3.6.45 \(\int (a+b \sin ^2(c+d x))^p \tan (c+d x) \, dx\) [545]

Optimal. Leaf size=59 \[ \frac {\, _2F_1\left (1,1+p;2+p;\frac {a+b \sin ^2(c+d x)}{a+b}\right ) \left (a+b \sin ^2(c+d x)\right )^{1+p}}{2 (a+b) d (1+p)} \]

[Out]

1/2*hypergeom([1, 1+p],[2+p],(a+b*sin(d*x+c)^2)/(a+b))*(a+b*sin(d*x+c)^2)^(1+p)/(a+b)/d/(1+p)

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Rubi [A]
time = 0.03, antiderivative size = 59, normalized size of antiderivative = 1.00, number of steps used = 2, number of rules used = 2, integrand size = 21, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.095, Rules used = {3273, 70} \begin {gather*} \frac {\left (a+b \sin ^2(c+d x)\right )^{p+1} \, _2F_1\left (1,p+1;p+2;\frac {b \sin ^2(c+d x)+a}{a+b}\right )}{2 d (p+1) (a+b)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(Hypergeometric2F1[1, 1 + p, 2 + p, (a + b*Sin[c + d*x]^2)/(a + b)]*(a + b*Sin[c + d*x]^2)^(1 + p))/(2*(a + b)
*d*(1 + p))

Rule 70

Int[((a_) + (b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_), x_Symbol] :> Simp[(b*c - a*d)^n*((a + b*x)^(m + 1)/(b^(
n + 1)*(m + 1)))*Hypergeometric2F1[-n, m + 1, m + 2, (-d)*((a + b*x)/(b*c - a*d))], x] /; FreeQ[{a, b, c, d, m
}, x] && NeQ[b*c - a*d, 0] &&  !IntegerQ[m] && IntegerQ[n]

Rule 3273

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

Rubi steps

\begin {align*} \int \left (a+b \sin ^2(c+d x)\right )^p \tan (c+d x) \, dx &=\frac {\text {Subst}\left (\int \frac {(a+b x)^p}{1-x} \, dx,x,\sin ^2(c+d x)\right )}{2 d}\\ &=\frac {\, _2F_1\left (1,1+p;2+p;\frac {a+b \sin ^2(c+d x)}{a+b}\right ) \left (a+b \sin ^2(c+d x)\right )^{1+p}}{2 (a+b) d (1+p)}\\ \end {align*}

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Mathematica [A]
time = 0.05, size = 61, normalized size = 1.03 \begin {gather*} \frac {\left (a+b-b \cos ^2(c+d x)\right )^{1+p} \, _2F_1\left (1,1+p;2+p;1-\frac {b \cos ^2(c+d x)}{a+b}\right )}{2 (a+b) d (1+p)} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((a + b - b*Cos[c + d*x]^2)^(1 + p)*Hypergeometric2F1[1, 1 + p, 2 + p, 1 - (b*Cos[c + d*x]^2)/(a + b)])/(2*(a
+ b)*d*(1 + p))

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Maple [F]
time = 0.52, size = 0, normalized size = 0.00 \[\int \left (a +\left (\sin ^{2}\left (d x +c \right )\right ) b \right )^{p} \tan \left (d x +c \right )\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

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

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Fricas [F]
time = 0.39, size = 25, normalized size = 0.42 \begin {gather*} {\rm integral}\left ({\left (-b \cos \left (d x + c\right )^{2} + a + b\right )}^{p} \tan \left (d x + c\right ), 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),x, algorithm="fricas")

[Out]

integral((-b*cos(d*x + c)^2 + a + b)^p*tan(d*x + c), x)

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

[Out]

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

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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),x, algorithm="giac")

[Out]

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

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.02 \begin {gather*} \int \mathrm {tan}\left (c+d\,x\right )\,{\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)*(a + b*sin(c + d*x)^2)^p,x)

[Out]

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

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