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

Optimal. Leaf size=123 \[ \frac{\tan (e+f x) \cos ^2(e+f x)^{p+\frac{1}{2}} (d \sin (e+f x))^m \left (\frac{-a \sin ^2(e+f x)+a+b}{a+b}\right )^{-p} \left (a+b \sec ^2(e+f x)\right )^p F_1\left (\frac{m+1}{2};p+\frac{1}{2},-p;\frac{m+3}{2};\sin ^2(e+f x),\frac{a \sin ^2(e+f x)}{a+b}\right )}{f (m+1)} \]

[Out]

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

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Rubi [F]  time = 0.0454144, antiderivative size = 0, normalized size of antiderivative = 0., number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0., Rules used = {} \[ \int \left (a+b \sec ^2(e+f x)\right )^p (d \sin (e+f x))^m \, dx \]

Verification is Not applicable to the result.

[In]

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

[Out]

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

Rubi steps

\begin{align*} \int \left (a+b \sec ^2(e+f x)\right )^p (d \sin (e+f x))^m \, dx &=\int \left (a+b \sec ^2(e+f x)\right )^p (d \sin (e+f x))^m \, dx\\ \end{align*}

Mathematica [B]  time = 4.05454, size = 286, normalized size = 2.33 \[ \frac{\sin (e+f x) \cos (e+f x) (d \sin (e+f x))^m \left (a+b \sec ^2(e+f x)\right )^p F_1\left (\frac{m+1}{2};\frac{m+2}{2},-p;\frac{m+3}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a+b}\right )}{f (m+1) \left (F_1\left (\frac{m+1}{2};\frac{m+2}{2},-p;\frac{m+3}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a+b}\right )-\frac{\tan ^2(e+f x) \left ((m+2) (a+b) F_1\left (\frac{m+3}{2};\frac{m+4}{2},-p;\frac{m+5}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a+b}\right )-2 b p F_1\left (\frac{m+3}{2};\frac{m+2}{2},1-p;\frac{m+5}{2};-\tan ^2(e+f x),-\frac{b \tan ^2(e+f x)}{a+b}\right )\right )}{(m+3) (a+b)}\right )} \]

Warning: Unable to verify antiderivative.

[In]

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

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Maple [F]  time = 1.088, size = 0, normalized size = 0. \begin{align*} \int \left ( a+b \left ( \sec \left ( fx+e \right ) \right ) ^{2} \right ) ^{p} \left ( d\sin \left ( fx+e \right ) \right ) ^{m}\, dx \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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

[Out]

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

[Out]

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

[Out]

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