3.496 \(\int (d \csc (e+f x))^m (b \tan ^2(e+f x))^p \, dx\)

Optimal. Leaf size=98 \[ \frac{\tan (e+f x) \cos ^2(e+f x)^{p+\frac{1}{2}} \left (b \tan ^2(e+f x)\right )^p (d \csc (e+f x))^m \text{Hypergeometric2F1}\left (\frac{1}{2} (2 p+1),\frac{1}{2} (-m+2 p+1),\frac{1}{2} (-m+2 p+3),\sin ^2(e+f x)\right )}{f (-m+2 p+1)} \]

[Out]

((Cos[e + f*x]^2)^(1/2 + p)*(d*Csc[e + f*x])^m*Hypergeometric2F1[(1 + 2*p)/2, (1 - m + 2*p)/2, (3 - m + 2*p)/2
, Sin[e + f*x]^2]*Tan[e + f*x]*(b*Tan[e + f*x]^2)^p)/(f*(1 - m + 2*p))

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Rubi [A]  time = 0.189325, antiderivative size = 98, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 4, integrand size = 23, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.174, Rules used = {3658, 2618, 2602, 2577} \[ \frac{\tan (e+f x) \cos ^2(e+f x)^{p+\frac{1}{2}} \left (b \tan ^2(e+f x)\right )^p (d \csc (e+f x))^m \, _2F_1\left (\frac{1}{2} (2 p+1),\frac{1}{2} (-m+2 p+1);\frac{1}{2} (-m+2 p+3);\sin ^2(e+f x)\right )}{f (-m+2 p+1)} \]

Antiderivative was successfully verified.

[In]

Int[(d*Csc[e + f*x])^m*(b*Tan[e + f*x]^2)^p,x]

[Out]

((Cos[e + f*x]^2)^(1/2 + p)*(d*Csc[e + f*x])^m*Hypergeometric2F1[(1 + 2*p)/2, (1 - m + 2*p)/2, (3 - m + 2*p)/2
, Sin[e + f*x]^2]*Tan[e + f*x]*(b*Tan[e + f*x]^2)^p)/(f*(1 - m + 2*p))

Rule 3658

Int[(u_.)*((b_.)*tan[(e_.) + (f_.)*(x_)]^(n_))^(p_), x_Symbol] :> With[{ff = FreeFactors[Tan[e + f*x], x]}, Di
st[((b*ff^n)^IntPart[p]*(b*Tan[e + f*x]^n)^FracPart[p])/(Tan[e + f*x]/ff)^(n*FracPart[p]), Int[ActivateTrig[u]
*(Tan[e + f*x]/ff)^(n*p), x], x]] /; FreeQ[{b, e, f, n, p}, x] &&  !IntegerQ[p] && IntegerQ[n] && (EqQ[u, 1] |
| MatchQ[u, ((d_.)*(trig_)[e + f*x])^(m_.) /; FreeQ[{d, m}, x] && MemberQ[{sin, cos, tan, cot, sec, csc}, trig
]])

Rule 2618

Int[(csc[(e_.) + (f_.)*(x_)]*(a_.))^(m_)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Csc[e + f*
x])^FracPart[m]*(Sin[e + f*x]/a)^FracPart[m], Int[(b*Tan[e + f*x])^n/(Sin[e + f*x]/a)^m, x], x] /; FreeQ[{a, b
, e, f, m, n}, x] &&  !IntegerQ[m] &&  !IntegerQ[n]

Rule 2602

Int[((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((b_.)*tan[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Dist[(a*Cos[e + f
*x]^(n + 1)*(b*Tan[e + f*x])^(n + 1))/(b*(a*Sin[e + f*x])^(n + 1)), 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]

Rule 2577

Int[(cos[(e_.) + (f_.)*(x_)]*(b_.))^(n_)*((a_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(b^(2*IntPart
[(n - 1)/2] + 1)*(b*Cos[e + f*x])^(2*FracPart[(n - 1)/2])*(a*Sin[e + f*x])^(m + 1)*Hypergeometric2F1[(1 + m)/2
, (1 - n)/2, (3 + m)/2, Sin[e + f*x]^2])/(a*f*(m + 1)*(Cos[e + f*x]^2)^FracPart[(n - 1)/2]), x] /; FreeQ[{a, b
, e, f, m, n}, x]

