3.140 \(\int (a+a \sec (c+d x))^{3/2} (e \sin (c+d x))^m \, dx\)
Optimal. Leaf size=106 \[ \frac{2 a e \sqrt{a \sec (c+d x)+a} (1-\cos (c+d x))^{\frac{1-m}{2}} (\cos (c+d x)+1)^{-m/2} F_1\left (-\frac{1}{2};\frac{1-m}{2},\frac{1}{2} (-m-2);\frac{1}{2};\cos (c+d x),-\cos (c+d x)\right ) (e \sin (c+d x))^{m-1}}{d} \]
[Out]
(2*a*e*AppellF1[-1/2, (1 - m)/2, (-2 - m)/2, 1/2, Cos[c + d*x], -Cos[c + d*x]]*(1 - Cos[c + d*x])^((1 - m)/2)*
Sqrt[a + a*Sec[c + d*x]]*(e*Sin[c + d*x])^(-1 + m))/(d*(1 + Cos[c + d*x])^(m/2))
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Rubi [A] time = 0.37476, antiderivative size = 106, normalized size of antiderivative = 1.,
number of steps used = 5, number of rules used = 4, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.16, Rules used =
{3876, 2886, 135, 133} \[ \frac{2 a e \sqrt{a \sec (c+d x)+a} (1-\cos (c+d x))^{\frac{1-m}{2}} (\cos (c+d x)+1)^{-m/2} F_1\left (-\frac{1}{2};\frac{1-m}{2},\frac{1}{2} (-m-2);\frac{1}{2};\cos (c+d x),-\cos (c+d x)\right ) (e \sin (c+d x))^{m-1}}{d} \]
Antiderivative was successfully verified.
[In]
Int[(a + a*Sec[c + d*x])^(3/2)*(e*Sin[c + d*x])^m,x]
[Out]
(2*a*e*AppellF1[-1/2, (1 - m)/2, (-2 - m)/2, 1/2, Cos[c + d*x], -Cos[c + d*x]]*(1 - Cos[c + d*x])^((1 - m)/2)*
Sqrt[a + a*Sec[c + d*x]]*(e*Sin[c + d*x])^(-1 + m))/(d*(1 + Cos[c + d*x])^(m/2))
Rule 3876
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_.)*(csc[(e_.) + (f_.)*(x_)]*(b_.) + (a_))^(m_), x_Symbol] :> Dist[(Sin[
e + f*x]^FracPart[m]*(a + b*Csc[e + f*x])^FracPart[m])/(b + a*Sin[e + f*x])^FracPart[m], Int[((g*Cos[e + f*x])
^p*(b + a*Sin[e + f*x])^m)/Sin[e + f*x]^m, x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && (EqQ[a^2 - b^2, 0] ||
IntegersQ[2*m, p])
Rule 2886
Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((d_.)*sin[(e_.) + (f_.)*(x_)])^(n_)*((a_) + (b_.)*sin[(e_.) + (f_.)*
(x_)])^(m_), x_Symbol] :> Dist[(g*(g*Cos[e + f*x])^(p - 1))/(f*(a + b*Sin[e + f*x])^((p - 1)/2)*(a - b*Sin[e +
f*x])^((p - 1)/2)), Subst[Int[(d*x)^n*(a + b*x)^(m + (p - 1)/2)*(a - b*x)^((p - 1)/2), x], x, Sin[e + f*x]],
x] /; FreeQ[{a, b, d, e, f, m, n, p}, x] && EqQ[a^2 - b^2, 0] && !IntegerQ[m]
