3.183 \(\int \frac {(a+a \sec (c+d x))^3}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\)

Optimal. Leaf size=131 \[ \frac {2 a^3 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a^3 \sin (c+d x)}{d \sqrt {\sec (c+d x)}}+\frac {4 a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {36 a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d} \]

[Out]

2/5*a^3*sin(d*x+c)/d/sec(d*x+c)^(3/2)+2*a^3*sin(d*x+c)/d/sec(d*x+c)^(1/2)+36/5*a^3*(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))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)/d+4*a^3*(cos(1/2*
d*x+1/2*c)^2)^(1/2)/cos(1/2*d*x+1/2*c)*EllipticF(sin(1/2*d*x+1/2*c),2^(1/2))*cos(d*x+c)^(1/2)*sec(d*x+c)^(1/2)
/d

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Rubi [A]  time = 0.14, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 12, number of rules used = 5, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.217, Rules used = {3791, 3769, 3771, 2639, 2641} \[ \frac {2 a^3 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a^3 \sin (c+d x)}{d \sqrt {\sec (c+d x)}}+\frac {4 a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{d}+\frac {36 a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^3/Sec[c + d*x]^(5/2),x]

[Out]

(36*a^3*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (4*a^3*Sqrt[Cos[c + d*x]]*Ell
ipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/d + (2*a^3*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (2*a^3*Sin[c +
d*x])/(d*Sqrt[Sec[c + d*x]])

Rule 2639

Int[Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticE[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ[{
c, d}, x]

Rule 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rule 3769

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Simp[(Cos[c + d*x]*(b*Csc[c + d*x])^(n + 1))/(b*d*n), x
] + Dist[(n + 1)/(b^2*n), Int[(b*Csc[c + d*x])^(n + 2), x], x] /; FreeQ[{b, c, d}, x] && LtQ[n, -1] && Integer
Q[2*n]

Rule 3771

Int[(csc[(c_.) + (d_.)*(x_)]*(b_.))^(n_), x_Symbol] :> Dist[(b*Csc[c + d*x])^n*Sin[c + d*x]^n, Int[1/Sin[c + d
*x]^n, x], x] /; FreeQ[{b, c, d}, x] && EqQ[n^2, 1/4]

Rule 3791

Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Int[Expand
Trig[(a + b*csc[e + f*x])^m*(d*csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0]
&& IGtQ[m, 0] && RationalQ[n]

Rubi steps

\begin {align*} \int \frac {(a+a \sec (c+d x))^3}{\sec ^{\frac {5}{2}}(c+d x)} \, dx &=\int \left (\frac {a^3}{\sec ^{\frac {5}{2}}(c+d x)}+\frac {3 a^3}{\sec ^{\frac {3}{2}}(c+d x)}+\frac {3 a^3}{\sqrt {\sec (c+d x)}}+a^3 \sqrt {\sec (c+d x)}\right ) \, dx\\ &=a^3 \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x)} \, dx+a^3 \int \sqrt {\sec (c+d x)} \, dx+\left (3 a^3\right ) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx+\left (3 a^3\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 a^3 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a^3 \sin (c+d x)}{d \sqrt {\sec (c+d x)}}+\frac {1}{5} \left (3 a^3\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+a^3 \int \sqrt {\sec (c+d x)} \, dx+\left (a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+\left (3 a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=\frac {6 a^3 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {2 a^3 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {2 a^3 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a^3 \sin (c+d x)}{d \sqrt {\sec (c+d x)}}+\frac {1}{5} \left (3 a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\left (a^3 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {36 a^3 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {4 a^3 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {2 a^3 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a^3 \sin (c+d x)}{d \sqrt {\sec (c+d x)}}\\ \end {align*}

