3.298 \(\int (d \sec (e+f x))^m (a+b \sec ^2(e+f x))^p \, dx\)
Optimal. Leaf size=111 \[ \frac {\sqrt {-\tan ^2(e+f x)} \cot (e+f x) (d \sec (e+f x))^m \left (a+b \sec ^2(e+f x)\right )^p \left (\frac {b \sec ^2(e+f x)}{a}+1\right )^{-p} F_1\left (\frac {m}{2};\frac {1}{2},-p;\frac {m+2}{2};\sec ^2(e+f x),-\frac {b \sec ^2(e+f x)}{a}\right )}{f m} \]
[Out]
AppellF1(1/2*m,1/2,-p,1+1/2*m,sec(f*x+e)^2,-b*sec(f*x+e)^2/a)*cos(f*x+e)*(d*sec(f*x+e))^m*(a+b*sec(f*x+e)^2)^p
*(-tan(f*x+e)^2)^(1/2)/f/m/((1+b*sec(f*x+e)^2/a)^p)/sin(f*x+e)
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Rubi [F] time = 0.05, antiderivative size = 0, normalized size of antiderivative = 0.00,
number of steps used = 0, number of rules used = 0, integrand size = 0, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.000, Rules used =
{} \[ \int (d \sec (e+f x))^m \left (a+b \sec ^2(e+f x)\right )^p \, dx \]
Verification is Not applicable to the result.
[In]
Int[(d*Sec[e + f*x])^m*(a + b*Sec[e + f*x]^2)^p,x]
[Out]
Defer[Int][(d*Sec[e + f*x])^m*(a + b*Sec[e + f*x]^2)^p, x]
Rubi steps
\begin {align*} \int (d \sec (e+f x))^m \left (a+b \sec ^2(e+f x)\right )^p \, dx &=\int (d \sec (e+f x))^m \left (a+b \sec ^2(e+f x)\right )^p \, dx\\ \end {align*}
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Mathematica [B] time = 18.90, size = 2195, normalized size = 19.77 \[ \text {Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(d*Sec[e + f*x])^m*(a + b*Sec[e + f*x]^2)^p,x]
[Out]
(3*(a + b)*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))]*(a + 2*b + a*Cos[2*
(e + f*x)])^p*(d*Sec[e + f*x])^m*(Sec[e + f*x]^2)^(-1 + m/2 + p)*(a + b*Sec[e + f*x]^2)^p*Tan[e + f*x])/(f*(3*
(a + b)*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] + (2*b*p*AppellF1[3/2,
1 - m/2, 1 - p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] + (a + b)*(-2 + m)*AppellF1[3/2, 2 - m/2
, -p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))])*Tan[e + f*x]^2)*((3*(a + b)*AppellF1[1/2, 1 - m/2,
-p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))]*(a + 2*b + a*Cos[2*(e + f*x)])^p*(Sec[e + f*x]^2)^(m
/2 + p))/(3*(a + b)*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] + (2*b*p*A
ppellF1[3/2, 1 - m/2, 1 - p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] + (a + b)*(-2 + m)*AppellF1[
3/2, 2 - m/2, -p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))])*Tan[e + f*x]^2) - (6*a*(a + b)*p*Appel
lF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))]*(a + 2*b + a*Cos[2*(e + f*x)])^(-1
+ p)*(Sec[e + f*x]^2)^(-1 + m/2 + p)*Sin[2*(e + f*x)]*Tan[e + f*x])/(3*(a + b)*AppellF1[1/2, 1 - m/2, -p, 3/2,
-Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] + (2*b*p*AppellF1[3/2, 1 - m/2, 1 - p, 5/2, -Tan[e + f*x]^2,
-((b*Tan[e + f*x]^2)/(a + b))] + (a + b)*(-2 + m)*AppellF1[3/2, 2 - m/2, -p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e
+ f*x]^2)/(a + b))])*Tan[e + f*x]^2) + (6*(a + b)*(-1 + m/2 + p)*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]
^2, -((b*Tan[e + f*x]^2)/(a + b))]*(a + 2*b + a*Cos[2*(e + f*x)])^p*(Sec[e + f*x]^2)^(-1 + m/2 + p)*Tan[e + f*
x]^2)/(3*(a + b)*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] + (2*b*p*Appe
llF1[3/2, 1 - m/2, 1 - p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] + (a + b)*(-2 + m)*AppellF1[3/2
, 2 - m/2, -p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))])*Tan[e + f*x]^2) + (3*(a + b)*(a + 2*b + a
*Cos[2*(e + f*x)])^p*(Sec[e + f*x]^2)^(-1 + m/2 + p)*Tan[e + f*x]*((2*b*p*AppellF1[3/2, 1 - m/2, 1 - p, 5/2, -
Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))]*Sec[e + f*x]^2*Tan[e + f*x])/(3*(a + b)) - (2*(1 - m/2)*AppellF
1[3/2, 2 - m/2, -p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))]*Sec[e + f*x]^2*Tan[e + f*x])/3))/(3*(
