3.272 \(\int \frac{(d \csc (a+b x))^{7/2}}{(c \sec (a+b x))^{5/2}} \, dx\)
Optimal. Leaf size=135 \[ \frac{6 d^4 E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{5 b c^2 \sqrt{\sin (2 a+2 b x)} \sqrt{c \sec (a+b x)} \sqrt{d \csc (a+b x)}}+\frac{6 d^3 \sqrt{d \csc (a+b x)}}{5 b c (c \sec (a+b x))^{3/2}}-\frac{2 d (d \csc (a+b x))^{5/2}}{5 b c (c \sec (a+b x))^{3/2}} \]
[Out]
(6*d^3*Sqrt[d*Csc[a + b*x]])/(5*b*c*(c*Sec[a + b*x])^(3/2)) - (2*d*(d*Csc[a + b*x])^(5/2))/(5*b*c*(c*Sec[a + b
*x])^(3/2)) + (6*d^4*EllipticE[a - Pi/4 + b*x, 2])/(5*b*c^2*Sqrt[d*Csc[a + b*x]]*Sqrt[c*Sec[a + b*x]]*Sqrt[Sin
[2*a + 2*b*x]])
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Rubi [A] time = 0.202232, antiderivative size = 135, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used =
{2623, 2625, 2630, 2572, 2639} \[ \frac{6 d^4 E\left (\left .a+b x-\frac{\pi }{4}\right |2\right )}{5 b c^2 \sqrt{\sin (2 a+2 b x)} \sqrt{c \sec (a+b x)} \sqrt{d \csc (a+b x)}}+\frac{6 d^3 \sqrt{d \csc (a+b x)}}{5 b c (c \sec (a+b x))^{3/2}}-\frac{2 d (d \csc (a+b x))^{5/2}}{5 b c (c \sec (a+b x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Int[(d*Csc[a + b*x])^(7/2)/(c*Sec[a + b*x])^(5/2),x]
[Out]
(6*d^3*Sqrt[d*Csc[a + b*x]])/(5*b*c*(c*Sec[a + b*x])^(3/2)) - (2*d*(d*Csc[a + b*x])^(5/2))/(5*b*c*(c*Sec[a + b
*x])^(3/2)) + (6*d^4*EllipticE[a - Pi/4 + b*x, 2])/(5*b*c^2*Sqrt[d*Csc[a + b*x]]*Sqrt[c*Sec[a + b*x]]*Sqrt[Sin
[2*a + 2*b*x]])
Rule 2623
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> -Simp[(a*(a*Csc[e
+ f*x])^(m - 1)*(b*Sec[e + f*x])^(n + 1))/(f*b*(m - 1)), x] + Dist[(a^2*(n + 1))/(b^2*(m - 1)), Int[(a*Csc[e +
f*x])^(m - 2)*(b*Sec[e + f*x])^(n + 2), x], x] /; FreeQ[{a, b, e, f}, x] && GtQ[m, 1] && LtQ[n, -1] && Intege
rsQ[2*m, 2*n]
Rule 2625
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> -Simp[(a*b*(a*Csc
[e + f*x])^(m - 1)*(b*Sec[e + f*x])^(n - 1))/(f*(m - 1)), x] + Dist[(a^2*(m + n - 2))/(m - 1), Int[(a*Csc[e +
f*x])^(m - 2)*(b*Sec[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && GtQ[m, 1] && IntegersQ[2*m, 2*n] &&
!GtQ[n, m]
Rule 2630
Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Csc[e + f*
x])^m*(b*Sec[e + f*x])^n*(a*Sin[e + f*x])^m*(b*Cos[e + f*x])^n, Int[1/((a*Sin[e + f*x])^m*(b*Cos[e + f*x])^n),
x], x] /; FreeQ[{a, b, e, f, m, n}, x] && IntegerQ[m - 1/2] && IntegerQ[n - 1/2]
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{(d \csc (a+b x))^{7/2}}{(c \sec (a+b x))^{5/2}} \, dx &=-\frac{2 d (d \csc (a+b x))^{5/2}}{5 b c (c \sec (a+b x))^{3/2}}-\frac{\left (3 d^2\right ) \int \frac{(d \csc (a+b x))^{3/2}}{\sqrt{c \sec (a+b x)}} \, dx}{5 c^2}\\ &=\frac{6 d^3 \sqrt{d \csc (a+b x)}}{5 b c (c \sec (a+b x))^{3/2}}-\frac{2 d (d \csc (a+b x))^{5/2}}{5 b c (c \sec (a+b x))^{3/2}}+\frac{\left (6 d^4\right ) \int \frac{1}{\sqrt{d \csc (a+b x)} \sqrt{c \sec (a+b x)}} \, dx}{5 c^2}\\ &=\frac{6 d^3 \sqrt{d \csc (a+b x)}}{5 b c (c \sec (a+b x))^{3/2}}-\frac{2 d (d \csc (a+b x))^{5/2}}{5 b c (c \sec (a+b x))^{3/2}}+\frac{\left (6 d^4\right ) \int \sqrt{c \cos (a+b x)} \sqrt{d \sin (a+b x)} \, dx}{5 c^2 \sqrt{c \cos (a+b x)} \sqrt{d \csc (a+b x)} \sqrt{c \sec (a+b x)} \sqrt{d \sin (a+b x)}}\\ &=\frac{6 d^3 \sqrt{d \csc (a+b x)}}{5 b c (c \sec (a+b x))^{3/2}}-\frac{2 d (d \csc (a+b x))^{5/2}}{5 b c (c \sec (a+b x))^{3/2}}+\frac{\left (6 d^4\right ) \int \sqrt{\sin (2 a+2 b x)} \, dx}{5 c^2 \sqrt{d \csc (a+b x)} \sqrt{c \sec (a+b x)} \sqrt{\sin (2 a+2 b x)}}\\ &=\frac{6 d^3 \sqrt{d \csc (a+b x)}}{5 b c (c \sec (a+b x))^{3/2}}-\frac{2 d (d \csc (a+b x))^{5/2}}{5 b c (c \sec (a+b x))^{3/2}}+\frac{6 d^4 E\left (\left .a-\frac{\pi }{4}+b x\right |2\right )}{5 b c^2 \sqrt{d \csc (a+b x)} \sqrt{c \sec (a+b x)} \sqrt{\sin (2 a+2 b x)}}\\ \end{align*}
