3.2.90 \(\int \frac {(a+a \sec (c+d x))^4}{\sec ^{\frac {5}{2}}(c+d x)} \, dx\) [190]

Optimal. Leaf size=159 \[ \frac {56 a^4 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{5 d}+\frac {32 a^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 a^4 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {8 a^4 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 a^4 \sqrt {\sec (c+d x)} \sin (c+d x)}{d} \]

[Out]

2/5*a^4*sin(d*x+c)/d/sec(d*x+c)^(3/2)+8/3*a^4*sin(d*x+c)/d/sec(d*x+c)^(1/2)+2*a^4*sin(d*x+c)*sec(d*x+c)^(1/2)/
d+56/5*a^4*(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+32/3*a^4*(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 = 0.15, antiderivative size = 159, normalized size of antiderivative = 1.00, number of steps used = 15, number of rules used = 6, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.261, Rules used = {3876, 3854, 3856, 2719, 2720, 3853} \begin {gather*} \frac {2 a^4 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {2 a^4 \sin (c+d x) \sqrt {\sec (c+d x)}}{d}+\frac {8 a^4 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {32 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{3 d}+\frac {56 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right )}{5 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + a*Sec[c + d*x])^4/Sec[c + d*x]^(5/2),x]

[Out]

(56*a^4*Sqrt[Cos[c + d*x]]*EllipticE[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(5*d) + (32*a^4*Sqrt[Cos[c + d*x]]*El
lipticF[(c + d*x)/2, 2]*Sqrt[Sec[c + d*x]])/(3*d) + (2*a^4*Sin[c + d*x])/(5*d*Sec[c + d*x]^(3/2)) + (8*a^4*Sin
[c + d*x])/(3*d*Sqrt[Sec[c + d*x]]) + (2*a^4*Sqrt[Sec[c + d*x]]*Sin[c + d*x])/d

Rule 2719

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

Rule 2720

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

Rule 3853

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

Rule 3854

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 3856

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 3876

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))^4}{\sec ^{\frac {5}{2}}(c+d x)} \, dx &=\int \left (\frac {a^4}{\sec ^{\frac {5}{2}}(c+d x)}+\frac {4 a^4}{\sec ^{\frac {3}{2}}(c+d x)}+\frac {6 a^4}{\sqrt {\sec (c+d x)}}+4 a^4 \sqrt {\sec (c+d x)}+a^4 \sec ^{\frac {3}{2}}(c+d x)\right ) \, dx\\ &=a^4 \int \frac {1}{\sec ^{\frac {5}{2}}(c+d x)} \, dx+a^4 \int \sec ^{\frac {3}{2}}(c+d x) \, dx+\left (4 a^4\right ) \int \frac {1}{\sec ^{\frac {3}{2}}(c+d x)} \, dx+\left (4 a^4\right ) \int \sqrt {\sec (c+d x)} \, dx+\left (6 a^4\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx\\ &=\frac {2 a^4 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {8 a^4 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 a^4 \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {1}{5} \left (3 a^4\right ) \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx-a^4 \int \frac {1}{\sqrt {\sec (c+d x)}} \, dx+\frac {1}{3} \left (4 a^4\right ) \int \sqrt {\sec (c+d x)} \, dx+\left (4 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx+\left (6 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx\\ &=\frac {12 a^4 \sqrt {\cos (c+d x)} E\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{d}+\frac {8 a^4 \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^4 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {8 a^4 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 a^4 \sqrt {\sec (c+d x)} \sin (c+d x)}{d}+\frac {1}{5} \left (3 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx-\left (a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \sqrt {\cos (c+d x)} \, dx+\frac {1}{3} \left (4 a^4 \sqrt {\cos (c+d x)} \sqrt {\sec (c+d x)}\right ) \int \frac {1}{\sqrt {\cos (c+d x)}} \, dx\\ &=\frac {56 a^4 \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 {32 a^4 \sqrt {\cos (c+d x)} F\left (\left .\frac {1}{2} (c+d x)\right |2\right ) \sqrt {\sec (c+d x)}}{3 d}+\frac {2 a^4 \sin (c+d x)}{5 d \sec ^{\frac {3}{2}}(c+d x)}+\frac {8 a^4 \sin (c+d x)}{3 d \sqrt {\sec (c+d x)}}+\frac {2 a^4 \sqrt {\sec (c+d x)} \sin (c+d x)}{d}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 1.21, size = 184, normalized size = 1.16 \begin {gather*} \frac {a^4 \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 (-336 i+\frac {672 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)}}}-320 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)+80 \sin (c+d x)+3 \sec (c+d x) \sin (3 (c+d x))+63 \tan (c+d x)\right )}{30 d \sqrt {\sec (c+d x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sec[c + d*x])^4/Sec[c + d*x]^(5/2),x]

[Out]

(a^4*(Cos[c/2] - I*Sin[c/2])*(Cos[c/2] + I*Sin[c/2])*(-336*I + ((672*I)*Hypergeometric2F1[-1/4, 1/2, 3/4, -E^(
(2*I)*(c + d*x))])/Sqrt[1 + E^((2*I)*(c + d*x))] - (320*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] + 80*Sin[c + d*x] + 3*Sec[c + d*x]*Sin[3*(c + d*x)] + 63*Tan[c
+ d*x]))/(30*d*Sqrt[Sec[c + d*x]])

