3.3.41 \(\int (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2} \, dx\) [241]

Optimal. Leaf size=125 \[ \frac {4 c d^3 \sqrt {c \sec (a+b x)}}{b (d \csc (a+b x))^{3/2}}-\frac {2 c d \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}}{b}-\frac {4 c^2 d^2 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{b \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)} \sqrt {\sin (2 a+2 b x)}} \]

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 125, 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, 2652, 2719} \begin {gather*} -\frac {4 c^2 d^2 E\left (\left .a+b x-\frac {\pi }{4}\right |2\right )}{b \sqrt {\sin (2 a+2 b x)} \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}+\frac {4 c d^3 \sqrt {c \sec (a+b x)}}{b (d \csc (a+b x))^{3/2}}-\frac {2 c d \sqrt {c \sec (a+b x)} \sqrt {d \csc (a+b x)}}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(4*c*d^3*Sqrt[c*Sec[a + b*x]])/(b*(d*Csc[a + b*x])^(3/2)) - (2*c*d*Sqrt[d*Csc[a + b*x]]*Sqrt[c*Sec[a + b*x]])/
b - (4*c^2*d^2*EllipticE[a - Pi/4 + b*x, 2])/(b*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 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 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 (d \csc (a+b x))^{3/2} (c \sec (a+b x))^{3/2} \, dx &=-\frac {2 c d \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}}{b}+\left (2 d^2\right ) \int \frac {(c \sec (a+b x))^{3/2}}{\sqrt {d \csc (a+b x)}} \, dx\\ &=\frac {4 c d^3 \sqrt {c \sec (a+b x)}}{b (d \csc (a+b x))^{3/2}}-\frac {2 c d \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}}{b}-\left (4 c^2 d^2\right ) \int \frac {1}{\sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}} \, dx\\ &=\frac {4 c d^3 \sqrt {c \sec (a+b x)}}{b (d \csc (a+b x))^{3/2}}-\frac {2 c d \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}}{b}-\frac {\left (4 c^2 d^2\right ) \int \sqrt {c \cos (a+b x)} \sqrt {d \sin (a+b x)} \, dx}{\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 {4 c d^3 \sqrt {c \sec (a+b x)}}{b (d \csc (a+b x))^{3/2}}-\frac {2 c d \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}}{b}-\frac {\left (4 c^2 d^2\right ) \int \sqrt {\sin (2 a+2 b x)} \, dx}{\sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)} \sqrt {\sin (2 a+2 b x)}}\\ &=\frac {4 c d^3 \sqrt {c \sec (a+b x)}}{b (d \csc (a+b x))^{3/2}}-\frac {2 c d \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)}}{b}-\frac {4 c^2 d^2 E\left (\left .a-\frac {\pi }{4}+b x\right |2\right )}{b \sqrt {d \csc (a+b x)} \sqrt {c \sec (a+b x)} \sqrt {\sin (2 a+2 b x)}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 10.55, size = 99, normalized size = 0.79 \begin {gather*} -\frac {2 c d \sqrt {d \csc (a+b x)} \left (\cos (2 (a+b x)) \cot ^2(a+b x)+2 \cos ^2(a+b x) \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)} \tan ^2(a+b x)}{b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(-2*c*d*Sqrt[d*Csc[a + b*x]]*(Cos[2*(a + b*x)]*Cot[a + b*x]^2 + 2*Cos[a + b*x]^2*(-Cot[a + b*x]^2)^(1/4)*Hyper
geometric2F1[-1/2, 1/4, 1/2, Csc[a + b*x]^2])*Sqrt[c*Sec[a + b*x]]*Tan[a + b*x]^2)/b

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(497\) vs. \(2(136)=272\).
time = 58.21, size = 498, normalized size = 3.98

method	result	size
default	\(-\frac {\left (-4 \cos \left (b x +a \right ) \sqrt {\frac {1-\cos \left (b x +a \right )+\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 )}}\, \EllipticE \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )+2 \cos \left (b x +a \right ) \sqrt {\frac {1-\cos \left (b x +a \right )+\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 {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )-4 \sqrt {\frac {1-\cos \left (b x +a \right )+\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 )}}\, \EllipticE \left (\sqrt {\frac {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )+2 \sqrt {\frac {1-\cos \left (b x +a \right )+\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 {1-\cos \left (b x +a \right )+\sin \left (b x +a \right )}{\sin \left (b x +a \right )}}, \frac {\sqrt {2}}{2}\right )+2 \sqrt {2}\, \cos \left (b x +a \right )-\sqrt {2}\right ) \cos \left (b x +a \right ) \left (\frac {d}{\sin \left (b x +a \right )}\right )^{\frac {3}{2}} \left (\frac {c}{\cos \left (b x +a \right )}\right )^{\frac {3}{2}} \sin \left (b x +a \right ) \sqrt {2}}{b}\)	\(498\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*csc(b*x+a))^(3/2)*(c*sec(b*x+a))^(3/2),x,method=_RETURNVERBOSE)

[Out]

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

[Out]

integrate((d*csc(b*x + a))^(3/2)*(c*sec(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((d*csc(b*x+a))^(3/2)*(c*sec(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(-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))**(3/2)*(c*sec(b*x+a))**(3/2),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))^(3/2)*(c*sec(b*x+a))^(3/2),x, algorithm="giac")

[Out]

integrate((d*csc(b*x + a))^(3/2)*(c*sec(b*x + a))^(3/2), 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 )}^{3/2}\,{\left (\frac {d}{\sin \left (a+b\,x\right )}\right )}^{3/2} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((c/cos(a + b*x))^(3/2)*(d/sin(a + b*x))^(3/2),x)

[Out]

int((c/cos(a + b*x))^(3/2)*(d/sin(a + b*x))^(3/2), x)

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