3.5.97 \(\int (d \csc (e+f x))^m (a+b \tan ^2(e+f x))^p \, dx\) [497]

Optimal. Leaf size=127 \[ \frac {F_1\left (\frac {1-m}{2};1-\frac {m}{2},-p;\frac {3-m}{2};-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right ) (d \csc (e+f x))^m \sec ^2(e+f x)^{-m/2} \tan (e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a}\right )^{-p}}{f (1-m)} \]

[Out]

AppellF1(1/2-1/2*m,1-1/2*m,-p,3/2-1/2*m,-tan(f*x+e)^2,-b*tan(f*x+e)^2/a)*(d*csc(f*x+e))^m*tan(f*x+e)*(a+b*tan(
f*x+e)^2)^p/f/(1-m)/((sec(f*x+e)^2)^(1/2*m))/((1+b*tan(f*x+e)^2/a)^p)

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Rubi [A]
time = 0.12, antiderivative size = 127, normalized size of antiderivative = 1.00, number of steps used = 4, number of rules used = 4, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.160, Rules used = {3762, 3748, 525, 524} \begin {gather*} \frac {\tan (e+f x) \sec ^2(e+f x)^{-m/2} (d \csc (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \left (\frac {b \tan ^2(e+f x)}{a}+1\right )^{-p} F_1\left (\frac {1-m}{2};1-\frac {m}{2},-p;\frac {3-m}{2};-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right )}{f (1-m)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 524

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Simp[a^p*c^q*
((e*x)^(m + 1)/(e*(m + 1)))*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, (-b)*(x^n/a), (-d)*(x^n/c)], x] /; Free
Q[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] && (IntegerQ[p] || GtQ[a
, 0]) && (IntegerQ[q] || GtQ[c, 0])

Rule 525

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[a^IntPar
t[p]*((a + b*x^n)^FracPart[p]/(1 + b*(x^n/a))^FracPart[p]), Int[(e*x)^m*(1 + b*(x^n/a))^p*(c + d*x^n)^q, x], x
] /; FreeQ[{a, b, c, d, e, m, n, p, q}, x] && NeQ[b*c - a*d, 0] && NeQ[m, -1] && NeQ[m, n - 1] &&  !(IntegerQ[
p] || GtQ[a, 0])

Rule 3748

Int[((d_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((a_) + (b_.)*tan[(e_.) + (f_.)*(x_)]^2)^(p_), x_Symbol] :> With[{ff
= FreeFactors[Tan[e + f*x], x]}, Dist[ff*(d*Sin[e + f*x])^m*((Sec[e + f*x]^2)^(m/2)/(f*Tan[e + f*x]^m)), Subst
[Int[(ff*x)^m*((a + b*ff^2*x^2)^p/(1 + ff^2*x^2)^(m/2 + 1)), x], x, Tan[e + f*x]/ff], x]] /; FreeQ[{a, b, d, e
, f, m, p}, x] &&  !IntegerQ[m]

Rule 3762

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

Rubi steps

\begin {align*} \int (d \csc (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \, dx &=\left ((d \csc (e+f x))^m \left (\frac {\sin (e+f x)}{d}\right )^m\right ) \int \left (\frac {\sin (e+f x)}{d}\right )^{-m} \left (a+b \tan ^2(e+f x)\right )^p \, dx\\ &=\frac {\left ((d \csc (e+f x))^m \sec ^2(e+f x)^{-m/2} \tan ^m(e+f x)\right ) \text {Subst}\left (\int x^{-m} \left (1+x^2\right )^{-1+\frac {m}{2}} \left (a+b x^2\right )^p \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {\left ((d \csc (e+f x))^m \sec ^2(e+f x)^{-m/2} \tan ^m(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a}\right )^{-p}\right ) \text {Subst}\left (\int x^{-m} \left (1+x^2\right )^{-1+\frac {m}{2}} \left (1+\frac {b x^2}{a}\right )^p \, dx,x,\tan (e+f x)\right )}{f}\\ &=\frac {F_1\left (\frac {1-m}{2};1-\frac {m}{2},-p;\frac {3-m}{2};-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right ) (d \csc (e+f x))^m \sec ^2(e+f x)^{-m/2} \tan (e+f x) \left (a+b \tan ^2(e+f x)\right )^p \left (1+\frac {b \tan ^2(e+f x)}{a}\right )^{-p}}{f (1-m)}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(292\) vs. \(2(127)=254\).
time = 2.40, size = 292, normalized size = 2.30 \begin {gather*} -\frac {a (-3+m) F_1\left (\frac {1}{2}-\frac {m}{2};1-\frac {m}{2},-p;\frac {3}{2}-\frac {m}{2};-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right ) \cos ^2(e+f x) \cot (e+f x) (d \csc (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p}{f (-1+m) \left (-2 b p F_1\left (\frac {3}{2}-\frac {m}{2};1-\frac {m}{2},1-p;\frac {5}{2}-\frac {m}{2};-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right )-a (-2+m) F_1\left (\frac {3}{2}-\frac {m}{2};2-\frac {m}{2},-p;\frac {5}{2}-\frac {m}{2};-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right )+a (-3+m) F_1\left (\frac {1}{2}-\frac {m}{2};1-\frac {m}{2},-p;\frac {3}{2}-\frac {m}{2};-\tan ^2(e+f x),-\frac {b \tan ^2(e+f x)}{a}\right ) \cot ^2(e+f x)\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

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

[Out]

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

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Maple [F]
time = 0.41, size = 0, normalized size = 0.00 \[\int \left (d \csc \left (f x +e \right )\right )^{m} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

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Fricas [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(f*x+e))^m*(a+b*tan(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

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

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

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int {\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^p\,{\left (\frac {d}{\sin \left (e+f\,x\right )}\right )}^m \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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