + f*x])^(5/2)) - ((11*c - 3*d)*Tan[e + f*x])/(16*a*f*(a + a*Sec[e + f*x])^(3/2))
Rule 3922
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] + Dist[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]
Rule 3920
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.) + (c_))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[c/a,
Int[Sqrt[a + b*Csc[e + f*x]], x], x] - Dist[(b*c - a*d)/a, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] /
; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0]
Rule 3774
Int[Sqrt[csc[(c_.) + (d_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b)/d, Subst[Int[1/(a + x^2), x], x, (b*C
ot[c + d*x])/Sqrt[a + b*Csc[c + d*x]]], x] /; FreeQ[{a, b, c, d}, x] && EqQ[a^2 - b^2, 0]
Rule 203
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])
Rule 3795
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
]
Rubi steps
\begin{align*} \int \frac{c+d \sec (e+f x)}{(a+a \sec (e+f x))^{5/2}} \, dx &=-\frac{(c-d) \tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2}}-\frac{\int \frac{-4 a c+\frac{3}{2} a (c-d) \sec (e+f x)}{(a+a \sec (e+f x))^{3/2}} \, dx}{4 a^2}\\ &=-\frac{(c-d) \tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2}}-\frac{(11 c-3 d) \tan (e+f x)}{16 a f (a+a \sec (e+f x))^{3/2}}+\frac{\int \frac{8 a^2 c-\frac{1}{4} a^2 (11 c-3 d) \sec (e+f x)}{\sqrt{a+a \sec (e+f x)}} \, dx}{8 a^4}\\ &=-\frac{(c-d) \tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2}}-\frac{(11 c-3 d) \tan (e+f x)}{16 a f (a+a \sec (e+f x))^{3/2}}+\frac{c \int \sqrt{a+a \sec (e+f x)} \, dx}{a^3}-\frac{(43 c-3 d) \int \frac{\sec (e+f x)}{\sqrt{a+a \sec (e+f x)}} \, dx}{32 a^2}\\ &=-\frac{(c-d) \tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2}}-\frac{(11 c-3 d) \tan (e+f x)}{16 a f (a+a \sec (e+f x))^{3/2}}-\frac{(2 c) \operatorname{Subst}\left (\int \frac{1}{a+x^2} \, dx,x,-\frac{a \tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{a^2 f}+\frac{(43 c-3 d) \operatorname{Subst}\left (\int \frac{1}{2 a+x^2} \, dx,x,-\frac{a \tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{16 a^2 f}\\ &=\frac{2 c \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{a+a \sec (e+f x)}}\right )}{a^{5/2} f}-\frac{(43 c-3 d) \tan ^{-1}\left (\frac{\sqrt{a} \tan (e+f x)}{\sqrt{2} \sqrt{a+a \sec (e+f x)}}\right )}{16 \sqrt{2} a^{5/2} f}-\frac{(c-d) \tan (e+f x)}{4 f (a+a \sec (e+f x))^{5/2}}-\frac{(11 c-3 d) \tan (e+f x)}{16 a f (a+a \sec (e+f x))^{3/2}}\\ \end{align*}
Mathematica [C] time = 26.8806, size = 10177, normalized size = 62.05 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(c + d*Sec[e + f*x])/(a + a*Sec[e + f*x])^(5/2),x]
[Out]
Result too large to show
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Maple [B] time = 0.185, size = 824, normalized size = 5. \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(5/2),x)
[Out]
-1/32/f/a^3*(1/cos(f*x+e)*a*(1+cos(f*x+e)))^(1/2)*(32*2^(1/2)*sin(f*x+e)*cos(f*x+e)^2*(-2*cos(f*x+e)/(1+cos(f*
x+e)))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*x+e))*c+64*2^(1/2)*sin(
f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e
)/cos(f*x+e))*c*cos(f*x+e)+43*sin(f*x+e)*cos(f*x+e)^2*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*ln(-(-(-2*cos(f*x+e
