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(1/2+1/2*m,1+1/2*m,-p,3/2+1/2*m,-tan(f*x+e)^2,-b*tan(f*x+e)^2/a)*(sec(f*x+e)^2)^(1/2*m)*(d*sin(f*x+e))
^m*tan(f*x+e)*(a+b*tan(f*x+e)^2)^p/f/(1+m)/((1+b*tan(f*x+e)^2/a)^p)

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Rubi [A]  time = 0.14, antiderivative size = 121, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 25, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.120, 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 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])

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 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]

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*}

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Mathematica [B]  time = 2.43, 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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fricas [F]  time = 1.38, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sin \left (f x + e\right )\right )^{m}, x\right ) \]

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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giac [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sin \left (f x + e\right )\right )^{m}\,{d x} \]

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)

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maple [F]  time = 2.14, size = 0, normalized size = 0.00 \[ \int \left (d \sin \left (f x +e \right )\right )^{m} \left (a +b \left (\tan ^{2}\left (f x +e \right )\right )\right )^{p}\, dx \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((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 = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \tan \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sin \left (f x + e\right )\right )^{m}\,{d x} \]

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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mupad [F]  time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (d\,\sin \left (e+f\,x\right )\right )}^m\,{\left (b\,{\mathrm {tan}\left (e+f\,x\right )}^2+a\right )}^p \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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sympy [F(-1)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Timed out} \]

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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