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Maple [A]
time = 0.06, size = 194, normalized size = 1.22

method	result	size
default	\(\frac {8 a^{4} \left (6 \left (\sin ^{6}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-26 \left (\sin ^{4}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )+19 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right ) \cos \left (\frac {d x}{2}+\frac {c}{2}\right )-20 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticF \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )+21 \sqrt {2 \left (\sin ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, \sqrt {\frac {1}{2}-\frac {\cos \left (d x +c \right )}{2}}\, \EllipticE \left (\cos \left (\frac {d x}{2}+\frac {c}{2}\right ), \sqrt {2}\right )\right )}{15 \sin \left (\frac {d x}{2}+\frac {c}{2}\right ) \sqrt {2 \left (\cos ^{2}\left (\frac {d x}{2}+\frac {c}{2}\right )\right )-1}\, d}\)	\(194\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sec(d*x+c))^4/sec(d*x+c)^(5/2),x,method=_RETURNVERBOSE)

[Out]

8/15*a^4*(6*sin(1/2*d*x+1/2*c)^6*cos(1/2*d*x+1/2*c)-26*sin(1/2*d*x+1/2*c)^4*cos(1/2*d*x+1/2*c)+19*sin(1/2*d*x+
1/2*c)^2*cos(1/2*d*x+1/2*c)-20*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticF(cos(1/2
*d*x+1/2*c),2^(1/2))+21*(2*sin(1/2*d*x+1/2*c)^2-1)^(1/2)*(sin(1/2*d*x+1/2*c)^2)^(1/2)*EllipticE(cos(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 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^4/sec(d*x+c)^(5/2),x, algorithm="maxima")

[Out]

integrate((a*sec(d*x + c) + a)^4/sec(d*x + c)^(5/2), x)

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Fricas [C] Result contains higher order function than in optimal. Order 9 vs. order 4.
time = 1.24, size = 162, normalized size = 1.02 \begin {gather*} -\frac {2 \, {\left (40 i \, \sqrt {2} a^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right ) - 40 i \, \sqrt {2} a^{4} {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right ) - 42 i \, \sqrt {2} a^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) + i \, \sin \left (d x + c\right )\right )\right ) + 42 i \, \sqrt {2} a^{4} {\rm weierstrassZeta}\left (-4, 0, {\rm weierstrassPInverse}\left (-4, 0, \cos \left (d x + c\right ) - i \, \sin \left (d x + c\right )\right )\right ) - \frac {{\left (3 \, a^{4} \cos \left (d x + c\right )^{2} + 20 \, a^{4} \cos \left (d x + c\right ) + 15 \, a^{4}\right )} \sin \left (d x + c\right )}{\sqrt {\cos \left (d x + c\right )}}\right )}}{15 \, d} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))^4/sec(d*x+c)^(5/2),x, algorithm="fricas")

[Out]

-2/15*(40*I*sqrt(2)*a^4*weierstrassPInverse(-4, 0, cos(d*x + c) + I*sin(d*x + c)) - 40*I*sqrt(2)*a^4*weierstra
ssPInverse(-4, 0, cos(d*x + c) - I*sin(d*x + c)) - 42*I*sqrt(2)*a^4*weierstrassZeta(-4, 0, weierstrassPInverse
(-4, 0, cos(d*x + c) + I*sin(d*x + c))) + 42*I*sqrt(2)*a^4*weierstrassZeta(-4, 0, weierstrassPInverse(-4, 0, c
os(d*x + c) - I*sin(d*x + c))) - (3*a^4*cos(d*x + c)^2 + 20*a^4*cos(d*x + c) + 15*a^4)*sin(d*x + c)/sqrt(cos(d
*x + c)))/d

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} a^{4} \left (\int \frac {1}{\sec ^{\frac {5}{2}}{\left (c + d x \right )}}\, dx + \int \frac {4}{\sec ^{\frac {3}{2}}{\left (c + d x \right )}}\, dx + \int \frac {6}{\sqrt {\sec {\left (c + d x \right )}}}\, dx + \int 4 \sqrt {\sec {\left (c + d x \right )}}\, dx + \int \sec ^{\frac {3}{2}}{\left (c + d x \right )}\, dx\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sec(d*x+c))**4/sec(d*x+c)**(5/2),x)

[Out]

a**4*(Integral(sec(c + d*x)**(-5/2), x) + Integral(4/sec(c + d*x)**(3/2), x) + Integral(6/sqrt(sec(c + d*x)),
x) + Integral(4*sqrt(sec(c + d*x)), x) + Integral(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((a+a*sec(d*x+c))^4/sec(d*x+c)^(5/2),x, algorithm="giac")

[Out]

integrate((a*sec(d*x + c) + a)^4/sec(d*x + c)^(5/2), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (a+\frac {a}{\cos \left (c+d\,x\right )}\right )}^4}{{\left (\frac {1}{\cos \left (c+d\,x\right )}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a/cos(c + d*x))^4/(1/cos(c + d*x))^(5/2),x)

[Out]

int((a + a/cos(c + d*x))^4/(1/cos(c + d*x))^(5/2), x)

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