3.3.63 \(\int \frac {\sec ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx\) [263]

Optimal. Leaf size=138 \[ -\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}-\frac {24 \cos (a+b x) E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sqrt {d \tan (a+b x)}}{5 b d^2 \sqrt {\sin (2 a+2 b x)}}+\frac {24 \cos (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}+\frac {12 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3} \]

[Out]

-2*sec(b*x+a)^3/b/d/(d*tan(b*x+a))^(1/2)+24/5*cos(b*x+a)*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*Ellipti
cE(cos(a+1/4*Pi+b*x),2^(1/2))*(d*tan(b*x+a))^(1/2)/b/d^2/sin(2*b*x+2*a)^(1/2)+24/5*cos(b*x+a)*(d*tan(b*x+a))^(
3/2)/b/d^3+12/5*sec(b*x+a)*(d*tan(b*x+a))^(3/2)/b/d^3

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Rubi [A]
time = 0.12, antiderivative size = 138, 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 = {2688, 2693, 2695, 2652, 2719} \begin {gather*} \frac {24 \cos (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}+\frac {12 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac {24 \cos (a+b x) E\left (\left .a+b x-\frac {\pi }{4}\right |2\right ) \sqrt {d \tan (a+b x)}}{5 b d^2 \sqrt {\sin (2 a+2 b x)}}-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*Sec[a + b*x]^3)/(b*d*Sqrt[d*Tan[a + b*x]]) - (24*Cos[a + b*x]*EllipticE[a - Pi/4 + b*x, 2]*Sqrt[d*Tan[a +
b*x]])/(5*b*d^2*Sqrt[Sin[2*a + 2*b*x]]) + (24*Cos[a + b*x]*(d*Tan[a + b*x])^(3/2))/(5*b*d^3) + (12*Sec[a + b*x
]*(d*Tan[a + b*x])^(3/2))/(5*b*d^3)

Rule 2652

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

Rule 2688

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*(n + 1))), x] - Dist[a^2*((m - 2)/(b^2*(n + 1))), Int[(a*Sec[e
+ f*x])^(m - 2)*(b*Tan[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && LtQ[n, -1] && (GtQ[m, 1] || (Eq
Q[m, 1] && EqQ[n, -3/2])) && IntegersQ[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 2695

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

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]

Rubi steps

\begin {align*} \int \frac {\sec ^5(a+b x)}{(d \tan (a+b x))^{3/2}} \, dx &=-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}+\frac {6 \int \sec ^3(a+b x) \sqrt {d \tan (a+b x)} \, dx}{d^2}\\ &=-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}+\frac {12 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}+\frac {12 \int \sec (a+b x) \sqrt {d \tan (a+b x)} \, dx}{5 d^2}\\ &=-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}+\frac {24 \cos (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}+\frac {12 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac {24 \int \cos (a+b x) \sqrt {d \tan (a+b x)} \, dx}{5 d^2}\\ &=-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}+\frac {24 \cos (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}+\frac {12 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac {\left (24 \sqrt {\cos (a+b x)} \sqrt {d \tan (a+b x)}\right ) \int \sqrt {\cos (a+b x)} \sqrt {\sin (a+b x)} \, dx}{5 d^2 \sqrt {\sin (a+b x)}}\\ &=-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}+\frac {24 \cos (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}+\frac {12 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}-\frac {\left (24 \cos (a+b x) \sqrt {d \tan (a+b x)}\right ) \int \sqrt {\sin (2 a+2 b x)} \, dx}{5 d^2 \sqrt {\sin (2 a+2 b x)}}\\ &=-\frac {2 \sec ^3(a+b x)}{b d \sqrt {d \tan (a+b x)}}-\frac {24 \cos (a+b x) E\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sqrt {d \tan (a+b x)}}{5 b d^2 \sqrt {\sin (2 a+2 b x)}}+\frac {24 \cos (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}+\frac {12 \sec (a+b x) (d \tan (a+b x))^{3/2}}{5 b d^3}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 0.96, size = 104, normalized size = 0.75 \begin {gather*} \frac {2 \csc (a+b x) \sqrt {d \tan (a+b x)} \left (-8 \, _2F_1\left (\frac {3}{4},\frac {3}{2};\frac {7}{4};-\tan ^2(a+b x)\right ) \tan ^2(a+b x)+\sqrt {\sec ^2(a+b x)} \left (-5+12 \sin ^2(a+b x)+\tan ^2(a+b x)\right )\right )}{5 b d^2 \sqrt {\sec ^2(a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(2*Csc[a + b*x]*Sqrt[d*Tan[a + b*x]]*(-8*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[a + b*x]^2]*Tan[a + b*x]^2 + Sq
rt[Sec[a + b*x]^2]*(-5 + 12*Sin[a + b*x]^2 + Tan[a + b*x]^2)))/(5*b*d^2*Sqrt[Sec[a + b*x]^2])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(534\) vs. \(2(147)=294\).
time = 0.32, size = 535, normalized size = 3.88

method	result	size
default	\(\frac {\left (24 \EllipticE \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 ) \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \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 )}}\, \left (\cos ^{3}\left (b x +a \right )\right )-12 \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \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 )}}\, \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 ) \left (\cos ^{3}\left (b x +a \right )\right )+24 \EllipticE \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 ) \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \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 )}}\, \left (\cos ^{2}\left (b x +a \right )\right )-12 \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \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 )}}\, \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 ) \left (\cos ^{2}\left (b x +a \right )\right )-12 \left (\cos ^{3}\left (b x +a \right )\right ) \sqrt {2}+6 \left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}+\sqrt {2}\right ) \sin \left (b x +a \right ) \sqrt {2}}{5 b \cos \left (b x +a \right )^{4} \left (\frac {d \sin \left (b x +a \right )}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}}}\)	\(535\)

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]

1/5/b*(24*EllipticE((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((cos(b*x+a)-1+sin(b*x+a))/sin(
b*x+a))^(1/2)*((-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)^3-12
*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*(-(cos(b*x+a)-1-sin(b*x+a))/s
in(b*x+a))^(1/2)*EllipticF((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*cos(b*x+a)^3+24*Elliptic
E((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*((cos(b*x+a)-1+sin(b*x+a))/sin(b*x+a))^(1/2)*((-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)^2-12*((cos(b*x+a)-1+si
n(b*x+a))/sin(b*x+a))^(1/2)*((-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)*E
llipticF((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))*cos(b*x+a)^2-12*cos(b*x+a)^3*2^(1/2)+6*cos
(b*x+a)^2*2^(1/2)+2^(1/2))*sin(b*x+a)/cos(b*x+a)^4/(d*sin(b*x+a)/cos(b*x+a))^(3/2)*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(sec(b*x + a)^5/(d*tan(b*x + a))^(3/2), x)

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Fricas [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: TypeError} \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]

Exception raised: TypeError >> Symbolic function elliptic_ec takes exactly 1 arguments (2 given)

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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