Rule 135
Int[((b_.)*(x_))^(m_)*((c_) + (d_.)*(x_))^(n_)*((e_) + (f_.)*(x_))^(p_), x_Symbol] :> Dist[(c^IntPart[n]*(c +
d*x)^FracPart[n])/(1 + (d*x)/c)^FracPart[n], Int[(b*x)^m*(1 + (d*x)/c)^n*(e + f*x)^p, x], x] /; FreeQ[{b, c, d
, e, f, m, n, p}, x] && !IntegerQ[m] && !IntegerQ[n] && !GtQ[c, 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 (a+a \sec (c+d x))^{3/2} (e \sin (c+d x))^m \, dx &=\frac{\left (\sqrt{-\cos (c+d x)} \sqrt{a+a \sec (c+d x)}\right ) \int \frac{(-a-a \cos (c+d x))^{3/2} (e \sin (c+d x))^m}{(-\cos (c+d x))^{3/2}} \, dx}{\sqrt{-a-a \cos (c+d x)}}\\ &=-\frac{\left (e \sqrt{-\cos (c+d x)} (-a-a \cos (c+d x))^{-\frac{1}{2}+\frac{1-m}{2}} (-a+a \cos (c+d x))^{\frac{1-m}{2}} \sqrt{a+a \sec (c+d x)} (e \sin (c+d x))^{-1+m}\right ) \operatorname{Subst}\left (\int \frac{(-a-a x)^{\frac{3}{2}+\frac{1}{2} (-1+m)} (-a+a x)^{\frac{1}{2} (-1+m)}}{(-x)^{3/2}} \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac{\left (a e \sqrt{-\cos (c+d x)} (1+\cos (c+d x))^{-m/2} (-a-a \cos (c+d x))^{-\frac{1}{2}+\frac{1-m}{2}+\frac{m}{2}} (-a+a \cos (c+d x))^{\frac{1-m}{2}} \sqrt{a+a \sec (c+d x)} (e \sin (c+d x))^{-1+m}\right ) \operatorname{Subst}\left (\int \frac{(1+x)^{\frac{3}{2}+\frac{1}{2} (-1+m)} (-a+a x)^{\frac{1}{2} (-1+m)}}{(-x)^{3/2}} \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac{\left (a e (1-\cos (c+d x))^{\frac{1}{2}-\frac{m}{2}} \sqrt{-\cos (c+d x)} (1+\cos (c+d x))^{-m/2} (-a-a \cos (c+d x))^{-\frac{1}{2}+\frac{1-m}{2}+\frac{m}{2}} (-a+a \cos (c+d x))^{-\frac{1}{2}+\frac{1-m}{2}+\frac{m}{2}} \sqrt{a+a \sec (c+d x)} (e \sin (c+d x))^{-1+m}\right ) \operatorname{Subst}\left (\int \frac{(1-x)^{\frac{1}{2} (-1+m)} (1+x)^{\frac{3}{2}+\frac{1}{2} (-1+m)}}{(-x)^{3/2}} \, dx,x,\cos (c+d x)\right )}{d}\\ &=\frac{2 a e F_1\left (-\frac{1}{2};\frac{1-m}{2},\frac{1}{2} (-2-m);\frac{1}{2};\cos (c+d x),-\cos (c+d x)\right ) (1-\cos (c+d x))^{\frac{1-m}{2}} (1+\cos (c+d x))^{-m/2} \sqrt{a+a \sec (c+d x)} (e \sin (c+d x))^{-1+m}}{d}\\ \end{align*}
Mathematica [B] time = 9.70993, size = 1243, normalized size = 11.73 \[ \text{result too large to display} \]
Warning: Unable to verify antiderivative.
