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

Optimal. Leaf size=126 \[ \frac {2 a \sin (c+d x) \sqrt {\sec (c+d x)}}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {2 \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}} \]

[Out]

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

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Rubi [A]  time = 0.16, antiderivative size = 126, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {3844, 21, 3856, 2655, 2653} \[ \frac {2 a \sin (c+d x) \sqrt {\sec (c+d x)}}{d \left (a^2-b^2\right ) \sqrt {a+b \sec (c+d x)}}-\frac {2 \sqrt {a+b \sec (c+d x)} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )}{d \left (a^2-b^2\right ) \sqrt {\sec (c+d x)} \sqrt {\frac {a \cos (c+d x)+b}{a+b}}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 2653

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*Sqrt[a + b]*EllipticE[(1*(c - Pi/2 + d*x)
)/2, (2*b)/(a + b)])/d, x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 - b^2, 0] && GtQ[a + b, 0]

Rule 2655

Int[Sqrt[(a_) + (b_.)*sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Dist[Sqrt[a + b*Sin[c + d*x]]/Sqrt[(a + b*Sin[c +
d*x])/(a + b)], Int[Sqrt[a/(a + b) + (b*Sin[c + d*x])/(a + b)], x], x] /; FreeQ[{a, b, c, d}, x] && NeQ[a^2 -
b^2, 0] &&  !GtQ[a + b, 0]

Rule 3844

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Simp[(a*d^2
*Cot[e + f*x]*(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2))/(f*(m + 1)*(a^2 - b^2)), x] - Dist[d^2/((
m + 1)*(a^2 - b^2)), Int[(a + b*Csc[e + f*x])^(m + 1)*(d*Csc[e + f*x])^(n - 2)*(a*(n - 2) + b*(m + 1)*Csc[e +
f*x] - a*(m + n)*Csc[e + f*x]^2), x], x] /; FreeQ[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0] && LtQ[m, -1] && Lt
Q[1, n, 2] && IntegersQ[2*m, 2*n]

Rule 3856

Int[Sqrt[csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_)]/Sqrt[csc[(e_.) + (f_.)*(x_)]*(d_.)], x_Symbol] :> Dist[Sqrt[a +
b*Csc[e + f*x]]/(Sqrt[d*Csc[e + f*x]]*Sqrt[b + a*Sin[e + f*x]]), Int[Sqrt[b + a*Sin[e + f*x]], x], x] /; Free
Q[{a, b, d, e, f}, x] && NeQ[a^2 - b^2, 0]

Rubi steps

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

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Mathematica [A]  time = 0.32, size = 103, normalized size = 0.82 \[ -\frac {2 \sec ^{\frac {3}{2}}(c+d x) (a \cos (c+d x)+b) \left ((a+b) \sqrt {\frac {a \cos (c+d x)+b}{a+b}} E\left (\frac {1}{2} (c+d x)|\frac {2 a}{a+b}\right )-a \sin (c+d x)\right )}{d (a-b) (a+b) (a+b \sec (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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fricas [F]  time = 0.77, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {\sqrt {b \sec \left (d x + c\right ) + a} \sec \left (d x + c\right )^{\frac {3}{2}}}{b^{2} \sec \left (d x + c\right )^{2} + 2 \, a b \sec \left (d x + c\right ) + a^{2}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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maple [B]  time = 1.92, size = 501, normalized size = 3.98 \[ -\frac {2 \left (\EllipticF \left (\frac {\left (-1+\cos \left (d x +c \right )\right ) \sqrt {\frac {a -b}{a +b}}}{\sin \left (d x +c \right )}, \sqrt {-\frac {a +b}{a -b}}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \left (a +b \right )}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}-\EllipticE \left (\frac {\left (-1+\cos \left (d x +c \right )\right ) \sqrt {\frac {a -b}{a +b}}}{\sin \left (d x +c \right )}, \sqrt {-\frac {a +b}{a -b}}\right ) \cos \left (d x +c \right ) \sin \left (d x +c \right ) \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \left (a +b \right )}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}+\EllipticF \left (\frac {\left (-1+\cos \left (d x +c \right )\right ) \sqrt {\frac {a -b}{a +b}}}{\sin \left (d x +c \right )}, \sqrt {-\frac {a +b}{a -b}}\right ) \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \left (a +b \right )}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )-\EllipticE \left (\frac {\left (-1+\cos \left (d x +c \right )\right ) \sqrt {\frac {a -b}{a +b}}}{\sin \left (d x +c \right )}, \sqrt {-\frac {a +b}{a -b}}\right ) \sqrt {\frac {b +a \cos \left (d x +c \right )}{\left (1+\cos \left (d x +c \right )\right ) \left (a +b \right )}}\, \sqrt {\frac {1}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right )+\cos \left (d x +c \right ) \sqrt {\frac {a -b}{a +b}}-\sqrt {\frac {a -b}{a +b}}\right ) \sqrt {\frac {b +a \cos \left (d x +c \right )}{\cos \left (d x +c \right )}}\, \left (\frac {1}{\cos \left (d x +c \right )}\right )^{\frac {3}{2}} \left (\cos ^{2}\left (d x +c \right )\right )}{d \left (b +a \cos \left (d x +c \right )\right ) \sin \left (d x +c \right ) \sqrt {\frac {a -b}{a +b}}\, \left (a +b \right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{3/2}}{{\left (a+\frac {b}{\cos \left (c+d\,x\right )}\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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