3.3.48 \(\int (d \csc (a+b x))^{5/2} (c \sec (a+b x))^{5/2} \, dx\) [248]

Optimal. Leaf size=131 \[ \frac {4 c d^3 (c \sec (a+b x))^{3/2}}{3 b \sqrt {d \csc (a+b x)}}-\frac {2 c d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}}{3 b}+\frac {4 c^2 d^2 \sqrt {d \csc (a+b x)} F\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sqrt {c \sec (a+b x)} \sqrt {\sin (2 a+2 b x)}}{3 b} \]

[Out]

-2/3*c*d*(d*csc(b*x+a))^(3/2)*(c*sec(b*x+a))^(3/2)/b+4/3*c*d^3*(c*sec(b*x+a))^(3/2)/b/(d*csc(b*x+a))^(1/2)-4/3
*c^2*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))*(d*csc(b*x+a))^(1/
2)*(c*sec(b*x+a))^(1/2)*sin(2*b*x+2*a)^(1/2)/b

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Rubi [A]
time = 0.14, antiderivative size = 131, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.200, Rules used = {2705, 2706, 2710, 2653, 2720} \begin {gather*} \frac {4 c^2 d^2 \sqrt {\sin (2 a+2 b x)} F\left (\left .a+b x-\frac {\pi }{4}\right |2\right ) \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}{3 b}+\frac {4 c d^3 (c \sec (a+b x))^{3/2}}{3 b \sqrt {d \csc (a+b x)}}-\frac {2 c d (c \sec (a+b x))^{3/2} (d \csc (a+b x))^{3/2}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*c*d^3*(c*Sec[a + b*x])^(3/2))/(3*b*Sqrt[d*Csc[a + b*x]]) - (2*c*d*(d*Csc[a + b*x])^(3/2)*(c*Sec[a + b*x])^(
3/2))/(3*b) + (4*c^2*d^2*Sqrt[d*Csc[a + b*x]]*EllipticF[a - Pi/4 + b*x, 2]*Sqrt[c*Sec[a + b*x]]*Sqrt[Sin[2*a +
2*b*x]])/(3*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 2705

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*sec[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[(-a)*b*(a*Cs
c[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 2706

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*(n - 1))), x] + Dist[b^2*((m + n - 2)/(n - 1)), Int[(a*Csc[e + f
*x])^m*(b*Sec[e + f*x])^(n - 2), x], x] /; FreeQ[{a, b, e, f, m}, x] && GtQ[n, 1] && IntegersQ[2*m, 2*n]

Rule 2710

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 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 (d \csc (a+b x))^{5/2} (c \sec (a+b x))^{5/2} \, dx &=-\frac {2 c d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}}{3 b}+\left (2 d^2\right ) \int \sqrt {d \csc (a+b x)} (c \sec (a+b x))^{5/2} \, dx\\ &=\frac {4 c d^3 (c \sec (a+b x))^{3/2}}{3 b \sqrt {d \csc (a+b x)}}-\frac {2 c d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}}{3 b}+\frac {1}{3} \left (4 c^2 d^2\right ) \int \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)} \, dx\\ &=\frac {4 c d^3 (c \sec (a+b x))^{3/2}}{3 b \sqrt {d \csc (a+b x)}}-\frac {2 c d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}}{3 b}+\frac {1}{3} \left (4 c^2 d^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)}\right ) \int \frac {1}{\sqrt {c \cos (a+b x)} \sqrt {d \sin (a+b x)}} \, dx\\ &=\frac {4 c d^3 (c \sec (a+b x))^{3/2}}{3 b \sqrt {d \csc (a+b x)}}-\frac {2 c d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}}{3 b}+\frac {1}{3} \left (4 c^2 d^2 \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)} \sqrt {\sin (2 a+2 b x)}\right ) \int \frac {1}{\sqrt {\sin (2 a+2 b x)}} \, dx\\ &=\frac {4 c d^3 (c \sec (a+b x))^{3/2}}{3 b \sqrt {d \csc (a+b x)}}-\frac {2 c d (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2}}{3 b}+\frac {4 c^2 d^2 \sqrt {d \csc (a+b x)} F\left (\left .a-\frac {\pi }{4}+b x\right |2\right ) \sqrt {c \sec (a+b x)} \sqrt {\sin (2 a+2 b x)}}{3 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 = 0.71, size = 87, normalized size = 0.66 \begin {gather*} -\frac {2 c^3 d (d \csc (a+b x))^{3/2} \left (-1+\cot ^2(a+b x)+2 \left (-\cot ^2(a+b x)\right )^{3/4} \, _2F_1\left (\frac {1}{2},\frac {3}{4};\frac {3}{2};\csc ^2(a+b x)\right )\right ) \tan ^2(a+b x)}{3 b \sqrt {c \sec (a+b x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c^3*d*(d*Csc[a + b*x])^(3/2)*(-1 + Cot[a + b*x]^2 + 2*(-Cot[a + b*x]^2)^(3/4)*Hypergeometric2F1[1/2, 3/4,
3/2, Csc[a + b*x]^2])*Tan[a + b*x]^2)/(3*b*Sqrt[c*Sec[a + b*x]])