)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)+cos(f*x+e)-1)/sin(f*x+e))*c-3*sin(f*x+e)*cos(f*x+e)^2*(-2*cos(f*x+e)/(1+cos
(f*x+e)))^(1/2)*ln(-(-(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)+cos(f*x+e)-1)/sin(f*x+e))*d+32*(-2*cos(f
*x+e)/(1+cos(f*x+e)))^(1/2)*2^(1/2)*arctanh(1/2*2^(1/2)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)/cos(f*
x+e))*c*sin(f*x+e)+86*sin(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*ln(-(-(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/
2)*sin(f*x+e)+cos(f*x+e)-1)/sin(f*x+e))*c*cos(f*x+e)-6*sin(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*ln(-(-(
-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)+cos(f*x+e)-1)/sin(f*x+e))*d*cos(f*x+e)+43*sin(f*x+e)*(-2*cos(f*
x+e)/(1+cos(f*x+e)))^(1/2)*ln(-(-(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)+cos(f*x+e)-1)/sin(f*x+e))*c-3
*sin(f*x+e)*(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*ln(-(-(-2*cos(f*x+e)/(1+cos(f*x+e)))^(1/2)*sin(f*x+e)+cos(f*x
+e)-1)/sin(f*x+e))*d-30*cos(f*x+e)^3*c+14*cos(f*x+e)^3*d+8*cos(f*x+e)^2*c-8*cos(f*x+e)^2*d+22*c*cos(f*x+e)-6*d
*cos(f*x+e))/(1+cos(f*x+e))^2/sin(f*x+e)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{d \sec \left (f x + e\right ) + c}{{\left (a \sec \left (f x + e\right ) + a\right )}^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(5/2),x, algorithm="maxima")
[Out]
integrate((d*sec(f*x + e) + c)/(a*sec(f*x + e) + a)^(5/2), x)
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Fricas [B] time = 16.7046, size = 1754, normalized size = 10.7 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(5/2),x, algorithm="fricas")
[Out]
[1/64*(sqrt(2)*((43*c - 3*d)*cos(f*x + e)^3 + 3*(43*c - 3*d)*cos(f*x + e)^2 + 3*(43*c - 3*d)*cos(f*x + e) + 43
*c - 3*d)*sqrt(-a)*log((2*sqrt(2)*sqrt(-a)*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) +
3*a*cos(f*x + e)^2 + 2*a*cos(f*x + e) - a)/(cos(f*x + e)^2 + 2*cos(f*x + e) + 1)) - 64*(c*cos(f*x + e)^3 + 3*
c*cos(f*x + e)^2 + 3*c*cos(f*x + e) + c)*sqrt(-a)*log((2*a*cos(f*x + e)^2 + 2*sqrt(-a)*sqrt((a*cos(f*x + e) +
a)/cos(f*x + e))*cos(f*x + e)*sin(f*x + e) + a*cos(f*x + e) - a)/(cos(f*x + e) + 1)) - 4*((15*c - 7*d)*cos(f*x
+ e)^2 + (11*c - 3*d)*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x + e))/(a^3*f*cos(f*x + e)
^3 + 3*a^3*f*cos(f*x + e)^2 + 3*a^3*f*cos(f*x + e) + a^3*f), 1/32*(sqrt(2)*((43*c - 3*d)*cos(f*x + e)^3 + 3*(4
3*c - 3*d)*cos(f*x + e)^2 + 3*(43*c - 3*d)*cos(f*x + e) + 43*c - 3*d)*sqrt(a)*arctan(sqrt(2)*sqrt((a*cos(f*x +
e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e))) - 64*(c*cos(f*x + e)^3 + 3*c*cos(f*x + e)^2 + 3*c*
cos(f*x + e) + c)*sqrt(a)*arctan(sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*cos(f*x + e)/(sqrt(a)*sin(f*x + e)))
- 2*((15*c - 7*d)*cos(f*x + e)^2 + (11*c - 3*d)*cos(f*x + e))*sqrt((a*cos(f*x + e) + a)/cos(f*x + e))*sin(f*x
+ e))/(a^3*f*cos(f*x + e)^3 + 3*a^3*f*cos(f*x + e)^2 + 3*a^3*f*cos(f*x + e) + a^3*f)]
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Sympy [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sec(f*x+e))/(a+a*sec(f*x+e))**(5/2),x)
[Out]
Timed out
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Giac [F(-1)] time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((c+d*sec(f*x+e))/(a+a*sec(f*x+e))^(5/2),x, algorithm="giac")
[Out]
Timed out