3.531 \(\int \sec ^3(c+d x) \sqrt{a+b \sec (c+d x)} \, dx\)
Optimal. Leaf size=292 \[ \frac{2 (a-b) \sqrt{a+b} (2 a+9 b) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} \text{EllipticF}\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right ),\frac{a+b}{a-b}\right )}{15 b^2 d}+\frac{2 (a-b) \sqrt{a+b} \left (2 a^2-9 b^2\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b^3 d}+\frac{2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}-\frac{4 a \tan (c+d x) \sqrt{a+b \sec (c+d x)}}{15 b d} \]
[Out]
(2*(a - b)*Sqrt[a + b]*(2*a^2 - 9*b^2)*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))])/(15*b^3*d) + (2*(
a - b)*Sqrt[a + b]*(2*a + 9*b)*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))])/(15*b^2*d) - (4*a*Sqrt[a
+ b*Sec[c + d*x]]*Tan[c + d*x])/(15*b*d) + (2*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*b*d)
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Rubi [A] time = 0.441508, antiderivative size = 292, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.217, Rules used =
{3840, 4002, 4005, 3832, 4004} \[ \frac{2 (a-b) \sqrt{a+b} \left (2 a^2-9 b^2\right ) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} E\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b^3 d}+\frac{2 (a-b) \sqrt{a+b} (2 a+9 b) \cot (c+d x) \sqrt{\frac{b (1-\sec (c+d x))}{a+b}} \sqrt{-\frac{b (\sec (c+d x)+1)}{a-b}} F\left (\sin ^{-1}\left (\frac{\sqrt{a+b \sec (c+d x)}}{\sqrt{a+b}}\right )|\frac{a+b}{a-b}\right )}{15 b^2 d}+\frac{2 \tan (c+d x) (a+b \sec (c+d x))^{3/2}}{5 b d}-\frac{4 a \tan (c+d x) \sqrt{a+b \sec (c+d x)}}{15 b d} \]
Antiderivative was successfully verified.
[In]
Int[Sec[c + d*x]^3*Sqrt[a + b*Sec[c + d*x]],x]
[Out]
(2*(a - b)*Sqrt[a + b]*(2*a^2 - 9*b^2)*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))])/(15*b^3*d) + (2*(
a - b)*Sqrt[a + b]*(2*a + 9*b)*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))])/(15*b^2*d) - (4*a*Sqrt[a
+ b*Sec[c + d*x]]*Tan[c + d*x])/(15*b*d) + (2*(a + b*Sec[c + d*x])^(3/2)*Tan[c + d*x])/(5*b*d)
Rule 3840
Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> -Simp[(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*(b
*(m + 1) - a*Csc[e + f*x]), x], x] /; FreeQ[{a, b, e, f, m}, x] && NeQ[a^2 - b^2, 0] && !LtQ[m, -1]
Rule 4002
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[1/(m + 1), Int[Csc[e + f*x
]*(a + b*Csc[e + f*x])^(m - 1)*Simp[b*B*m + a*A*(m + 1) + (a*B*m + A*b*(m + 1))*Csc[e + f*x], x], x], x] /; Fr
eeQ[{a, b, A, B, e, f}, x] && NeQ[A*b - a*B, 0] && NeQ[a^2 - b^2, 0] && GtQ[m, 0]
Rule 4005
Int[(csc[(e_.) + (f_.)*(x_)]*(csc[(e_.) + (f_.)*(x_)]*(B_.) + (A_)))/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)
], x_Symbol] :> Dist[A - B, Int[Csc[e + f*x]/Sqrt[a + b*Csc[e + f*x]], x], x] + Dist[B, 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}, x] && NeQ[a^2 - b^2, 0] && NeQ[A
^2 - B^2, 0]
Rule 3832
Int[csc[(e_.) + (f_.)*(x_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)], x_Symbol] :> Simp[(-2*Rt[a + b, 2]*Sqr
t[(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)])/(b*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f}, x] && NeQ[a^2 - b^2, 0]
Rule 4004
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 + Cs
c[e + f*x]))/(a - b))]*EllipticE[ArcSin[Sqrt[a + b*Csc[e + f*x]]/Rt[a + (b*B)/A, 2]], (a*A + b*B)/(a*A - b*B)]