a + b)*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] + (2*b*p*AppellF1[3/2,
1 - m/2, 1 - p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] + (a + b)*(-2 + m)*AppellF1[3/2, 2 - m/2,
-p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))])*Tan[e + f*x]^2) - (3*(a + b)*AppellF1[1/2, 1 - m/2,
-p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))]*(a + 2*b + a*Cos[2*(e + f*x)])^p*(Sec[e + f*x]^2)^(-
1 + m/2 + p)*Tan[e + f*x]*(2*(2*b*p*AppellF1[3/2, 1 - m/2, 1 - p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(
a + b))] + (a + b)*(-2 + m)*AppellF1[3/2, 2 - m/2, -p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))])*S
ec[e + f*x]^2*Tan[e + f*x] + 3*(a + b)*((2*b*p*AppellF1[3/2, 1 - m/2, 1 - p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e
+ f*x]^2)/(a + b))]*Sec[e + f*x]^2*Tan[e + f*x])/(3*(a + b)) - (2*(1 - m/2)*AppellF1[3/2, 2 - m/2, -p, 5/2, -T
an[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))]*Sec[e + f*x]^2*Tan[e + f*x])/3) + Tan[e + f*x]^2*(2*b*p*((-6*b*(
1 - p)*AppellF1[5/2, 1 - m/2, 2 - p, 7/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))]*Sec[e + f*x]^2*Tan[e
+ f*x])/(5*(a + b)) - (6*(1 - m/2)*AppellF1[5/2, 2 - m/2, 1 - p, 7/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(
a + b))]*Sec[e + f*x]^2*Tan[e + f*x])/5) + (a + b)*(-2 + m)*((6*b*p*AppellF1[5/2, 2 - m/2, 1 - p, 7/2, -Tan[e
+ f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))]*Sec[e + f*x]^2*Tan[e + f*x])/(5*(a + b)) - (6*(2 - m/2)*AppellF1[5/2,
3 - m/2, -p, 7/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))]*Sec[e + f*x]^2*Tan[e + f*x])/5))))/(3*(a +
b)*AppellF1[1/2, 1 - m/2, -p, 3/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] + (2*b*p*AppellF1[3/2, 1 -
m/2, 1 - p, 5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))] + (a + b)*(-2 + m)*AppellF1[3/2, 2 - m/2, -p,
5/2, -Tan[e + f*x]^2, -((b*Tan[e + f*x]^2)/(a + b))])*Tan[e + f*x]^2)^2))
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fricas [F] time = 0.57, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sec \left (f x + e\right )\right )^{m}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))^m*(a+b*sec(f*x+e)^2)^p,x, algorithm="fricas")
[Out]
integral((b*sec(f*x + e)^2 + a)^p*(d*sec(f*x + e))^m, x)
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sec \left (f x + e\right )\right )^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))^m*(a+b*sec(f*x+e)^2)^p,x, algorithm="giac")
[Out]
integrate((b*sec(f*x + e)^2 + a)^p*(d*sec(f*x + e))^m, x)
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maple [F] time = 4.43, size = 0, normalized size = 0.00 \[ \int \left (d \sec \left (f x +e \right )\right )^{m} \left (a +b \left (\sec ^{2}\left (f x +e \right )\right )\right )^{p}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*sec(f*x+e))^m*(a+b*sec(f*x+e)^2)^p,x)
[Out]
int((d*sec(f*x+e))^m*(a+b*sec(f*x+e)^2)^p,x)
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (b \sec \left (f x + e\right )^{2} + a\right )}^{p} \left (d \sec \left (f x + e\right )\right )^{m}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))^m*(a+b*sec(f*x+e)^2)^p,x, algorithm="maxima")
[Out]
integrate((b*sec(f*x + e)^2 + a)^p*(d*sec(f*x + e))^m, x)
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mupad [F] time = 0.00, size = -1, normalized size = -0.01 \[ \int {\left (a+\frac {b}{{\cos \left (e+f\,x\right )}^2}\right )}^p\,{\left (\frac {d}{\cos \left (e+f\,x\right )}\right )}^m \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a + b/cos(e + f*x)^2)^p*(d/cos(e + f*x))^m,x)
[Out]
int((a + b/cos(e + f*x)^2)^p*(d/cos(e + f*x))^m, 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*sec(f*x+e))**m*(a+b*sec(f*x+e)**2)**p,x)
[Out]
Timed out
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