3.614 \(\int \frac{\sec ^{\frac{3}{2}}(c+d x) (A+B \sec (c+d x)+C \sec ^2(c+d x))}{(a+a \sec (c+d x))^{3/2}} \, dx\)
Optimal. Leaf size=202 \[ \frac{(A-5 B+9 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(2 B-3 C) \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{a^{3/2} d}-\frac{(A-B+C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}+\frac{(A-B+3 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{2 a d \sqrt{a \sec (c+d x)+a}} \]
[Out]
((2*B - 3*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(3/2)*d) + ((A - 5*B + 9*C)*ArcTanh[
(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((A - B
+ C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) + ((A - B + 3*C)*Sec[c + d*x]^(3/2)*Si
n[c + d*x])/(2*a*d*Sqrt[a + a*Sec[c + d*x]])
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Rubi [A] time = 0.603515, antiderivative size = 202, normalized size of antiderivative = 1.,
number of steps used = 7, number of rules used = 7, integrand size = 45, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.156, Rules used =
{4084, 4021, 4023, 3808, 206, 3801, 215} \[ \frac{(A-5 B+9 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sin (c+d x) \sqrt{\sec (c+d x)}}{\sqrt{2} \sqrt{a \sec (c+d x)+a}}\right )}{2 \sqrt{2} a^{3/2} d}+\frac{(2 B-3 C) \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a \sec (c+d x)+a}}\right )}{a^{3/2} d}-\frac{(A-B+C) \sin (c+d x) \sec ^{\frac{5}{2}}(c+d x)}{2 d (a \sec (c+d x)+a)^{3/2}}+\frac{(A-B+3 C) \sin (c+d x) \sec ^{\frac{3}{2}}(c+d x)}{2 a d \sqrt{a \sec (c+d x)+a}} \]
Antiderivative was successfully verified.
[In]
Int[(Sec[c + d*x]^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(3/2),x]
[Out]
((2*B - 3*C)*ArcSinh[(Sqrt[a]*Tan[c + d*x])/Sqrt[a + a*Sec[c + d*x]]])/(a^(3/2)*d) + ((A - 5*B + 9*C)*ArcTanh[
(Sqrt[a]*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/(Sqrt[2]*Sqrt[a + a*Sec[c + d*x]])])/(2*Sqrt[2]*a^(3/2)*d) - ((A - B
+ C)*Sec[c + d*x]^(5/2)*Sin[c + d*x])/(2*d*(a + a*Sec[c + d*x])^(3/2)) + ((A - B + 3*C)*Sec[c + d*x]^(3/2)*Si
n[c + d*x])/(2*a*d*Sqrt[a + a*Sec[c + d*x]])
Rule 4084
Int[((A_.) + csc[(e_.) + (f_.)*(x_)]*(B_.) + csc[(e_.) + (f_.)*(x_)]^2*(C_.))*(csc[(e_.) + (f_.)*(x_)]*(d_.))^
(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[((a*A - b*B + a*C)*Cot[e + f*x]*(a + b*Cs
c[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[a*B*n - b*C*n - A*b*(2*m + n + 1) - (b*B*(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, B, C, n}, x] && EqQ[a^2 - b^2, 0] && LtQ[m, -2^(-1)]
Rule 4021
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 4023
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_)*(csc[(e_.) + (f_.)*(x_)]*
(B_.) + (A_)), x_Symbol] :> Dist[(A*b - a*B)/b, Int[(a + b*Csc[e + f*x])^m*(d*Csc[e + f*x])^n, x], x] + Dist[B
/b, Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, A, B, m}, x] && NeQ[A
*b - a*B, 0] && EqQ[a^2 - b^2, 0]
Rule 3808
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*b*d)
/(a*f), Subst[Int[1/(2*b - d*x^2), x], x, (b*Cot[e + f*x])/(Sqrt[a + b*Csc[e + f*x]]*Sqrt[d*Csc[e + f*x]])], x
] /; FreeQ[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0]