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Mathematica [C]  time = 1.07, size = 171, normalized size = 1.31 \[ \frac {a^3 \left (\cos \left (\frac {c}{2}\right )-i \sin \left (\frac {c}{2}\right )\right ) \left (\cos \left (\frac {c}{2}\right )+i \sin \left (\frac {c}{2}\right )\right ) \left (\frac {144 i \, _2F_1\left (-\frac {1}{4},\frac {1}{2};\frac {3}{4};-e^{2 i (c+d x)}\right )}{\sqrt {1+e^{2 i (c+d x)}}}+2 \left (-20 i \sqrt {1+e^{2 i (c+d x)}} \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-e^{2 i (c+d x)}\right ) \sec (c+d x)+10 \sin (c+d x)+\sin (2 (c+d x))-36 i\right )\right )}{10 d \sqrt {\sec (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^3/Sec[c + d*x]^(5/2),x]

[Out]

(a^3*(Cos[c/2] - I*Sin[c/2])*(Cos[c/2] + I*Sin[c/2])*(((144*I)*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^((2*I)*(c
+ d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] + 2*(-36*I - (20*I)*Sqrt[1 + E^((2*I)*(c + d*x))]*Hypergeometric2F1[1/
4, 1/2, 5/4, -E^((2*I)*(c + d*x))]*Sec[c + d*x] + 10*Sin[c + d*x] + Sin[2*(c + d*x)])))/(10*d*Sqrt[Sec[c + d*x
]])

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fricas [F]  time = 1.02, size = 0, normalized size = 0.00 \[ {\rm integral}\left (\frac {a^{3} \sec \left (d x + c\right )^{3} + 3 \, a^{3} \sec \left (d x + c\right )^{2} + 3 \, a^{3} \sec \left (d x + c\right ) + a^{3}}{\sec \left (d x + c\right )^{\frac {5}{2}}}, x\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3/sec(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

integral((a^3*sec(d*x + c)^3 + 3*a^3*sec(d*x + c)^2 + 3*a^3*sec(d*x + c) + a^3)/sec(d*x + c)^(5/2), x)

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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{3}}{\sec \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^3/sec(d*x + c)^(5/2), x)

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maple [A]  time = 3.76, size = 250, normalized size = 1.91 \[ -\frac {4 \sqrt {\left (2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1\right ) \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )}\, a^{3} \left (-4 \cos \left (\frac {d x}{2}+\frac {c}{2}\right ) \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+14 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+5 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-9 \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )-6 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )\right )}{5 \sqrt {-2 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )+\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )}\, \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^3/sec(d*x+c)^(5/2),x)

[Out]

-4/5*((2*cos(1/2*d*x+1/2*c)^2-1)*sin(1/2*d*x+1/2*c)^2)^(1/2)*a^3*(-4*cos(1/2*d*x+1/2*c)*sin(1/2*d*x+1/2*c)^6+1
4*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+5*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*Elli
pticF(cos(1/2*d*x+1/2*c),2^(1/2))-9*(sin(1/2*d*x+1/2*c)^2)^(1/2)*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*EllipticE(co
s(1/2*d*x+1/2*c),2^(1/2))-6*sin(1/2*d*x+1/2*c)^2*cos(1/2*d*x+1/2*c))/(-2*sin(1/2*d*x+1/2*c)^4+sin(1/2*d*x+1/2*
c)^2)^(1/2)/sin(1/2*d*x+1/2*c)/(2*cos(1/2*d*x+1/2*c)^2-1)^(1/2)/d

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sec \left (d x + c\right ) + a\right )}^{3}}{\sec \left (d x + c\right )^{\frac {5}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^3/sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^3/sec(d*x + c)^(5/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^3}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^3/(1/cos(c + d*x))^(5/2),x)

[Out]

int((a + a/cos(c + d*x))^3/(1/cos(c + d*x))^(5/2), x)

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sympy [F]  time = 0.00, size = 0, normalized size = 0.00 \[ a^{3} \left (\int \frac {1}{\sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \frac {3}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {3}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int \sqrt {\sec {\left (c + d x \right )}}\, dx\right ) \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**3/sec(d*x+c)**(5/2),x)

[Out]

a**3*(Integral(sec(c + d*x)**(-5/2), x) + Integral(3/sec(c + d*x)**(3/2), x) + Integral(3/sqrt(sec(c + d*x)),
x) + Integral(sqrt(sec(c + d*x)), x))

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