3.543 \(\int (a+b \sin ^2(e+f x))^p (d \tan (e+f x))^m \, dx\)

Optimal. Leaf size=120 \[ \frac{\cos ^2(e+f x)^{\frac{m+1}{2}} (d \tan (e+f x))^{m+1} \left (a+b \sin ^2(e+f x)\right )^p \left (\frac{b \sin ^2(e+f x)}{a}+1\right )^{-p} F_1\left (\frac{m+1}{2};\frac{m+1}{2},-p;\frac{m+3}{2};\sin ^2(e+f x),-\frac{b \sin ^2(e+f x)}{a}\right )}{d f (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.121811, antiderivative size = 120, normalized size of antiderivative = 1., number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.12, Rules used = {3197, 511, 510} \[ \frac{\cos ^2(e+f x)^{\frac{m+1}{2}} (d \tan (e+f x))^{m+1} \left (a+b \sin ^2(e+f x)\right )^p \left (\frac{b \sin ^2(e+f x)}{a}+1\right )^{-p} F_1\left (\frac{m+1}{2};\frac{m+1}{2},-p;\frac{m+3}{2};\sin ^2(e+f x),-\frac{b \sin ^2(e+f x)}{a}\right )}{d f (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3197

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

Rule 511

Int[((e_.)*(x_))^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_))^(q_), x_Symbol] :> Dist[(a^IntPa
rt[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 510

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)*AppellF1[(m + 1)/n, -p, -q, 1 + (m + 1)/n, -((b*x^n)/a), -((d*x^n)/c)])/(e*(m + 1)), 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])

Rubi steps

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

Mathematica [A]  time = 0.501859, size = 121, normalized size = 1.01 \[ \frac{\tan (e+f x) \cos ^2(e+f x)^{\frac{m+1}{2}} (d \tan (e+f x))^m \left (a+b \sin ^2(e+f x)\right )^p \left (\frac{b \sin ^2(e+f x)}{a}+1\right )^{-p} F_1\left (\frac{m+1}{2};\frac{m+1}{2},-p;\frac{m+3}{2};\sin ^2(e+f x),-\frac{b \sin ^2(e+f x)}{a}\right )}{f (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [F]  time = 1.605, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b \left ( \sin \left ( fx+e \right ) \right ) ^{2} \right ) ^{p} \left ( d\tan \left ( fx+e \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((b*sin(f*x + e)^2 + a)^p*(d*tan(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 \cos \left (f x + e\right )^{2} + a + b\right )}^{p} \left (d \tan \left (f x + e\right )\right )^{m}, x\right ) \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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