Rule 206
Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTanh[(Rt[-b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[-b, 2]), x]
/; FreeQ[{a, b}, x] && NegQ[a/b] && (GtQ[a, 0] || LtQ[b, 0])
Rule 3801
Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)]*Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Dist[(-2*a*Sq
rt[(a*d)/b])/(b*f), Subst[Int[1/Sqrt[1 + x^2/a], x], x, (b*Cot[e + f*x])/Sqrt[a + b*Csc[e + f*x]]], x] /; Free
Q[{a, b, d, e, f}, x] && EqQ[a^2 - b^2, 0] && GtQ[(a*d)/b, 0]
Rule 215
Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]
Rubi steps
\begin{align*} \int \frac{\sec ^{\frac{3}{2}}(c+d x) \left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right )}{(a+a \sec (c+d x))^{3/2}} \, dx &=-\frac{(A-B+C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{\int \frac{\sec ^{\frac{3}{2}}(c+d x) \left (\frac{1}{2} a (A+3 B-3 C)+a (A-B+3 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{2 a^2}\\ &=-\frac{(A-B+C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(A-B+3 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 a d \sqrt{a+a \sec (c+d x)}}+\frac{\int \frac{\sqrt{\sec (c+d x)} \left (\frac{1}{2} a^2 (A-B+3 C)+a^2 (2 B-3 C) \sec (c+d x)\right )}{\sqrt{a+a \sec (c+d x)}} \, dx}{2 a^3}\\ &=-\frac{(A-B+C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(A-B+3 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 a d \sqrt{a+a \sec (c+d x)}}+\frac{(2 B-3 C) \int \sqrt{\sec (c+d x)} \sqrt{a+a \sec (c+d x)} \, dx}{2 a^2}+\frac{(A-5 B+9 C) \int \frac{\sqrt{\sec (c+d x)}}{\sqrt{a+a \sec (c+d x)}} \, dx}{4 a}\\ &=-\frac{(A-B+C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(A-B+3 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 a d \sqrt{a+a \sec (c+d x)}}-\frac{(2 B-3 C) \operatorname{Subst}\left (\int \frac{1}{\sqrt{1+\frac{x^2}{a}}} \, dx,x,-\frac{a \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{a^2 d}-\frac{(A-5 B+9 C) \operatorname{Subst}\left (\int \frac{1}{2 a-x^2} \, dx,x,-\frac{a \sqrt{\sec (c+d x)} \sin (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{2 a d}\\ &=\frac{(2 B-3 C) \sinh ^{-1}\left (\frac{\sqrt{a} \tan (c+d x)}{\sqrt{a+a \sec (c+d x)}}\right )}{a^{3/2} d}+\frac{(A-5 B+9 C) \tanh ^{-1}\left (\frac{\sqrt{a} \sqrt{\sec (c+d x)} \sin (c+d x)}{\sqrt{2} \sqrt{a+a \sec (c+d x)}}\right )}{2 \sqrt{2} a^{3/2} d}-\frac{(A-B+C) \sec ^{\frac{5}{2}}(c+d x) \sin (c+d x)}{2 d (a+a \sec (c+d x))^{3/2}}+\frac{(A-B+3 C) \sec ^{\frac{3}{2}}(c+d x) \sin (c+d x)}{2 a d \sqrt{a+a \sec (c+d x)}}\\ \end{align*}
Mathematica [A] time = 2.1853, size = 177, normalized size = 0.88 \[ \frac{\left (A+B \sec (c+d x)+C \sec ^2(c+d x)\right ) \left (\tan \left (\frac{1}{2} (c+d x)\right ) (A-B+2 C \sec (c+d x)+3 C)+(A-5 B+9 C) \cos \left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sin \left (\frac{1}{2} (c+d x)\right )\right )+2 \sqrt{2} (2 B-3 C) \cos \left (\frac{1}{2} (c+d x)\right ) \tanh ^{-1}\left (\sqrt{2} \sin \left (\frac{1}{2} (c+d x)\right )\right )\right )}{a d \sec ^{\frac{3}{2}}(c+d x) \sqrt{a (\sec (c+d x)+1)} (A \cos (2 (c+d x))+A+2 B \cos (c+d x)+2 C)} \]
Antiderivative was successfully verified.
