3.2.58 \(\int \csc ^3(e+f x) (a+b \tan ^2(e+f x))^p \, dx\) [158]

Optimal. Leaf size=92 \[ \frac {F_1\left (\frac {3}{2};2,-p;\frac {5}{2};\sec ^2(e+f x),-\frac {b \sec ^2(e+f x)}{a-b}\right ) \sec ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^p \left (1+\frac {b \sec ^2(e+f x)}{a-b}\right )^{-p}}{3 f} \]

[Out]

1/3*AppellF1(3/2,2,-p,5/2,sec(f*x+e)^2,-b*sec(f*x+e)^2/(a-b))*sec(f*x+e)^3*(a-b+b*sec(f*x+e)^2)^p/f/((1+b*sec(
f*x+e)^2/(a-b))^p)

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Rubi [A]
time = 0.07, antiderivative size = 92, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.130, Rules used = {3745, 525, 524} \begin {gather*} \frac {\sec ^3(e+f x) \left (a+b \sec ^2(e+f x)-b\right )^p \left (\frac {b \sec ^2(e+f x)}{a-b}+1\right )^{-p} F_1\left (\frac {3}{2};2,-p;\frac {5}{2};\sec ^2(e+f x),-\frac {b \sec ^2(e+f x)}{a-b}\right )}{3 f} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[Csc[e + f*x]^3*(a + b*Tan[e + f*x]^2)^p,x]

[Out]

(AppellF1[3/2, 2, -p, 5/2, Sec[e + f*x]^2, -((b*Sec[e + f*x]^2)/(a - b))]*Sec[e + f*x]^3*(a - b + b*Sec[e + f*
x]^2)^p)/(3*f*(1 + (b*Sec[e + f*x]^2)/(a - b))^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 3745

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

Rubi steps

\begin {align*} \int \csc ^3(e+f x) \left (a+b \tan ^2(e+f x)\right )^p \, dx &=\frac {\text {Subst}\left (\int \frac {x^2 \left (a-b+b x^2\right )^p}{\left (-1+x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac {\left (\left (a-b+b \sec ^2(e+f x)\right )^p \left (1+\frac {b \sec ^2(e+f x)}{a-b}\right )^{-p}\right ) \text {Subst}\left (\int \frac {x^2 \left (1+\frac {b x^2}{a-b}\right )^p}{\left (-1+x^2\right )^2} \, dx,x,\sec (e+f x)\right )}{f}\\ &=\frac {F_1\left (\frac {3}{2};2,-p;\frac {5}{2};\sec ^2(e+f x),-\frac {b \sec ^2(e+f x)}{a-b}\right ) \sec ^3(e+f x) \left (a-b+b \sec ^2(e+f x)\right )^p \left (1+\frac {b \sec ^2(e+f x)}{a-b}\right )^{-p}}{3 f}\\ \end {align*}

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Mathematica [B] Leaf count is larger than twice the leaf count of optimal. \(252\) vs. \(2(92)=184\).
time = 15.05, size = 252, normalized size = 2.74 \begin {gather*} \frac {b (-3+2 p) F_1\left (\frac {1}{2}-p;-\frac {1}{2},-p;\frac {3}{2}-p;-\cot ^2(e+f x),-\frac {a \cot ^2(e+f x)}{b}\right ) \cot (e+f x) \csc (e+f x) \left (a+b \tan ^2(e+f x)\right )^p}{f (-1+2 p) \left (b (-3+2 p) F_1\left (\frac {1}{2}-p;-\frac {1}{2},-p;\frac {3}{2}-p;-\cot ^2(e+f x),-\frac {a \cot ^2(e+f x)}{b}\right )-\left (2 a p F_1\left (\frac {3}{2}-p;-\frac {1}{2},1-p;\frac {5}{2}-p;-\cot ^2(e+f x),-\frac {a \cot ^2(e+f x)}{b}\right )+b F_1\left (\frac {3}{2}-p;\frac {1}{2},-p;\frac {5}{2}-p;-\cot ^2(e+f x),-\frac {a \cot ^2(e+f x)}{b}\right )\right ) \cot ^2(e+f x)\right )} \end {gather*}

Warning: Unable to verify antiderivative.

[In]

Integrate[Csc[e + f*x]^3*(a + b*Tan[e + f*x]^2)^p,x]

[Out]

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

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Maple [F]
time = 0.22, size = 0, normalized size = 0.00 \[\int \left (\csc ^{3}\left (f x +e \right )\right ) \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{p}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^p,x)

[Out]

int(csc(f*x+e)^3*(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(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^p,x, algorithm="maxima")

[Out]

integrate((b*tan(f*x + e)^2 + a)^p*csc(f*x + e)^3, 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(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^p,x, algorithm="fricas")

[Out]

integral((b*tan(f*x + e)^2 + a)^p*csc(f*x + e)^3, 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(csc(f*x+e)**3*(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(csc(f*x+e)^3*(a+b*tan(f*x+e)^2)^p,x, algorithm="giac")

[Out]

integrate((b*tan(f*x + e)^2 + a)^p*csc(f*x + e)^3, x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^p}{{\sin \left (e+f\,x\right )}^3} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + b*tan(e + f*x)^2)^p/sin(e + f*x)^3,x)

[Out]

int((a + b*tan(e + f*x)^2)^p/sin(e + f*x)^3, x)

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