)/(b^2*f*Cot[e + f*x]), x] /; FreeQ[{a, b, e, f, A, B}, x] && NeQ[a^2 - b^2, 0] && EqQ[A^2 - B^2, 0]
Rubi steps
\begin{align*} \int \sec ^3(c+d x) \sqrt{a+b \sec (c+d x)} \, dx &=\frac{2 (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 b d}+\frac{2 \int \sec (c+d x) \left (\frac{3 b}{2}-a \sec (c+d x)\right ) \sqrt{a+b \sec (c+d x)} \, dx}{5 b}\\ &=-\frac{4 a \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{15 b d}+\frac{2 (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 b d}+\frac{4 \int \frac{\sec (c+d x) \left (\frac{7 a b}{4}-\frac{1}{4} \left (2 a^2-9 b^2\right ) \sec (c+d x)\right )}{\sqrt{a+b \sec (c+d x)}} \, dx}{15 b}\\ &=-\frac{4 a \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{15 b d}+\frac{2 (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 b d}+\frac{((a-b) (2 a+9 b)) \int \frac{\sec (c+d x)}{\sqrt{a+b \sec (c+d x)}} \, dx}{15 b}+\frac{\left (-2 a^2+9 b^2\right ) \int \frac{\sec (c+d x) (1+\sec (c+d x))}{\sqrt{a+b \sec (c+d x)}} \, dx}{15 b}\\ &=\frac{2 (a-b) \sqrt{a+b} \left (2 a^2-9 b^2\right ) \cot (c+d x) E\left (\sin ^{-1}\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}}}{15 b^3 d}+\frac{2 (a-b) \sqrt{a+b} (2 a+9 b) \cot (c+d x) F\left (\sin ^{-1}\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}}}{15 b^2 d}-\frac{4 a \sqrt{a+b \sec (c+d x)} \tan (c+d x)}{15 b d}+\frac{2 (a+b \sec (c+d x))^{3/2} \tan (c+d x)}{5 b d}\\ \end{align*}
Mathematica [A] time = 13.595, size = 401, normalized size = 1.37 \[ \frac{2 \sqrt{\cos ^2\left (\frac{1}{2} (c+d x)\right ) \sec (c+d x)} \sqrt{a+b \sec (c+d x)} \left (2 b \left (-2 a^2+7 a b+9 b^2\right ) \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sqrt{\frac{a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} \text{EllipticF}\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right ),\frac{a-b}{a+b}\right )+\left (2 a^2-9 b^2\right ) \cos (c+d x) \tan \left (\frac{1}{2} (c+d x)\right ) \sec ^2\left (\frac{1}{2} (c+d x)\right ) (a \cos (c+d x)+b)+2 \left (2 a^2 b+2 a^3-9 a b^2-9 b^3\right ) \sqrt{\frac{\cos (c+d x)}{\cos (c+d x)+1}} \sqrt{\frac{a \cos (c+d x)+b}{(a+b) (\cos (c+d x)+1)}} E\left (\sin ^{-1}\left (\tan \left (\frac{1}{2} (c+d x)\right )\right )|\frac{a-b}{a+b}\right )\right )}{15 b^2 d \sqrt{\sec ^2\left (\frac{1}{2} (c+d x)\right )} \sqrt{\sec (c+d x)} (a \cos (c+d x)+b)}+\frac{\sqrt{a+b \sec (c+d x)} \left (\frac{2 \left (9 b^2-2 a^2\right ) \sin (c+d x)}{15 b^2}+\frac{2 a \tan (c+d x)}{15 b}+\frac{2}{5} \tan (c+d x) \sec (c+d x)\right )}{d} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[Sec[c + d*x]^3*Sqrt[a + b*Sec[c + d*x]],x]
[Out]
(2*Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*Sqrt[a + b*Sec[c + d*x]]*(2*(2*a^3 + 2*a^2*b - 9*a*b^2 - 9*b^3)*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*(-2*a^2 + 7*a*b + 9*b^2)*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)] + (2*a^2 -
9*b^2)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2]))/(15*b^2*d*(b + a*Cos[c + d*x])
*Sqrt[Sec[(c + d*x)/2]^2]*Sqrt[Sec[c + d*x]]) + (Sqrt[a + b*Sec[c + d*x]]*((2*(-2*a^2 + 9*b^2)*Sin[c + d*x])/(
15*b^2) + (2*a*Tan[c + d*x])/(15*b) + (2*Sec[c + d*x]*Tan[c + d*x])/5))/d
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Maple [B] time = 0.614, size = 1582, normalized size = 5.4 \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*(a+b*sec(d*x+c))^(1/2),x)