[In]
Integrate[(a + a*Sec[c + d*x])^(3/2)*(e*Sin[c + d*x])^m,x]
[Out]
(4*(3 + m)*(AppellF1[(1 + m)/2, -1/2, 1 + m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + AppellF1[(1
+ m)/2, 1/2, m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 2*AppellF1[(1 + m)/2, 3/2, m, (3 + m)/2
, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2])*Cos[(c + d*x)/2]^3*(a*(1 + Sec[c + d*x]))^(3/2)*Sin[(c + d*x)/2]*(
e*Sin[c + d*x])^m)/(d*(1 + m)*(6*AppellF1[(1 + m)/2, 3/2, m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^
2] + 2*m*AppellF1[(1 + m)/2, 3/2, m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*AppellF1[(3 + m)/
2, -1/2, 2 + m, (5 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 2*m*AppellF1[(3 + m)/2, -1/2, 2 + m, (5
+ m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - AppellF1[(3 + m)/2, 1/2, 1 + m, (5 + m)/2, Tan[(c + d*x)/2]
^2, -Tan[(c + d*x)/2]^2] - 2*m*AppellF1[(3 + m)/2, 1/2, 1 + m, (5 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2
]^2] + AppellF1[(3 + m)/2, 3/2, m, (5 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] - 4*m*AppellF1[(3 + m)/
2, 3/2, 1 + m, (5 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 6*AppellF1[(3 + m)/2, 5/2, m, (5 + m)/2,
Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2] + 6*AppellF1[(1 + m)/2, 3/2, m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(
c + d*x)/2]^2]*Cos[c + d*x] + 2*m*AppellF1[(1 + m)/2, 3/2, m, (3 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]
^2]*Cos[c + d*x] + 2*AppellF1[(3 + m)/2, -1/2, 2 + m, (5 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[
c + d*x] + 2*m*AppellF1[(3 + m)/2, -1/2, 2 + m, (5 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[c + d*
x] + AppellF1[(3 + m)/2, 1/2, 1 + m, (5 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[c + d*x] + 2*m*Ap
pellF1[(3 + m)/2, 1/2, 1 + m, (5 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[c + d*x] - AppellF1[(3 +
m)/2, 3/2, m, (5 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[c + d*x] + 4*m*AppellF1[(3 + m)/2, 3/2,
1 + m, (5 + m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[c + d*x] - 6*AppellF1[(3 + m)/2, 5/2, m, (5 +
m)/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*Cos[c + d*x] + (3 + m)*AppellF1[(1 + m)/2, -1/2, 1 + m, (3 + m)
/2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + Cos[c + d*x]) + (3 + m)*AppellF1[(1 + m)/2, 1/2, m, (3 + m)/
2, Tan[(c + d*x)/2]^2, -Tan[(c + d*x)/2]^2]*(1 + Cos[c + d*x])))
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Maple [F] time = 0.191, size = 0, normalized size = 0. \begin{align*} \int \left ( a+a\sec \left ( dx+c \right ) \right ) ^{{\frac{3}{2}}} \left ( e\sin \left ( dx+c \right ) \right ) ^{m}\, dx \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
int((a+a*sec(d*x+c))^(3/2)*(e*sin(d*x+c))^m,x)
[Out]
int((a+a*sec(d*x+c))^(3/2)*(e*sin(d*x+c))^m,x)
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Maxima [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \left (e \sin \left (d x + c\right )\right )^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))^(3/2)*(e*sin(d*x+c))^m,x, algorithm="maxima")
[Out]
integrate((a*sec(d*x + c) + a)^(3/2)*(e*sin(d*x + c))^m, x)
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Fricas [F] time = 0., size = 0, normalized size = 0. \begin{align*}{\rm integral}\left ({\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \left (e \sin \left (d x + c\right )\right )^{m}, x\right ) \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))^(3/2)*(e*sin(d*x+c))^m,x, algorithm="fricas")
[Out]
integral((a*sec(d*x + c) + a)^(3/2)*(e*sin(d*x + c))^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+a*sec(d*x+c))**(3/2)*(e*sin(d*x+c))**m,x)
[Out]
Timed out
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Giac [F] time = 0., size = 0, normalized size = 0. \begin{align*} \int{\left (a \sec \left (d x + c\right ) + a\right )}^{\frac{3}{2}} \left (e \sin \left (d x + c\right )\right )^{m}\,{d x} \end{align*}
Verification of antiderivative is not currently implemented for this CAS.
[In]
integrate((a+a*sec(d*x+c))^(3/2)*(e*sin(d*x+c))^m,x, algorithm="giac")
[Out]
integrate((a*sec(d*x + c) + a)^(3/2)*(e*sin(d*x + c))^m, x)