[In]
Integrate[(Sec[c + d*x]^(3/2)*(A + B*Sec[c + d*x] + C*Sec[c + d*x]^2))/(a + a*Sec[c + d*x])^(3/2),x]
[Out]
((A + B*Sec[c + d*x] + C*Sec[c + d*x]^2)*((A - 5*B + 9*C)*ArcTanh[Sin[(c + d*x)/2]]*Cos[(c + d*x)/2] + 2*Sqrt[
2]*(2*B - 3*C)*ArcTanh[Sqrt[2]*Sin[(c + d*x)/2]]*Cos[(c + d*x)/2] + (A - B + 3*C + 2*C*Sec[c + d*x])*Tan[(c +
d*x)/2]))/(a*d*(A + 2*C + 2*B*Cos[c + d*x] + A*Cos[2*(c + d*x)])*Sec[c + d*x]^(3/2)*Sqrt[a*(1 + Sec[c + d*x])]
)
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Maple [B] time = 0.389, size = 561, normalized size = 2.8 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int(sec(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x)
[Out]
1/4/d/a^2*(-2*B*2^(1/2)*sin(d*x+c)*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))*cos
(d*x+c)+2*B*2^(1/2)*sin(d*x+c)*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))*cos(d*x
+c)+3*C*sin(d*x+c)*2^(1/2)*cos(d*x+c)*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1-sin(d*x+c)))-
3*C*sin(d*x+c)*2^(1/2)*cos(d*x+c)*arctan(1/4*2^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)*(cos(d*x+c)+1+sin(d*x+c)))+A*si
n(d*x+c)*cos(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))-A*cos(d*x+c)^2*(-2/(cos(d*x+c)+1))^(1/2)-
5*B*sin(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)+B*(-2/(cos(d*x+c)+1))^(1/2)*cos(d*x
+c)^2+9*C*sin(d*x+c)*cos(d*x+c)*arctan(1/2*sin(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2))-3*C*cos(d*x+c)^2*(-2/(cos(d*x
+c)+1))^(1/2)+A*cos(d*x+c)*(-2/(cos(d*x+c)+1))^(1/2)-B*(-2/(cos(d*x+c)+1))^(1/2)*cos(d*x+c)+C*cos(d*x+c)*(-2/(
cos(d*x+c)+1))^(1/2)+2*C*(-2/(cos(d*x+c)+1))^(1/2))*cos(d*x+c)*(1/cos(d*x+c))^(3/2)*(a*(cos(d*x+c)+1)/cos(d*x+
c))^(1/2)*(-2/(cos(d*x+c)+1))^(1/2)/sin(d*x+c)^3*(cos(d*x+c)^2-1)
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Maxima [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(sec(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="maxima")
[Out]
Timed out
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Fricas [A] time = 1.03804, size = 1793, normalized size = 8.88 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="fricas")
[Out]
[1/8*(sqrt(2)*((A - 5*B + 9*C)*cos(d*x + c)^2 + 2*(A - 5*B + 9*C)*cos(d*x + c) + A - 5*B + 9*C)*sqrt(a)*log(-(
a*cos(d*x + c)^2 - 2*sqrt(2)*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))*sin(d*x + c) -
2*a*cos(d*x + c) - 3*a)/(cos(d*x + c)^2 + 2*cos(d*x + c) + 1)) - 2*((2*B - 3*C)*cos(d*x + c)^2 + 2*(2*B - 3*C
)*cos(d*x + c) + 2*B - 3*C)*sqrt(a)*log((a*cos(d*x + c)^3 - 7*a*cos(d*x + c)^2 + 4*(cos(d*x + c)^2 - 2*cos(d*x
+ c))*sqrt(a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)) + 8*a)/(cos(d*x + c)^3
+ cos(d*x + c)^2)) + 4*((A - B + 3*C)*cos(d*x + c) + 2*C)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sin(d*x + c)
/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2*d), -1/4*(sqrt(2)*((A - 5*B + 9*C)*cos
(d*x + c)^2 + 2*(A - 5*B + 9*C)*cos(d*x + c) + A - 5*B + 9*C)*sqrt(-a)*arctan(sqrt(2)*sqrt(-a)*sqrt((a*cos(d*x
+ c) + a)/cos(d*x + c))*sqrt(cos(d*x + c))/(a*sin(d*x + c))) - 2*((2*B - 3*C)*cos(d*x + c)^2 + 2*(2*B - 3*C)*
cos(d*x + c) + 2*B - 3*C)*sqrt(-a)*arctan(2*sqrt(-a)*sqrt((a*cos(d*x + c) + a)/cos(d*x + c))*sqrt(cos(d*x + c)
)*sin(d*x + c)/(a*cos(d*x + c)^2 - a*cos(d*x + c) - 2*a)) - 2*((A - B + 3*C)*cos(d*x + c) + 2*C)*sqrt((a*cos(d
*x + c) + a)/cos(d*x + c))*sin(d*x + c)/sqrt(cos(d*x + c)))/(a^2*d*cos(d*x + c)^2 + 2*a^2*d*cos(d*x + c) + a^2
*d)]
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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(sec(d*x+c)**(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)**2)/(a+a*sec(d*x+c))**(3/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{{\left (C \sec \left (d x + c\right )^{2} + B \sec \left (d x + c\right ) + A\right )} \sec \left (d x + c\right )^{\frac{3}{2}}}{{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^(3/2)*(A+B*sec(d*x+c)+C*sec(d*x+c)^2)/(a+a*sec(d*x+c))^(3/2),x, algorithm="giac")
[Out]
integrate((C*sec(d*x + c)^2 + B*sec(d*x + c) + A)*sec(d*x + c)^(3/2)/(a*sec(d*x + c) + a)^(3/2), x)