3.347 \(\int (d \sec (e+f x))^{3/2} (a+a \sec (e+f x))^m \, dx\)
Optimal. Leaf size=98 \[ -\frac{2 \tan (e+f x) (d \sec (e+f x))^{3/2} (\sec (e+f x)+1)^{-m-\frac{1}{2}} (a \sec (e+f x)+a)^m F_1\left (\frac{3}{2};\frac{1}{2},\frac{1}{2}-m;\frac{5}{2};\sec (e+f x),-\sec (e+f x)\right )}{3 f \sqrt{1-\sec (e+f x)}} \]
[Out]
(-2*AppellF1[3/2, 1/2, 1/2 - m, 5/2, Sec[e + f*x], -Sec[e + f*x]]*(d*Sec[e + f*x])^(3/2)*(1 + Sec[e + f*x])^(-
1/2 - m)*(a + a*Sec[e + f*x])^m*Tan[e + f*x])/(3*f*Sqrt[1 - Sec[e + f*x]])
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Rubi [A] time = 0.133955, antiderivative size = 98, 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 =
{3828, 3827, 133} \[ -\frac{2 \tan (e+f x) (d \sec (e+f x))^{3/2} (\sec (e+f x)+1)^{-m-\frac{1}{2}} (a \sec (e+f x)+a)^m F_1\left (\frac{3}{2};\frac{1}{2},\frac{1}{2}-m;\frac{5}{2};\sec (e+f x),-\sec (e+f x)\right )}{3 f \sqrt{1-\sec (e+f x)}} \]
Antiderivative was successfully verified.
[In]
Int[(d*Sec[e + f*x])^(3/2)*(a + a*Sec[e + f*x])^m,x]
[Out]
(-2*AppellF1[3/2, 1/2, 1/2 - m, 5/2, Sec[e + f*x], -Sec[e + f*x]]*(d*Sec[e + f*x])^(3/2)*(1 + Sec[e + f*x])^(-
1/2 - m)*(a + a*Sec[e + f*x])^m*Tan[e + f*x])/(3*f*Sqrt[1 - Sec[e + f*x]])
Rule 3828
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(a^In
tPart[m]*(a + b*Csc[e + f*x])^FracPart[m])/(1 + (b*Csc[e + f*x])/a)^FracPart[m], Int[(1 + (b*Csc[e + f*x])/a)^
m*(d*Csc[e + f*x])^n, x], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[m] && !GtQ
[a, 0]
Rule 3827
Int[(csc[(e_.) + (f_.)*(x_)]*(d_.))^(n_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(a^2*
d*Cot[e + f*x])/(f*Sqrt[a + b*Csc[e + f*x]]*Sqrt[a - b*Csc[e + f*x]]), Subst[Int[((d*x)^(n - 1)*(a + b*x)^(m -
1/2))/Sqrt[a - b*x], x], x, Csc[e + f*x]], x] /; FreeQ[{a, b, d, e, f, m, n}, x] && EqQ[a^2 - b^2, 0] && !In
tegerQ[m] && GtQ[a, 0]
Rule 133
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Simp[(c^n*e^p*(b*x)^(m +
1)*AppellF1[m + 1, -n, -p, m + 2, -((d*x)/c), -((f*x)/e)])/(b*(m + 1)), x] /; FreeQ[{b, c, d, e, f, m, n, p},
x] && !IntegerQ[m] && !IntegerQ[n] && GtQ[c, 0] && (IntegerQ[p] || GtQ[e, 0])
Rubi steps
\begin{align*} \int (d \sec (e+f x))^{3/2} (a+a \sec (e+f x))^m \, dx &=\left ((1+\sec (e+f x))^{-m} (a+a \sec (e+f x))^m\right ) \int (d \sec (e+f x))^{3/2} (1+\sec (e+f x))^m \, dx\\ &=-\frac{\left (d (1+\sec (e+f x))^{-\frac{1}{2}-m} (a+a \sec (e+f x))^m \tan (e+f x)\right ) \operatorname{Subst}\left (\int \frac{\sqrt{d x} (1+x)^{-\frac{1}{2}+m}}{\sqrt{1-x}} \, dx,x,\sec (e+f x)\right )}{f \sqrt{1-\sec (e+f x)}}\\ &=-\frac{2 F_1\left (\frac{3}{2};\frac{1}{2},\frac{1}{2}-m;\frac{5}{2};\sec (e+f x),-\sec (e+f x)\right ) (d \sec (e+f x))^{3/2} (1+\sec (e+f x))^{-\frac{1}{2}-m} (a+a \sec (e+f x))^m \tan (e+f x)}{3 f \sqrt{1-\sec (e+f x)}}\\ \end{align*}
Mathematica [B] time = 14.9408, size = 2529, normalized size = 25.81 \[ \text{Result too large to show} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(d*Sec[e + f*x])^(3/2)*(a + a*Sec[e + f*x])^m,x]
