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

Optimal. Leaf size=121 \[ \frac{\tan (e+f x) \sec ^2(e+f x)^{m/2} (d \sin (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \left (\frac{b \tan ^2(e+f x)}{a}+1\right )^{-p} F_1\left (\frac{m+1}{2};\frac{m+2}{2},-p;\frac{m+3}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a}\right )}{f (m+1)} \]

[Out]

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

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Rubi [A]  time = 0.141187, antiderivative size = 121, 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 = {3667, 511, 510} \[ \frac{\tan (e+f x) \sec ^2(e+f x)^{m/2} (d \sin (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \left (\frac{b \tan ^2(e+f x)}{a}+1\right )^{-p} F_1\left (\frac{m+1}{2};\frac{m+2}{2},-p;\frac{m+3}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a}\right )}{f (m+1)} \]

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 3667

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 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 (d \sin (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p \, dx &=\frac{\left (\sec ^2(e+f x)^{m/2} (d \sin (e+f x))^m \tan ^{-m}(e+f x)\right ) \operatorname{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 (\sec ^2(e+f x)^{m/2} (d \sin (e+f x))^m \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 ) \operatorname{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};\frac{2+m}{2},-p;\frac{3+m}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a}\right ) \sec ^2(e+f x)^{m/2} (d \sin (e+f x))^m \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*}

Mathematica [B]  time = 2.3441, size = 275, normalized size = 2.27 \[ \frac{a (m+3) \sin (e+f x) \cos (e+f x) (d \sin (e+f x))^m \left (a+b \tan ^2(e+f x)\right )^p F_1\left (\frac{m+1}{2};\frac{m+2}{2},-p;\frac{m+3}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a}\right )}{f (m+1) \left (\tan ^2(e+f x) \left (2 b p F_1\left (\frac{m+3}{2};\frac{m+2}{2},1-p;\frac{m+5}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a}\right )-a (m+2) F_1\left (\frac{m+3}{2};\frac{m+4}{2},-p;\frac{m+5}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a}\right )\right )+a (m+3) F_1\left (\frac{m+1}{2};\frac{m+2}{2},-p;\frac{m+3}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a}\right )\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

[Out]

(a*(3 + m)*AppellF1[(1 + m)/2, (2 + m)/2, -p, (3 + m)/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)]*Cos[e + f*x
]*Sin[e + f*x]*(d*Sin[e + f*x])^m*(a + b*Tan[e + f*x]^2)^p)/(f*(1 + m)*(a*(3 + m)*AppellF1[(1 + m)/2, (2 + m)/
2, -p, (3 + m)/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)] + (2*b*p*AppellF1[(3 + m)/2, (2 + m)/2, 1 - p, (5
+ m)/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)] - a*(2 + m)*AppellF1[(3 + m)/2, (4 + m)/2, -p, (5 + m)/2, -T
an[e + f*x]^2, -((b*Tan[e + f*x]^2)/a)])*Tan[e + f*x]^2))

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Maple [F]  time = 0.658, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sin \left ( fx+e \right ) \right ) ^{m} \left ( a+b \left ( \tan \left ( fx+e \right ) \right ) ^{2} \right ) ^{p}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((d*sin(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*sin(f*x + e))^m, x)