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

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Rubi [A]  time = 0.12, antiderivative size = 120, 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 = {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 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 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]

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*}

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Mathematica [A]  time = 0.50, 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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fricas [F]  time = 0.81, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (-b \cos \left (f x + e\right )^{2} + a + b\right )}^{p} \left (d \tan \left (f x + e\right )\right )^{m}, x\right ) \]

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

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)

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maple [F]  time = 4.01, size = 0, normalized size = 0.00 \[ \int \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )^{p} \left (d \tan \left (f x +e \right )\right )^{m}\, dx \]

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

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

[Out]

Timed out

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