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(305\) vs. \(2(136)=272\).
time = 56.53, size = 306, normalized size = 2.34

method	result	size
default	\(-\frac {\left (-4 \left (\cos ^{2}\left (b x +a \right )\right ) \sin \left (b x +a \right ) \sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticF \left (\sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )-4 \cos \left (b x +a \right ) \sin \left (b x +a \right ) \sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {\cos \left (b x +a \right )-1+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \sqrt {\frac {-1+\cos \left (b x +a \right )}{\sin \left (b x +a \right )}}\, \EllipticF \left (\sqrt {-\frac {\cos \left (b x +a \right )-1-\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )+2 \left (\cos ^{2}\left (b x +a \right )\right ) \sqrt {2}-\sqrt {2}\right ) \cos \left (b x +a \right ) \left (\frac {d}{\sin \left (b x +a \right )}\right )^{\frac {5}{2}} \left (\frac {c}{\cos \left (b x +a \right )}\right )^{\frac {5}{2}} \sin \left (b x +a \right ) \sqrt {2}}{3 b}\)	\(306\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/3/b*(-4*cos(b*x+a)^2*sin(b*x+a)*(-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2)*((cos(b*x+a)-1+sin(b*x+a))/si
n(b*x+a))^(1/2)*((-1+cos(b*x+a))/sin(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))-4*cos(b*x+a)*sin(b*x+a)*(-(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)*((-1+cos(b*x+a))/sin(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))+2*cos(b*x+a)^2*2^(1/2)-2^(1/2))*cos(b*x+a)*(d/sin(b*x+a))^(5/2)*(c/cos(b*x+a))^(5/2)*sin(b*x+a)*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((d*csc(b*x+a))^(5/2)*(c*sec(b*x+a))^(5/2),x, algorithm="maxima")

[Out]

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

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Fricas [C] Result contains complex when optimal does not.
time = 0.62, size = 158, normalized size = 1.21 \begin {gather*} -\frac {2 \, {\left (i \, \sqrt {-4 i \, c d} c^{2} d^{2} \cos \left (b x + a\right ) {\rm ellipticF}\left (\cos \left (b x + a\right ) + i \, \sin \left (b x + a\right ), -1\right ) \sin \left (b x + a\right ) - i \, \sqrt {4 i \, c d} c^{2} d^{2} \cos \left (b x + a\right ) {\rm ellipticF}\left (\cos \left (b x + a\right ) - i \, \sin \left (b x + a\right ), -1\right ) \sin \left (b x + a\right ) + {\left (2 \, c^{2} d^{2} \cos \left (b x + a\right )^{2} - c^{2} d^{2}\right )} \sqrt {\frac {c}{\cos \left (b x + a\right )}} \sqrt {\frac {d}{\sin \left (b x + a\right )}}\right )}}{3 \, b \cos \left (b x + a\right ) \sin \left (b x + a\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*(I*sqrt(-4*I*c*d)*c^2*d^2*cos(b*x + a)*ellipticF(cos(b*x + a) + I*sin(b*x + a), -1)*sin(b*x + a) - I*sqrt
(4*I*c*d)*c^2*d^2*cos(b*x + a)*ellipticF(cos(b*x + a) - I*sin(b*x + a), -1)*sin(b*x + a) + (2*c^2*d^2*cos(b*x
+ a)^2 - c^2*d^2)*sqrt(c/cos(b*x + a))*sqrt(d/sin(b*x + a)))/(b*cos(b*x + a)*sin(b*x + a))

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Sympy [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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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((d*csc(b*x+a))^(5/2)*(c*sec(b*x+a))^(5/2),x, algorithm="giac")

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c/cos(a + b*x))^(5/2)*(d/sin(a + b*x))^(5/2),x)

[Out]

int((c/cos(a + b*x))^(5/2)*(d/sin(a + b*x))^(5/2), x)

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