3.3.41 \(\int \sec ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx\) [241]

Optimal. Leaf size=136 \[ -\frac {4 d^2 F\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sec (a+b x) \sqrt {\sin (2 a+2 b x)}}{77 b \sqrt {d \tan (a+b x)}}-\frac {4 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{77 b}-\frac {2 d \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{77 b}+\frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b} \]

[Out]

4/77*d^2*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticF(cos(a+1/4*Pi+b*x),2^(1/2))*sec(b*x+a)*sin(2*b
*x+2*a)^(1/2)/b/(d*tan(b*x+a))^(1/2)-4/77*d*sec(b*x+a)*(d*tan(b*x+a))^(1/2)/b-2/77*d*sec(b*x+a)^3*(d*tan(b*x+a
))^(1/2)/b+2/11*d*sec(b*x+a)^5*(d*tan(b*x+a))^(1/2)/b

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Rubi [A]
time = 0.13, antiderivative size = 136, normalized size of antiderivative = 1.00, number of steps used = 6, number of rules used = 5, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.238, Rules used = {2691, 2693, 2694, 2653, 2720} \begin {gather*} -\frac {4 d^2 \sqrt {\sin (2 a+2 b x)} \sec (a+b x) F\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{77 b \sqrt {d \tan (a+b x)}}+\frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {2 d \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{77 b}-\frac {4 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{77 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-4*d^2*EllipticF[a - Pi/4 + b*x, 2]*Sec[a + b*x]*Sqrt[Sin[2*a + 2*b*x]])/(77*b*Sqrt[d*Tan[a + b*x]]) - (4*d*S
ec[a + b*x]*Sqrt[d*Tan[a + b*x]])/(77*b) - (2*d*Sec[a + b*x]^3*Sqrt[d*Tan[a + b*x]])/(77*b) + (2*d*Sec[a + b*x
]^5*Sqrt[d*Tan[a + b*x]])/(11*b)

Rule 2653

Int[1/(Sqrt[cos[(e_.) + (f_.)*(x_)]*(b_.)]*Sqrt[(a_.)*sin[(e_.) + (f_.)*(x_)]]), x_Symbol] :> Dist[Sqrt[Sin[2*
e + 2*f*x]]/(Sqrt[a*Sin[e + f*x]]*Sqrt[b*Cos[e + f*x]]), Int[1/Sqrt[Sin[2*e + 2*f*x]], x], x] /; FreeQ[{a, b,
e, f}, x]

Rule 2691

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[b*(a*Sec[e +
f*x])^m*((b*Tan[e + f*x])^(n - 1)/(f*(m + n - 1))), x] - Dist[b^2*((n - 1)/(m + n - 1)), Int[(a*Sec[e + f*x])
^m*(b*Tan[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && NeQ[m + n - 1, 0] && Integers
Q[2*m, 2*n]

Rule 2693

Int[((a_.)*sec[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Simp[a^2*(a*Sec[e
+ f*x])^(m - 2)*((b*Tan[e + f*x])^(n + 1)/(b*f*(m + n - 1))), x] + Dist[a^2*((m - 2)/(m + n - 1)), Int[(a*Sec
[e + f*x])^(m - 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && (GtQ[m, 1] || (EqQ[m, 1] && EqQ[
n, 1/2])) && NeQ[m + n - 1, 0] && IntegersQ[2*m, 2*n]

Rule 2694

Int[sec[(e_.) + (f_.)*(x_)]/Sqrt[(b_.)*tan[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[Sqrt[Sin[e + f*x]]/(Sqrt[Co
s[e + f*x]]*Sqrt[b*Tan[e + f*x]]), Int[1/(Sqrt[Cos[e + f*x]]*Sqrt[Sin[e + f*x]]), x], x] /; FreeQ[{b, e, f}, 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]

Rubi steps

\begin {align*} \int \sec ^5(a+b x) (d \tan (a+b x))^{3/2} \, dx &=\frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {1}{11} d^2 \int \frac {\sec ^5(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx\\ &=-\frac {2 d \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{77 b}+\frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {1}{77} \left (6 d^2\right ) \int \frac {\sec ^3(a+b x)}{\sqrt {d \tan (a+b x)}} \, dx\\ &=-\frac {4 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{77 b}-\frac {2 d \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{77 b}+\frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {1}{77} \left (4 d^2\right ) \int \frac {\sec (a+b x)}{\sqrt {d \tan (a+b x)}} \, dx\\ &=-\frac {4 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{77 b}-\frac {2 d \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{77 b}+\frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {\left (4 d^2 \sqrt {\sin (a+b x)}\right ) \int \frac {1}{\sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)}} \, dx}{77 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}}\\ &=-\frac {4 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{77 b}-\frac {2 d \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{77 b}+\frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}-\frac {\left (4 d^2 \sec (a+b x) \sqrt {\sin (2 a+2 b x)}\right ) \int \frac {1}{\sqrt {\sin (2 a+2 b x)}} \, dx}{77 \sqrt {d \tan (a+b x)}}\\ &=-\frac {4 d^2 F\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sec (a+b x) \sqrt {\sin (2 a+2 b x)}}{77 b \sqrt {d \tan (a+b x)}}-\frac {4 d \sec (a+b x) \sqrt {d \tan (a+b x)}}{77 b}-\frac {2 d \sec ^3(a+b x) \sqrt {d \tan (a+b x)}}{77 b}+\frac {2 d \sec ^5(a+b x) \sqrt {d \tan (a+b x)}}{11 b}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.91, size = 90, normalized size = 0.66 \begin {gather*} -\frac {d \sec ^5(a+b x) \left (-23+6 \cos (2 (a+b x))+\cos (4 (a+b x))+16 \cos ^6(a+b x) \, _2F_1\left (\frac {1}{4},\frac {1}{2};\frac {5}{4};-\tan ^2(a+b x)\right ) \sqrt {\sec ^2(a+b x)}\right ) \sqrt {d \tan (a+b x)}}{154 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

