∫ sec3 (c+dx)___ (a+b sec(c+dx))3∕2 dx [564]

3.6.64 \(\int \frac {\sec ^3(c+d x)}{(a+b \sec (c+d x))^{3/2}} \, dx\) [564]

Optimal. Leaf size=257 \[ -\frac {2 \left (2 a^2-b^2\right ) \cot (c+d x) E\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{b^3 \sqrt {a+b} d}-\frac {2 (2 a+b) \cot (c+d x) F\left (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right ) \sqrt {\frac {b (1-\sec (c+d x))}{a+b}} \sqrt {-\frac {b (1+\sec (c+d x))}{a-b}}}{b^2 \sqrt {a+b} d}-\frac {2 a^2 \tan (c+d x)}{b \left (a^2-b^2\right ) d \sqrt {a+b \sec (c+d x)}} \]

[Out]

-2*(2*a^2-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)/b^3/d/(a+b)^(1/2)-2*(2*a+b)*cot(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)/b^2/d/(

a+b)^(1/2)-2*a^2*tan(d*x+c)/b/(a^2-b^2)/d/(a+b*sec(d*x+c))^(1/2)

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Rubi [A]
time = 0.21, antiderivative size = 257, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3924, 4090, 3917, 4089} \begin {gather*} -\frac {2 \left (2 a^2-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 (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^3 d \sqrt {a+b}}-\frac {2 a^2 \tan (c+d x)}{b d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {2 (2 a+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 (\text {ArcSin}\left (\frac {\sqrt {a+b \sec (c+d x)}}{\sqrt {a+b}}\right )|\frac {a+b}{a-b}\right )}{b^2 d \sqrt {a+b}} \end {gather*}

Antiderivative was successfully veriﬁed.

[In]

Int[Sec[c + d*x]^3/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

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

*Sqrt[a + b*Sec[c + d*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 3924

Int[csc[(e_.) + (f_.)*(x_)]^3*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(-a^2)*Cot[e + f*

x]*((a + b*Csc[e + f*x])^(m + 1)/(b*f*(m + 1)*(a^2 - b^2))), x] + Dist[1/(b*(m + 1)*(a^2 - b^2)), Int[Csc[e +

f*x]*(a + b*Csc[e + f*x])^(m + 1)*Simp[a*b*(m + 1) - (a^2 + b^2*(m + 1))*Csc[e + f*x], x], x], x] /; FreeQ[{a,

b, e, f}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1]

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 4090

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]

Rubi steps

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

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Mathematica [A]
time = 8.96, size = 395, normalized size = 1.54 \begin {gather*} -\frac {2 (b+a \cos (c+d x)) \sec ^{\frac {3}{2}}(c+d x) \left (\left (-a^2 b+b^3+\left (-2 a^3+a b^2\right ) \cos (c+d x)\right ) \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} \sqrt {\sec (c+d x)} \sin (c+d x)+\sqrt {\cos ^2\left (\frac {1}{2} (c+d x)\right ) \sec (c+d x)} \left (2 \left (2 a^3+2 a^2 b-a b^2-b^3\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 (\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+2 b \left (-2 a^2-a b+b^2\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))}} F\left (\text {ArcSin}\left (\tan \left (\frac {1}{2} (c+d x)\right )\right )|\frac {a-b}{a+b}\right )+\left (2 a^2-b^2\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 )\right )}{b^2 \left (a^2-b^2\right ) d \sqrt {\sec ^2\left (\frac {1}{2} (c+d x)\right )} (a+b \sec (c+d x))^{3/2}} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Sec[c + d*x]^3/(a + b*Sec[c + d*x])^(3/2),x]

[Out]

(-2*(b + a*Cos[c + d*x])*Sec[c + d*x]^(3/2)*((-(a^2*b) + b^3 + (-2*a^3 + a*b^2)*Cos[c + d*x])*Sqrt[Sec[(c + d*

x)/2]^2]*Sqrt[Sec[c + d*x]]*Sin[c + d*x] + Sqrt[Cos[(c + d*x)/2]^2*Sec[c + d*x]]*(2*(2*a^3 + 2*a^2*b - a*b^2 -

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 - a*b + b^2)*Sqrt[Cos[c + d*x]/(1 + Cos[c + d*x])]*Sq

rt[(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 - b^2)*Cos[c + d*x]*(b + a*Cos[c + d*x])*Sec[(c + d*x)/2]^2*Tan[(c + d*x)/2])))/(b^2*(a^2 - b^2)*d*Sqrt[

Sec[(c + d*x)/2]^2]*(a + b*Sec[c + d*x])^(3/2))

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(1450\) vs. \(2(237)=474\).
time = 0.22, size = 1451, normalized size = 5.65


method	result	size

default	\(\text {Expression too large to display}\)	\(1451\)

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(sec(d*x+c)^3/(a+b*sec(d*x+c))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/d*4^(1/2)*((b+a*cos(d*x+c))/cos(d*x+c))^(1/2)*(-2*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(

(b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3-2*s

in(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE

((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b+sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)

*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2

+sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ellipti

cE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3+2*sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/

2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2

*b+sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*Ellip

ticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2-sin(d*x+c)*cos(d*x+c)*(cos(d*x+c)/(1+cos(d*x+c)))^(

1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b

^3-2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticE((-1+cos(d*x+c))

/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^3*sin(d*x+c)-2*EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^

2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*sin(d*x+c)*b+EllipticE((-1+c

os(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+

c))/(a+b))^(1/2)*sin(d*x+c)*a+(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*

EllipticE((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3*sin(d*x+c)+2*(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*(

(b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a^2*b*s

in(d*x+c)+(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*

x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*a*b^2*sin(d*x+c)-(cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*((b+a*cos(d*x+c))/(1+

cos(d*x+c))/(a+b))^(1/2)*EllipticF((-1+cos(d*x+c))/sin(d*x+c),((a-b)/(a+b))^(1/2))*b^3*sin(d*x+c)+2*cos(d*x+c)

^2*a^3-cos(d*x+c)^2*a^2*b-cos(d*x+c)^2*a*b^2-2*cos(d*x+c)*a^3+2*cos(d*x+c)*a^2*b+cos(d*x+c)*a*b^2-cos(d*x+c)*b

^3-b*a^2+b^3)/(b+a*cos(d*x+c))/sin(d*x+c)/b^2/(a+b)/(a-b)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+b*sec(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate(sec(d*x + c)^3/(b*sec(d*x + c) + a)^(3/2), x)

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Fricas [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+b*sec(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

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

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\sec ^{3}{\left (c + d x \right )}}{\left (a + b \sec {\left (c + d x \right )}\right )^{\frac {3}{2}}}\, dx \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)**3/(a+b*sec(d*x+c))**(3/2),x)

[Out]

Integral(sec(c + d*x)**3/(a + b*sec(c + d*x))**(3/2), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(d*x+c)^3/(a+b*sec(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(sec(d*x + c)^3/(b*sec(d*x + c) + a)^(3/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {1}{{\cos \left (c+d\,x\right )}^3\,{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \end {gather*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/(cos(c + d*x)^3*(a + b/cos(c + d*x))^(3/2)),x)

[Out]

int(1/(cos(c + d*x)^3*(a + b/cos(c + d*x))^(3/2)), x)

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