Rubi steps

\begin{align*} \int (d \csc (e+f x))^m \left (b \tan ^2(e+f x)\right )^p \, dx &=\left (\tan ^{-2 p}(e+f x) \left (b \tan ^2(e+f x)\right )^p\right ) \int (d \csc (e+f x))^m \tan ^{2 p}(e+f x) \, dx\\ &=\left ((d \csc (e+f x))^m \left (\frac{\sin (e+f x)}{d}\right )^m \tan ^{-2 p}(e+f x) \left (b \tan ^2(e+f x)\right )^p\right ) \int \left (\frac{\sin (e+f x)}{d}\right )^{-m} \tan ^{2 p}(e+f x) \, dx\\ &=\frac{\left (\cos ^{2 p}(e+f x) (d \csc (e+f x))^{1+m} \sin (e+f x) \left (\frac{\sin (e+f x)}{d}\right )^{m-2 p} \left (b \tan ^2(e+f x)\right )^p\right ) \int \cos ^{-2 p}(e+f x) \left (\frac{\sin (e+f x)}{d}\right )^{-m+2 p} \, dx}{d}\\ &=\frac{\cos ^2(e+f x)^{\frac{1}{2}+p} (d \csc (e+f x))^{1+m} \, _2F_1\left (\frac{1}{2} (1+2 p),\frac{1}{2} (1-m+2 p);\frac{1}{2} (3-m+2 p);\sin ^2(e+f x)\right ) \sin (e+f x) \tan (e+f x) \left (b \tan ^2(e+f x)\right )^p}{d f (1-m+2 p)}\\ \end{align*}

Mathematica [C]  time = 1.92321, size = 299, normalized size = 3.05 \[ -\frac{d (m-2 p-3) \left (b \tan ^2(e+f x)\right )^p (d \csc (e+f x))^{m-1} F_1\left (-\frac{m}{2}+p+\frac{1}{2};2 p,1-m;-\frac{m}{2}+p+\frac{3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )}{f (m-2 p-1) \left (2 \tan ^2\left (\frac{1}{2} (e+f x)\right ) \left (-(m-1) F_1\left (-\frac{m}{2}+p+\frac{3}{2};2 p,2-m;-\frac{m}{2}+p+\frac{5}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )-2 p F_1\left (-\frac{m}{2}+p+\frac{3}{2};2 p+1,1-m;-\frac{m}{2}+p+\frac{5}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )+(m-2 p-3) F_1\left (-\frac{m}{2}+p+\frac{1}{2};2 p,1-m;-\frac{m}{2}+p+\frac{3}{2};\tan ^2\left (\frac{1}{2} (e+f x)\right ),-\tan ^2\left (\frac{1}{2} (e+f x)\right )\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

Integrate[(d*Csc[e + f*x])^m*(b*Tan[e + f*x]^2)^p,x]

[Out]

-((d*(-3 + m - 2*p)*AppellF1[1/2 - m/2 + p, 2*p, 1 - m, 3/2 - m/2 + p, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2
]*(d*Csc[e + f*x])^(-1 + m)*(b*Tan[e + f*x]^2)^p)/(f*(-1 + m - 2*p)*((-3 + m - 2*p)*AppellF1[1/2 - m/2 + p, 2*
p, 1 - m, 3/2 - m/2 + p, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + 2*(-((-1 + m)*AppellF1[3/2 - m/2 + p, 2*p,
2 - m, 5/2 - m/2 + p, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]) - 2*p*AppellF1[3/2 - m/2 + p, 1 + 2*p, 1 - m,
5/2 - m/2 + p, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)))

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Maple [F]  time = 0.924, size = 0, normalized size = 0. \begin{align*} \int \left ( d\csc \left ( fx+e \right ) \right ) ^{m} \left ( b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((d*csc(f*x+e))^m*(b*tan(f*x+e)^2)^p,x)

[Out]

int((d*csc(f*x+e))^m*(b*tan(f*x+e)^2)^p,x)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \tan \left (f x + e\right )^{2}\right )^{p} \left (d \csc \left (f x + e\right )\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*csc(f*x+e))^m*(b*tan(f*x+e)^2)^p,x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e)^2)^p*(d*csc(f*x + e))^m, x)

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Fricas [F]  time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\left (b \tan \left (f x + e\right )^{2}\right )^{p} \left (d \csc \left (f x + e\right )\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*csc(f*x+e))^m*(b*tan(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

integral((b*tan(f*x + e)^2)^p*(d*csc(f*x + e))^m, x)

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Sympy [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \tan ^{2}{\left (e + f x \right )}\right )^{p} \left (d \csc{\left (e + f x \right )}\right )^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*csc(f*x+e))**m*(b*tan(f*x+e)**2)**p,x)

[Out]

Integral((b*tan(e + f*x)**2)**p*(d*csc(e + f*x))**m, x)

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (b \tan \left (f x + e\right )^{2}\right )^{p} \left (d \csc \left (f x + e\right )\right )^{m}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*csc(f*x+e))^m*(b*tan(f*x+e)^2)^p,x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e)^2)^p*(d*csc(f*x + e))^m, x)