3.270 \(\int \frac{\sec ^3(a+b x)}{(d \tan (a+b x))^{7/2}} \, dx\)

Optimal. Leaf size=110 \[ -\frac{4 \cos (a+b x)}{5 b d^3 \sqrt{d \tan (a+b x)}}-\frac{4 \cos (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{5 b d^4 \sqrt{\sin (2 a+2 b x)}}-\frac{2 \sec (a+b x)}{5 b d (d \tan (a+b x))^{5/2}} \]

[Out]

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

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Rubi [A]  time = 0.137395, antiderivative size = 110, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 4, integrand size = 21, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.19, Rules used = {2608, 2615, 2572, 2639} \[ -\frac{4 \cos (a+b x)}{5 b d^3 \sqrt{d \tan (a+b x)}}-\frac{4 \cos (a+b x) E\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{5 b d^4 \sqrt{\sin (2 a+2 b x)}}-\frac{2 \sec (a+b x)}{5 b d (d \tan (a+b x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sec[a + b*x]^3/(d*Tan[a + b*x])^(7/2),x]

[Out]

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

Rule 2608

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 2615

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 2572

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 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]

Rubi steps

\begin{align*} \int \frac{\sec ^3(a+b x)}{(d \tan (a+b x))^{7/2}} \, dx &=-\frac{2 \sec (a+b x)}{5 b d (d \tan (a+b x))^{5/2}}+\frac{2 \int \frac{\sec (a+b x)}{(d \tan (a+b x))^{3/2}} \, dx}{5 d^2}\\ &=-\frac{2 \sec (a+b x)}{5 b d (d \tan (a+b x))^{5/2}}-\frac{4 \cos (a+b x)}{5 b d^3 \sqrt{d \tan (a+b x)}}-\frac{4 \int \cos (a+b x) \sqrt{d \tan (a+b x)} \, dx}{5 d^4}\\ &=-\frac{2 \sec (a+b x)}{5 b d (d \tan (a+b x))^{5/2}}-\frac{4 \cos (a+b x)}{5 b d^3 \sqrt{d \tan (a+b x)}}-\frac{\left (4 \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^4 \sqrt{\sin (a+b x)}}\\ &=-\frac{2 \sec (a+b x)}{5 b d (d \tan (a+b x))^{5/2}}-\frac{4 \cos (a+b x)}{5 b d^3 \sqrt{d \tan (a+b x)}}-\frac{\left (4 \cos (a+b x) \sqrt{d \tan (a+b x)}\right ) \int \sqrt{\sin (2 a+2 b x)} \, dx}{5 d^4 \sqrt{\sin (2 a+2 b x)}}\\ &=-\frac{2 \sec (a+b x)}{5 b d (d \tan (a+b x))^{5/2}}-\frac{4 \cos (a+b x)}{5 b d^3 \sqrt{d \tan (a+b x)}}-\frac{4 \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^4 \sqrt{\sin (2 a+2 b x)}}\\ \end{align*}

Mathematica [C]  time = 1.5268, size = 103, normalized size = 0.94 \[ -\frac{2 \sin (a+b x) \sqrt{d \tan (a+b x)} \left (4 \sec ^2(a+b x) \, _2F_1\left (\frac{3}{4},\frac{3}{2};\frac{7}{4};-\tan ^2(a+b x)\right )+3 \left (\csc ^4(a+b x)+\csc ^2(a+b x)-2\right ) \sqrt{\sec ^2(a+b x)}\right )}{15 b d^4 \sqrt{\sec ^2(a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sec[a + b*x]^3/(d*Tan[a + b*x])^(7/2),x]

[Out]

(-2*(4*Hypergeometric2F1[3/4, 3/2, 7/4, -Tan[a + b*x]^2]*Sec[a + b*x]^2 + 3*(-2 + Csc[a + b*x]^2 + Csc[a + b*x
]^4)*Sqrt[Sec[a + b*x]^2])*Sin[a + b*x]*Sqrt[d*Tan[a + b*x]])/(15*b*d^4*Sqrt[Sec[a + b*x]^2])

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Maple [B]  time = 0.177, size = 970, normalized size = 8.8 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(sec(b*x+a)^3/(d*tan(b*x+a))^(7/2),x)

[Out]

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

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (b x + a\right )^{3}}{\left (d \tan \left (b x + a\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3/(d*tan(b*x+a))^(7/2),x, algorithm="maxima")

[Out]

integrate(sec(b*x + a)^3/(d*tan(b*x + a))^(7/2), x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d \tan \left (b x + a\right )} \sec \left (b x + a\right )^{3}}{d^{4} \tan \left (b x + a\right )^{4}}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3/(d*tan(b*x+a))^(7/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*tan(b*x + a))*sec(b*x + a)^3/(d^4*tan(b*x + a)^4), x)

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)**3/(d*tan(b*x+a))**(7/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sec \left (b x + a\right )^{3}}{\left (d \tan \left (b x + a\right )\right )^{\frac{7}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(sec(b*x+a)^3/(d*tan(b*x+a))^(7/2),x, algorithm="giac")

[Out]

integrate(sec(b*x + a)^3/(d*tan(b*x + a))^(7/2), x)