[Out]
-2/15/d/b^2*(cos(d*x+c)+1)^2*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(-1+cos(d*x+c))^2*(-9*sin(d*x+c)*cos(d*x+c)^3
*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos
(d*x+c))/(cos(d*x+c)+1))^(1/2)*b^3+2*sin(d*x+c)*cos(d*x+c)^3*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b)
)^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^3-2*cos(d*x+c)^3*
a^2*b+9*cos(d*x+c)^3*b^3-2*cos(d*x+c)^4*a^3+2*cos(d*x+c)^3*a^3-6*cos(d*x+c)^2*b^3-3*b^3+9*sin(d*x+c)*cos(d*x+c
)^3*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*
cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b^3+2*sin(d*x+c)*cos(d*x+c)^2*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a
+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^3-9*sin(d*x+c)
*cos(d*x+c)^2*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(
a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b^3+9*sin(d*x+c)*cos(d*x+c)^2*EllipticF((-1+cos(d*x+c))/sin(d*x+c)
,((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*b^3+co
s(d*x+c)^4*a^2*b+9*cos(d*x+c)^4*a*b^2-5*cos(d*x+c)^3*a*b^2+cos(d*x+c)^2*a^2*b-4*cos(d*x+c)*a*b^2+2*sin(d*x+c)*
cos(d*x+c)^3*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a
+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b-9*sin(d*x+c)*cos(d*x+c)^3*EllipticE((-1+cos(d*x+c))/sin(d*x+c
),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a*b^2
-2*sin(d*x+c)*cos(d*x+c)^3*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1
))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b+7*sin(d*x+c)*cos(d*x+c)^3*EllipticF((-1+cos(d*x
+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1
))^(1/2)*a*b^2+2*sin(d*x+c)*cos(d*x+c)^2*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)
/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b-9*sin(d*x+c)*cos(d*x+c)^2*Ellipti
cE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))
/(cos(d*x+c)+1))^(1/2)*a*b^2-2*sin(d*x+c)*cos(d*x+c)^2*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2
))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a^2*b+7*sin(d*x+c)*cos(d*
x+c)^2*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*(cos(d*x+c)/(cos(d*x+c)+1))^(1/2)*(1/(a+b)*(b
+a*cos(d*x+c))/(cos(d*x+c)+1))^(1/2)*a*b^2)/(b+a*cos(d*x+c))/cos(d*x+c)^2/sin(d*x+c)^5
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))^(1/2),x, algorithm="maxima")
[Out]
integrate(sqrt(b*sec(d*x + c) + a)*sec(d*x + c)^3, x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{b \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{3}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))^(1/2),x, algorithm="fricas")
[Out]
integral(sqrt(b*sec(d*x + c) + a)*sec(d*x + c)^3, x)
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Sympy [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{a + b \sec{\left (c + d x \right )}} \sec ^{3}{\left (c + d x \right )}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)**3*(a+b*sec(d*x+c))**(1/2),x)
[Out]
Integral(sqrt(a + b*sec(c + d*x))*sec(c + d*x)**3, x)
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{b \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{3}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate(sec(d*x+c)^3*(a+b*sec(d*x+c))^(1/2),x, algorithm="giac")
[Out]
integrate(sqrt(b*sec(d*x + c) + a)*sec(d*x + c)^3, x)