Mathematica [C] time = 1.80213, size = 101, normalized size = 0.75 \[ \frac{d^5 \sqrt{c \sec (a+b x)} \left (6 \sqrt [4]{-\cot ^2(a+b x)} \text{Hypergeometric2F1}\left (-\frac{1}{2},\frac{1}{4},\frac{1}{2},\csc ^2(a+b x)\right )+(1-3 \cos (2 (a+b x))) \cot ^2(a+b x) \csc ^2(a+b x)\right )}{5 b c^3 (d \csc (a+b x))^{3/2}} \]
Antiderivative was successfully verified.
[In]
Integrate[(d*Csc[a + b*x])^(7/2)/(c*Sec[a + b*x])^(5/2),x]
[Out]
(d^5*((1 - 3*Cos[2*(a + b*x)])*Cot[a + b*x]^2*Csc[a + b*x]^2 + 6*(-Cot[a + b*x]^2)^(1/4)*Hypergeometric2F1[-1/
2, 1/4, 1/2, Csc[a + b*x]^2])*Sqrt[c*Sec[a + b*x]])/(5*b*c^3*(d*Csc[a + b*x])^(3/2))
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Maple [B] time = 0.196, size = 993, normalized size = 7.4 \begin{align*} \text{result too large to display} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*csc(b*x+a))^(7/2)/(c*sec(b*x+a))^(5/2),x)
[Out]
1/5/b*2^(1/2)*(6*cos(b*x+a)^3*(-(-1+cos(b*x+a)-sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b
*x+a))^(1/2)*((-1+cos(b*x+a))/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))-3*cos(b*x+a)^3*(-(-1+cos(b*x+a)-sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^
(1/2)*((-1+cos(b*x+a))/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))
+6*cos(b*x+a)^2*(-(-1+cos(b*x+a)-sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*(
(-1+cos(b*x+a))/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))-3*cos(
b*x+a)^2*(-(-1+cos(b*x+a)-sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos
(b*x+a))/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))-6*cos(b*x+a)*
(-(-1+cos(b*x+a)-sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/
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))+3*cos(b*x+a)*(-(-1+cos
(b*x+a)-sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/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))-3*cos(b*x+a)^3*2^(1/2)-6*(-(-1+
cos(b*x+a)-sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/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))+3*(-(-1+cos(b*x+a)-sin(b*x+a
))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a)+sin(b*x+a))/sin(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(b*x+a))^(1/2)*Elliptic
F((-(-1+cos(b*x+a)-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))-cos(b*x+a)^2*2^(1/2)+3*cos(b*x+a)*2^(1/2))*(d/si
n(b*x+a))^(7/2)*sin(b*x+a)/cos(b*x+a)^3/(c/cos(b*x+a))^(5/2)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \csc \left (b x + a\right )\right )^{\frac{7}{2}}}{\left (c \sec \left (b x + a\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*csc(b*x+a))^(7/2)/(c*sec(b*x+a))^(5/2),x, algorithm="maxima")
[Out]
integrate((d*csc(b*x + a))^(7/2)/(c*sec(b*x + a))^(5/2), x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\frac{\sqrt{d \csc \left (b x + a\right )} \sqrt{c \sec \left (b x + a\right )} d^{3} \csc \left (b x + a\right )^{3}}{c^{3} \sec \left (b x + a\right )^{3}}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*csc(b*x+a))^(7/2)/(c*sec(b*x+a))^(5/2),x, algorithm="fricas")
[Out]
integral(sqrt(d*csc(b*x + a))*sqrt(c*sec(b*x + a))*d^3*csc(b*x + a)^3/(c^3*sec(b*x + a)^3), 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((d*csc(b*x+a))**(7/2)/(c*sec(b*x+a))**(5/2),x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\left (d \csc \left (b x + a\right )\right )^{\frac{7}{2}}}{\left (c \sec \left (b x + a\right )\right )^{\frac{5}{2}}}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*csc(b*x+a))^(7/2)/(c*sec(b*x+a))^(5/2),x, algorithm="giac")
[Out]
integrate((d*csc(b*x + a))^(7/2)/(c*sec(b*x + a))^(5/2), x)