3.63 \(\int \csc ^5(a+b x) \sqrt{d \tan (a+b x)} \, dx\)

Optimal. Leaf size=105 \[ -\frac{2 d \csc ^3(a+b x)}{7 b \sqrt{d \tan (a+b x)}}-\frac{4 d \csc (a+b x)}{7 b \sqrt{d \tan (a+b x)}}+\frac{4 \sqrt{\sin (2 a+2 b x)} \csc (a+b x) F\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{7 b} \]

[Out]

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

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Rubi [A]  time = 0.142963, antiderivative size = 105, 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 = {2599, 2601, 2573, 2641} \[ -\frac{2 d \csc ^3(a+b x)}{7 b \sqrt{d \tan (a+b x)}}-\frac{4 d \csc (a+b x)}{7 b \sqrt{d \tan (a+b x)}}+\frac{4 \sqrt{\sin (2 a+2 b x)} \csc (a+b x) F\left (\left .a+b x-\frac{\pi }{4}\right |2\right ) \sqrt{d \tan (a+b x)}}{7 b} \]

Antiderivative was successfully verified.

[In]

Int[Csc[a + b*x]^5*Sqrt[d*Tan[a + b*x]],x]

[Out]

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

Rule 2599

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_.), x_Symbol] :> Simp[(b*(a*Sin[e
+ f*x])^(m + 2)*(b*Tan[e + f*x])^(n - 1))/(a^2*f*(m + n + 1)), x] + Dist[(m + 2)/(a^2*(m + n + 1)), Int[(a*Sin
[e + f*x])^(m + 2)*(b*Tan[e + f*x])^n, x], x] /; FreeQ[{a, b, e, f, n}, x] && LtQ[m, -1] && NeQ[m + n + 1, 0]
&& IntegersQ[2*m, 2*n]

Rule 2601

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(Cos[e + f*x
]^n*(b*Tan[e + f*x])^n)/(a*Sin[e + f*x])^n, Int[(a*Sin[e + f*x])^(m + n)/Cos[e + f*x]^n, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[n] && (ILtQ[m, 0] || (EqQ[m, 1] && EqQ[n, -2^(-1)]) || IntegersQ[m - 1/2, n -
1/2])

Rule 2573

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 2641

Int[1/Sqrt[sin[(c_.) + (d_.)*(x_)]], x_Symbol] :> Simp[(2*EllipticF[(1*(c - Pi/2 + d*x))/2, 2])/d, x] /; FreeQ
[{c, d}, x]

Rubi steps

\begin{align*} \int \csc ^5(a+b x) \sqrt{d \tan (a+b x)} \, dx &=-\frac{2 d \csc ^3(a+b x)}{7 b \sqrt{d \tan (a+b x)}}+\frac{6}{7} \int \csc ^3(a+b x) \sqrt{d \tan (a+b x)} \, dx\\ &=-\frac{4 d \csc (a+b x)}{7 b \sqrt{d \tan (a+b x)}}-\frac{2 d \csc ^3(a+b x)}{7 b \sqrt{d \tan (a+b x)}}+\frac{4}{7} \int \csc (a+b x) \sqrt{d \tan (a+b x)} \, dx\\ &=-\frac{4 d \csc (a+b x)}{7 b \sqrt{d \tan (a+b x)}}-\frac{2 d \csc ^3(a+b x)}{7 b \sqrt{d \tan (a+b x)}}+\frac{\left (4 \sqrt{\cos (a+b x)} \sqrt{d \tan (a+b x)}\right ) \int \frac{1}{\sqrt{\cos (a+b x)} \sqrt{\sin (a+b x)}} \, dx}{7 \sqrt{\sin (a+b x)}}\\ &=-\frac{4 d \csc (a+b x)}{7 b \sqrt{d \tan (a+b x)}}-\frac{2 d \csc ^3(a+b x)}{7 b \sqrt{d \tan (a+b x)}}+\frac{1}{7} \left (4 \csc (a+b x) \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}\right ) \int \frac{1}{\sqrt{\sin (2 a+2 b x)}} \, dx\\ &=-\frac{4 d \csc (a+b x)}{7 b \sqrt{d \tan (a+b x)}}-\frac{2 d \csc ^3(a+b x)}{7 b \sqrt{d \tan (a+b x)}}+\frac{4 \csc (a+b x) F\left (\left .a-\frac{\pi }{4}+b x\right |2\right ) \sqrt{\sin (2 a+2 b x)} \sqrt{d \tan (a+b x)}}{7 b}\\ \end{align*}

