3.6.49 \(\int (a+b \sin ^2(c+d x))^p \tan ^2(c+d x) \, dx\) [549]

Optimal. Leaf size=101 \[ \frac {F_1\left (\frac {3}{2};\frac {3}{2},-p;\frac {5}{2};\sin ^2(c+d x),-\frac {b \sin ^2(c+d x)}{a}\right ) \sqrt {\cos ^2(c+d x)} \sin ^2(c+d x) \left (a+b \sin ^2(c+d x)\right )^p \left (1+\frac {b \sin ^2(c+d x)}{a}\right )^{-p} \tan (c+d x)}{3 d} \]

[Out]

1/3*AppellF1(3/2,3/2,-p,5/2,sin(d*x+c)^2,-b*sin(d*x+c)^2/a)*sin(d*x+c)^2*(a+b*sin(d*x+c)^2)^p*(cos(d*x+c)^2)^(
1/2)*tan(d*x+c)/d/((1+b*sin(d*x+c)^2/a)^p)

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Rubi [A]
time = 0.07, antiderivative size = 101, 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 = {3275, 525, 524} \begin {gather*} \frac {\sin ^2(c+d x) \sqrt {\cos ^2(c+d x)} \tan (c+d x) \left (a+b \sin ^2(c+d x)\right )^p \left (\frac {b \sin ^2(c+d x)}{a}+1\right )^{-p} F_1\left (\frac {3}{2};\frac {3}{2},-p;\frac {5}{2};\sin ^2(c+d x),-\frac {b \sin ^2(c+d x)}{a}\right )}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[(a + b*Sin[c + d*x]^2)^p*Tan[c + d*x]^2,x]

[Out]

(AppellF1[3/2, 3/2, -p, 5/2, Sin[c + d*x]^2, -((b*Sin[c + d*x]^2)/a)]*Sqrt[Cos[c + d*x]^2]*Sin[c + d*x]^2*(a +
b*Sin[c + d*x]^2)^p*Tan[c + d*x])/(3*d*(1 + (b*Sin[c + d*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 3275

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

Rubi steps

\begin {align*} \int \left (a+b \sin ^2(c+d x)\right )^p \tan ^2(c+d x) \, dx &=\frac {\left (\sqrt {\cos ^2(c+d x)} \sec (c+d x)\right ) \text {Subst}\left (\int \frac {x^2 \left (a+b x^2\right )^p}{\left (1-x^2\right )^{3/2}} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {\left (\sqrt {\cos ^2(c+d x)} \sec (c+d x) \left (a+b \sin ^2(c+d x)\right )^p \left (1+\frac {b \sin ^2(c+d x)}{a}\right )^{-p}\right ) \text {Subst}\left (\int \frac {x^2 \left (1+\frac {b x^2}{a}\right )^p}{\left (1-x^2\right )^{3/2}} \, dx,x,\sin (c+d x)\right )}{d}\\ &=\frac {F_1\left (\frac {3}{2};\frac {3}{2},-p;\frac {5}{2};\sin ^2(c+d x),-\frac {b \sin ^2(c+d x)}{a}\right ) \sqrt {\cos ^2(c+d x)} \sin ^2(c+d x) \left (a+b \sin ^2(c+d x)\right )^p \left (1+\frac {b \sin ^2(c+d x)}{a}\right )^{-p} \tan (c+d x)}{3 d}\\ \end {align*}

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Mathematica [A]
time = 0.22, size = 102, normalized size = 1.01 \begin {gather*} \frac {F_1\left (\frac {3}{2};\frac {3}{2},-p;\frac {5}{2};\sin ^2(c+d x),-\frac {b \sin ^2(c+d x)}{a}\right ) \sqrt {\cos ^2(c+d x)} \sin ^2(c+d x) \left (a+b \sin ^2(c+d x)\right )^p \left (\frac {a+b \sin ^2(c+d x)}{a}\right )^{-p} \tan (c+d x)}{3 d} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[(a + b*Sin[c + d*x]^2)^p*Tan[c + d*x]^2,x]

[Out]

(AppellF1[3/2, 3/2, -p, 5/2, Sin[c + d*x]^2, -((b*Sin[c + d*x]^2)/a)]*Sqrt[Cos[c + d*x]^2]*Sin[c + d*x]^2*(a +
b*Sin[c + d*x]^2)^p*Tan[c + d*x])/(3*d*((a + b*Sin[c + d*x]^2)/a)^p)

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Maple [F]
time = 0.50, size = 0, normalized size = 0.00 \[\int \left (a +\left (\sin ^{2}\left (d x +c \right )\right ) b \right )^{p} \left (\tan ^{2}\left (d x +c \right )\right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+sin(d*x+c)^2*b)^p*tan(d*x+c)^2,x)

[Out]

int((a+sin(d*x+c)^2*b)^p*tan(d*x+c)^2,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((a+b*sin(d*x+c)^2)^p*tan(d*x+c)^2,x, algorithm="maxima")

[Out]

integrate((b*sin(d*x + c)^2 + a)^p*tan(d*x + c)^2, x)

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Fricas [F]
time = 0.39, size = 27, normalized size = 0.27 \begin {gather*} {\rm integral}\left ({\left (-b \cos \left (d x + c\right )^{2} + a + b\right )}^{p} \tan \left (d x + c\right )^{2}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+b*sin(d*x+c)^2)^p*tan(d*x+c)^2,x, algorithm="fricas")

[Out]

integral((-b*cos(d*x + c)^2 + a + b)^p*tan(d*x + c)^2, 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((a+b*sin(d*x+c)**2)**p*tan(d*x+c)**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((a+b*sin(d*x+c)^2)^p*tan(d*x+c)^2,x, algorithm="giac")

[Out]

integrate((b*sin(d*x + c)^2 + a)^p*tan(d*x + c)^2, x)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(tan(c + d*x)^2*(a + b*sin(c + d*x)^2)^p,x)

[Out]

int(tan(c + d*x)^2*(a + b*sin(c + d*x)^2)^p, x)

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