-1/154*(d*Sec[a + b*x]^5*(-23 + 6*Cos[2*(a + b*x)] + Cos[4*(a + b*x)] + 16*Cos[a + b*x]^6*Hypergeometric2F1[1/
4, 1/2, 5/4, -Tan[a + b*x]^2]*Sqrt[Sec[a + b*x]^2])*Sqrt[d*Tan[a + b*x]])/b

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Maple [A]
time = 0.54, size = 251, normalized size = 1.85

method	result	size
default	\(-\frac {\left (-1+\cos \left (b x +a \right )\right ) \left (-4 \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \left (\cos ^{5}\left (b x +a \right )\right ) \sin \left (b x +a \right ) \EllipticF \left (\sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )+2 \sqrt {2}\, \left (\cos ^{5}\left (b x +a \right )\right )-2 \left (\cos ^{4}\left (b x +a \right )\right ) \sqrt {2}+\left (\cos ^{3}\left (b x +a \right )\right ) \sqrt {2}-\left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}-7 \cos \left (b x +a \right ) \sqrt {2}+7 \sqrt {2}\right ) \left (\cos \left (b x +a \right )+1\right )^{2} \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}} \sqrt {2}}{77 b \sin \left (b x +a \right )^{5} \cos \left (b x +a \right )^{4}}\)	\(251\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^5*(d*tan(b*x+a))^(3/2),x,method=_RETURNVERBOSE)

[Out]

-1/77/b*(-1+cos(b*x+a))*(-4*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*(-
(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2)*cos(b*x+a)^5*sin(b*x+a)*EllipticF((-(cos(b*x+a)-1-sin(b*x+a))/sin(
b*x+a))^(1/2),1/2*2^(1/2))+2*2^(1/2)*cos(b*x+a)^5-2*cos(b*x+a)^4*2^(1/2)+cos(b*x+a)^3*2^(1/2)-cos(b*x+a)^2*2^(
1/2)-7*cos(b*x+a)*2^(1/2)+7*2^(1/2))*(cos(b*x+a)+1)^2*(d*sin(b*x+a)/cos(b*x+a))^(3/2)/sin(b*x+a)^5/cos(b*x+a)^
4*2^(1/2)

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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(b*x+a)^5*(d*tan(b*x+a))^(3/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains complex when optimal does not.
time = 0.11, size = 125, normalized size = 0.92 \begin {gather*} \frac {2 \, {\left (2 \, \sqrt {i \, d} d \cos \left (b x + a\right )^{5} {\rm ellipticF}\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ), -1\right ) + 2 \, \sqrt {-i \, d} d \cos \left (b x + a\right )^{5} {\rm ellipticF}\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ), -1\right ) - {\left (2 \, d \cos \left (b x + a\right )^{4} + d \cos \left (b x + a\right )^{2} - 7 \, d\right )} \sqrt {\frac {d \sin \left (b x + a\right )}{\cos \left (b x + a\right )}}\right )}}{77 \, b \cos \left (b x + a\right )^{5}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/77*(2*sqrt(I*d)*d*cos(b*x + a)^5*ellipticF(cos(b*x + a) + I*sin(b*x + a), -1) + 2*sqrt(-I*d)*d*cos(b*x + a)^
5*ellipticF(cos(b*x + a) - I*sin(b*x + a), -1) - (2*d*cos(b*x + a)^4 + d*cos(b*x + a)^2 - 7*d)*sqrt(d*sin(b*x
+ a)/cos(b*x + a)))/(b*cos(b*x + a)^5)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Integral((d*tan(a + b*x))**(3/2)*sec(a + b*x)**5, 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(b*x+a)^5*(d*tan(b*x+a))^(3/2),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (d\,\mathrm {tan}\left (a+b\,x\right )\right )}^{3/2}}{{\cos \left (a+b\,x\right )}^5} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*tan(a + b*x))^(3/2)/cos(a + b*x)^5,x)

[Out]

int((d*tan(a + b*x))^(3/2)/cos(a + b*x)^5, x)

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