Mathematica [C]  time = 1.42776, size = 124, normalized size = 1.18 \[ -\frac{2 d \cos (2 (a+b x)) \csc ^3(a+b x) \left ((\cos (2 (a+b x))-2) \sec ^2(a+b x)^{3/2}-4 \sqrt [4]{-1} \tan ^{\frac{7}{2}}(a+b x) F\left (\left .i \sinh ^{-1}\left (\sqrt [4]{-1} \sqrt{\tan (a+b x)}\right )\right |-1\right )\right )}{7 b \left (\tan ^2(a+b x)-1\right ) \sqrt{\sec ^2(a+b x)} \sqrt{d \tan (a+b x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[Csc[a + b*x]^5*Sqrt[d*Tan[a + b*x]],x]

[Out]

(-2*d*Cos[2*(a + b*x)]*Csc[a + b*x]^3*((-2 + Cos[2*(a + b*x)])*(Sec[a + b*x]^2)^(3/2) - 4*(-1)^(1/4)*EllipticF
[I*ArcSinh[(-1)^(1/4)*Sqrt[Tan[a + b*x]]], -1]*Tan[a + b*x]^(7/2)))/(7*b*Sqrt[Sec[a + b*x]^2]*Sqrt[d*Tan[a + b
*x]]*(-1 + Tan[a + b*x]^2))

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Maple [B]  time = 0.189, size = 558, normalized size = 5.3 \begin{align*} \text{result too large to display} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(b*x+a)^5*(d*tan(b*x+a))^(1/2),x)

[Out]

-1/7/b*2^(1/2)*(cos(b*x+a)-1)^2*(4*EllipticF((-(cos(b*x+a)-1-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)*((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)*cos(b*x+a)^3*sin(b*x+a)+4*EllipticF((-(cos(b*x+a)-1-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)*((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)*cos(b*x+a)^2*sin(b*x+a)-4*cos(b*x+a)*((cos(b*x+a)-1)/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))/sin(b*x+a))^(1/2)*sin(b*x+a)*EllipticF((-(cos(b*x+a)-1-sin(b*x+a))/
sin(b*x+a))^(1/2),1/2*2^(1/2))-4*sin(b*x+a)*EllipticF((-(cos(b*x+a)-1-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)*((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)-2*cos(b*x+a)^3*2^(1/2)+3*cos(b*x+a)*2^(1/2))*(cos(b*x+a)+1)^2*(d*sin(b*x+a)/cos(b*x+a))^(1/2
)/sin(b*x+a)^8

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{d \tan \left (b x + a\right )} \csc \left (b x + a\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^5*(d*tan(b*x+a))^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(d*tan(b*x + a))*csc(b*x + a)^5, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{d \tan \left (b x + a\right )} \csc \left (b x + a\right )^{5}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^5*(d*tan(b*x+a))^(1/2),x, algorithm="fricas")

[Out]

integral(sqrt(d*tan(b*x + a))*csc(b*x + a)^5, 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(csc(b*x+a)**5*(d*tan(b*x+a))**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \sqrt{d \tan \left (b x + a\right )} \csc \left (b x + a\right )^{5}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(csc(b*x+a)^5*(d*tan(b*x+a))^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(d*tan(b*x + a))*csc(b*x + a)^5, x)