3.3.72 \(\int \frac {(d \csc (a+b x))^{7/2}}{(c \sec (a+b x))^{5/2}} \, dx\) [272]

Optimal. Leaf size=135 \[ \frac {6 d^3 \sqrt {d \csc (a+b x)}}{5 b c (c \sec (a+b x))^{3/2}}-\frac {2 d (d \csc (a+b x))^{5/2}}{5 b c (c \sec (a+b x))^{3/2}}+\frac {6 d^4 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{5 b c^2 \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)} \sqrt {\sin (2 a+2 b x)}} \]

[Out]

-2/5*d*(d*csc(b*x+a))^(5/2)/b/c/(c*sec(b*x+a))^(3/2)+6/5*d^3*(d*csc(b*x+a))^(1/2)/b/c/(c*sec(b*x+a))^(3/2)-6/5
*d^4*(sin(a+1/4*Pi+b*x)^2)^(1/2)/sin(a+1/4*Pi+b*x)*EllipticE(cos(a+1/4*Pi+b*x),2^(1/2))/b/c^2/(d*csc(b*x+a))^(
1/2)/(c*sec(b*x+a))^(1/2)/sin(2*b*x+2*a)^(1/2)

________________________________________________________________________________________

Rubi [A]
time = 0.14, antiderivative size = 135, 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 = {2703, 2705, 2710, 2652, 2719} \begin {gather*} \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}} \end {gather*}

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 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 2703

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] && Inte
gersQ[2*m, 2*n]

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 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 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 {(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] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 11.86, size = 101, normalized size = 0.75 \begin {gather*} \frac {d^5 \left ((1-3 \cos (2 (a+b x))) \cot ^2(a+b x) \csc ^2(a+b x)+6 \sqrt [4]{-\cot ^2(a+b x)} \, _2F_1\left (-\frac {1}{2},\frac {1}{4};\frac {1}{2};\csc ^2(a+b x)\right )\right ) \sqrt {c \sec (a+b x)}}{5 b c^3 (d \csc (a+b x))^{3/2}} \end {gather*}

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))

________________________________________________________________________________________

Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(992\) vs. \(2(140)=280\).
time = 33.46, size = 993, normalized size = 7.36

method	result	size
default	\(\text {Expression too large to display}\)	\(993\)

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,method=_RETURNVERBOSE)

[Out]

1/5/b*(6*cos(b*x+a)^3*(-(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)*EllipticE((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))-3*c
os(b*x+a)^3*(-(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+co
s(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))+6*cos(b*x+a)^
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)*((-1+cos(b*x+a))/
sin(b*x+a))^(1/2)*EllipticE((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))-3*cos(b*x+a)^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)*((-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))-6*cos(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)*Elli
pticE((-(cos(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))+3*cos(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))-3*cos(b*x+a)^3*2^(1/2)-6*(-(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)*EllipticE((-(cos
(b*x+a)-1-sin(b*x+a))/sin(b*x+a))^(1/2),1/2*2^(1/2))+3*(-(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))/s
in(b*x+a))^(1/2),1/2*2^(1/2))-cos(b*x+a)^2*2^(1/2)+3*2^(1/2)*cos(b*x+a))*(d/sin(b*x+a))^(7/2)*sin(b*x+a)/cos(b
*x+a)^3/(c/cos(b*x+a))^(5/2)*2^(1/2)

________________________________________________________________________________________

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))^(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)

________________________________________________________________________________________

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((d*csc(b*x+a))^(7/2)/(c*sec(b*x+a))^(5/2),x, algorithm="fricas")

[Out]

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

________________________________________________________________________________________

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))**(7/2)/(c*sec(b*x+a))**(5/2),x)

[Out]

Timed out

________________________________________________________________________________________

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))^(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)

________________________________________________________________________________________

Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (\frac {d}{\sin \left (a+b\,x\right )}\right )}^{7/2}}{{\left (\frac {c}{\cos \left (a+b\,x\right )}\right )}^{5/2}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d/sin(a + b*x))^(7/2)/(c/cos(a + b*x))^(5/2),x)

[Out]

int((d/sin(a + b*x))^(7/2)/(c/cos(a + b*x))^(5/2), x)

________________________________________________________________________________________