[Out]
(-3*2^(1 + m)*AppellF1[1/2, 3/2 + m, -1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sqrt[Sec[e + f*x]]*(d
*Sec[e + f*x])^(3/2)*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^m*(a*(1 + Sec[e + f*x]))^m*Tan[(e + f*x)/2])/(f*(-1 + T
an[(e + f*x)/2]^2)*(3*AppellF1[1/2, 3/2 + m, -1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (AppellF1[3
/2, 3/2 + m, 1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (3 + 2*m)*AppellF1[3/2, 5/2 + m, -1/2, 5/2,
Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)*((3*2^(1 + m)*AppellF1[1/2, 3/2 + m, -1/2, 3/2,
Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Sqrt[Sec[e + f*x]]*(Cos[(e + f*x)/2]^2*Sec[e + f*x
])^m*Tan[(e + f*x)/2]^2)/((-1 + Tan[(e + f*x)/2]^2)^2*(3*AppellF1[1/2, 3/2 + m, -1/2, 3/2, Tan[(e + f*x)/2]^2,
-Tan[(e + f*x)/2]^2] + (AppellF1[3/2, 3/2 + m, 1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (3 + 2*m)
*AppellF1[3/2, 5/2 + m, -1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)) - (3*2^m*App
ellF1[1/2, 3/2 + m, -1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Sqrt[Sec[e + f*x]]*
(Cos[(e + f*x)/2]^2*Sec[e + f*x])^m)/((-1 + Tan[(e + f*x)/2]^2)*(3*AppellF1[1/2, 3/2 + m, -1/2, 3/2, Tan[(e +
f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (AppellF1[3/2, 3/2 + m, 1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] +
(3 + 2*m)*AppellF1[3/2, 5/2 + m, -1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)) -
(3*2^m*AppellF1[1/2, 3/2 + m, -1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[e + f*x]^(3/2)*(Cos[(e +
f*x)/2]^2*Sec[e + f*x])^m*Sin[e + f*x]*Tan[(e + f*x)/2])/((-1 + Tan[(e + f*x)/2]^2)*(3*AppellF1[1/2, 3/2 + m,
-1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (AppellF1[3/2, 3/2 + m, 1/2, 5/2, Tan[(e + f*x)/2]^2, -
Tan[(e + f*x)/2]^2] + (3 + 2*m)*AppellF1[3/2, 5/2 + m, -1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Ta
n[(e + f*x)/2]^2)) - (3*2^(1 + m)*Sqrt[Sec[e + f*x]]*(Cos[(e + f*x)/2]^2*Sec[e + f*x])^m*Tan[(e + f*x)/2]*((Ap
pellF1[3/2, 3/2 + m, 1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/6
+ ((3/2 + m)*AppellF1[3/2, 5/2 + m, -1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Ta
n[(e + f*x)/2])/3))/((-1 + Tan[(e + f*x)/2]^2)*(3*AppellF1[1/2, 3/2 + m, -1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(
e + f*x)/2]^2] + (AppellF1[3/2, 3/2 + m, 1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (3 + 2*m)*Appell
F1[3/2, 5/2 + m, -1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)) + (3*2^(1 + m)*Appe
llF1[1/2, 3/2 + m, -1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sqrt[Sec[e + f*x]]*(Cos[(e + f*x)/2]^2*
Sec[e + f*x])^m*Tan[(e + f*x)/2]*((AppellF1[3/2, 3/2 + m, 1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] +
(3 + 2*m)*AppellF1[3/2, 5/2 + m, -1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Sec[(e + f*x)/2]^2*Tan[
(e + f*x)/2] + 3*((AppellF1[3/2, 3/2 + m, 1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^
2*Tan[(e + f*x)/2])/6 + ((3/2 + m)*AppellF1[3/2, 5/2 + m, -1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*
Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/3) + Tan[(e + f*x)/2]^2*((-3*AppellF1[5/2, 3/2 + m, 3/2, 7/2, Tan[(e + f*
x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/10 + (3*(3/2 + m)*AppellF1[5/2, 5/2 + m, 1/
2, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/5 + (3 + 2*m)*((3*Appell
F1[5/2, 5/2 + m, 1/2, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan[(e + f*x)/2])/10 +
(3*(5/2 + m)*AppellF1[5/2, 7/2 + m, -1/2, 7/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sec[(e + f*x)/2]^2*Tan
[(e + f*x)/2])/5))))/((-1 + Tan[(e + f*x)/2]^2)*(3*AppellF1[1/2, 3/2 + m, -1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[
(e + f*x)/2]^2] + (AppellF1[3/2, 3/2 + m, 1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (3 + 2*m)*Appel
lF1[3/2, 5/2 + m, -1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2)^2) - (3*2^(1 + m)*m
*AppellF1[1/2, 3/2 + m, -1/2, 3/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2]*Sqrt[Sec[e + f*x]]*(Cos[(e + f*x)/
2]^2*Sec[e + f*x])^(-1 + m)*Tan[(e + f*x)/2]*(-(Cos[(e + f*x)/2]*Sec[e + f*x]*Sin[(e + f*x)/2]) + Cos[(e + f*x
)/2]^2*Sec[e + f*x]*Tan[e + f*x]))/((-1 + Tan[(e + f*x)/2]^2)*(3*AppellF1[1/2, 3/2 + m, -1/2, 3/2, Tan[(e + f*
x)/2]^2, -Tan[(e + f*x)/2]^2] + (AppellF1[3/2, 3/2 + m, 1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2] + (
3 + 2*m)*AppellF1[3/2, 5/2 + m, -1/2, 5/2, Tan[(e + f*x)/2]^2, -Tan[(e + f*x)/2]^2])*Tan[(e + f*x)/2]^2))))
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Maple [F] time = 0.195, size = 0, normalized size = 0. \begin{align*} \int \left ( d\sec \left ( fx+e \right ) \right ) ^{{\frac{3}{2}}} \left ( a+a\sec \left ( fx+e \right ) \right ) ^{m}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((d*sec(f*x+e))^(3/2)*(a+a*sec(f*x+e))^m,x)
[Out]
int((d*sec(f*x+e))^(3/2)*(a+a*sec(f*x+e))^m,x)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int \left (d \sec \left (f x + e\right )\right )^{\frac{3}{2}}{\left (a \sec \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))^(3/2)*(a+a*sec(f*x+e))^m,x, algorithm="maxima")
[Out]
integrate((d*sec(f*x + e))^(3/2)*(a*sec(f*x + e) + a)^m, x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left (\sqrt{d \sec \left (f x + e\right )}{\left (a \sec \left (f x + e\right ) + a\right )}^{m} d \sec \left (f x + e\right ), x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))^(3/2)*(a+a*sec(f*x+e))^m,x, algorithm="fricas")
[Out]
integral(sqrt(d*sec(f*x + e))*(a*sec(f*x + e) + a)^m*d*sec(f*x + e), 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*sec(f*x+e))**(3/2)*(a+a*sec(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 (d \sec \left (f x + e\right )\right )^{\frac{3}{2}}{\left (a \sec \left (f x + e\right ) + a\right )}^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((d*sec(f*x+e))^(3/2)*(a+a*sec(f*x+e))^m,x, algorithm="giac")
[Out]
integrate((d*sec(f*x + e))^(3/2)*(a*sec(f*x + e) + a)^m, x)