3.292 \(\int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=239 \[ \frac {5 a^3 (e \cos (c+d x))^{3/2}}{d e^3 \sqrt {a \sin (c+d x)+a}}-\frac {5 a^2 \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \tan ^{-1}\left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d e^{3/2} (\sin (c+d x)+\cos (c+d x)+1)}-\frac {5 a^2 \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \sinh ^{-1}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2} (\sin (c+d x)+\cos (c+d x)+1)}+\frac {4 a (a \sin (c+d x)+a)^{3/2}}{d e \sqrt {e \cos (c+d x)}} \]

[Out]

4*a*(a+a*sin(d*x+c))^(3/2)/d/e/(e*cos(d*x+c))^(1/2)+5*a^3*(e*cos(d*x+c))^(3/2)/d/e^3/(a+a*sin(d*x+c))^(1/2)-5*
a^2*arcsinh((e*cos(d*x+c))^(1/2)/e^(1/2))*(1+cos(d*x+c))^(1/2)*(a+a*sin(d*x+c))^(1/2)/d/e^(3/2)/(1+cos(d*x+c)+
sin(d*x+c))-5*a^2*arctan(sin(d*x+c)*e^(1/2)/(e*cos(d*x+c))^(1/2)/(1+cos(d*x+c))^(1/2))*(1+cos(d*x+c))^(1/2)*(a
+a*sin(d*x+c))^(1/2)/d/e^(3/2)/(1+cos(d*x+c)+sin(d*x+c))

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Rubi [A]  time = 0.36, antiderivative size = 239, normalized size of antiderivative = 1.00, number of steps used = 8, number of rules used = 8, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.296, Rules used = {2676, 2678, 2684, 2775, 203, 2833, 63, 215} \[ \frac {5 a^3 (e \cos (c+d x))^{3/2}}{d e^3 \sqrt {a \sin (c+d x)+a}}-\frac {5 a^2 \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \tan ^{-1}\left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {\cos (c+d x)+1} \sqrt {e \cos (c+d x)}}\right )}{d e^{3/2} (\sin (c+d x)+\cos (c+d x)+1)}-\frac {5 a^2 \sqrt {\cos (c+d x)+1} \sqrt {a \sin (c+d x)+a} \sinh ^{-1}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right )}{d e^{3/2} (\sin (c+d x)+\cos (c+d x)+1)}+\frac {4 a (a \sin (c+d x)+a)^{3/2}}{d e \sqrt {e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Int[(a + a*Sin[c + d*x])^(5/2)/(e*Cos[c + d*x])^(3/2),x]

[Out]

(5*a^3*(e*Cos[c + d*x])^(3/2))/(d*e^3*Sqrt[a + a*Sin[c + d*x]]) - (5*a^2*ArcSinh[Sqrt[e*Cos[c + d*x]]/Sqrt[e]]
*Sqrt[1 + Cos[c + d*x]]*Sqrt[a + a*Sin[c + d*x]])/(d*e^(3/2)*(1 + Cos[c + d*x] + Sin[c + d*x])) - (5*a^2*ArcTa
n[(Sqrt[e]*Sin[c + d*x])/(Sqrt[e*Cos[c + d*x]]*Sqrt[1 + Cos[c + d*x]])]*Sqrt[1 + Cos[c + d*x]]*Sqrt[a + a*Sin[
c + d*x]])/(d*e^(3/2)*(1 + Cos[c + d*x] + Sin[c + d*x])) + (4*a*(a + a*Sin[c + d*x])^(3/2))/(d*e*Sqrt[e*Cos[c
+ d*x]])

Rule 63

Int[((a_.) + (b_.)*(x_))^(m_)*((c_.) + (d_.)*(x_))^(n_), x_Symbol] :> With[{p = Denominator[m]}, Dist[p/b, Sub
st[Int[x^(p*(m + 1) - 1)*(c - (a*d)/b + (d*x^p)/b)^n, x], x, (a + b*x)^(1/p)], x]] /; FreeQ[{a, b, c, d}, x] &
& NeQ[b*c - a*d, 0] && LtQ[-1, m, 0] && LeQ[-1, n, 0] && LeQ[Denominator[n], Denominator[m]] && IntLinearQ[a,
b, c, d, m, n, x]

Rule 203

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(1*ArcTan[(Rt[b, 2]*x)/Rt[a, 2]])/(Rt[a, 2]*Rt[b, 2]), x] /;
FreeQ[{a, b}, x] && PosQ[a/b] && (GtQ[a, 0] || GtQ[b, 0])

Rule 215

Int[1/Sqrt[(a_) + (b_.)*(x_)^2], x_Symbol] :> Simp[ArcSinh[(Rt[b, 2]*x)/Sqrt[a]]/Rt[b, 2], x] /; FreeQ[{a, b},
x] && GtQ[a, 0] && PosQ[b]

Rule 2676

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> Simp[(-2*b*
(g*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(p + 1)), x] + Dist[(b^2*(2*m + p - 1))/(g^2*(p +
1)), Int[(g*Cos[e + f*x])^(p + 2)*(a + b*Sin[e + f*x])^(m - 2), x], x] /; FreeQ[{a, b, e, f, g}, x] && EqQ[a^2
- b^2, 0] && GtQ[m, 1] && LtQ[p, -1] && IntegersQ[2*m, 2*p]

Rule 2678

Int[(cos[(e_.) + (f_.)*(x_)]*(g_.))^(p_)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_), x_Symbol] :> -Simp[(b*(g
*Cos[e + f*x])^(p + 1)*(a + b*Sin[e + f*x])^(m - 1))/(f*g*(m + p)), x] + Dist[(a*(2*m + p - 1))/(m + p), Int[(
g*Cos[e + f*x])^p*(a + b*Sin[e + f*x])^(m - 1), x], x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^2, 0]
&& GtQ[m, 0] && NeQ[m + p, 0] && IntegersQ[2*m, 2*p]

Rule 2684

Int[Sqrt[cos[(e_.) + (f_.)*(x_)]*(g_.)]/Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[(g*Sqrt[
1 + Cos[e + f*x]]*Sqrt[a + b*Sin[e + f*x]])/(a + a*Cos[e + f*x] + b*Sin[e + f*x]), Int[Sqrt[1 + Cos[e + f*x]]/
Sqrt[g*Cos[e + f*x]], x], x] - Dist[(g*Sqrt[1 + Cos[e + f*x]]*Sqrt[a + b*Sin[e + f*x]])/(b + b*Cos[e + f*x] +
a*Sin[e + f*x]), Int[Sin[e + f*x]/(Sqrt[g*Cos[e + f*x]]*Sqrt[1 + Cos[e + f*x]]), x], x] /; FreeQ[{a, b, e, f,
g}, x] && EqQ[a^2 - b^2, 0]

Rule 2775

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/Sqrt[(c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)]], x_Symbol] :> Dist[
(-2*b)/f, Subst[Int[1/(b + d*x^2), x], x, (b*Cos[e + f*x])/(Sqrt[a + b*Sin[e + f*x]]*Sqrt[c + d*Sin[e + f*x]])
], x] /; FreeQ[{a, b, c, d, e, f}, x] && NeQ[b*c - a*d, 0] && EqQ[a^2 - b^2, 0] && NeQ[c^2 - d^2, 0]

Rule 2833

Int[cos[(e_.) + (f_.)*(x_)]*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_.)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)
])^(n_.), x_Symbol] :> Dist[1/(b*f), Subst[Int[(a + x)^m*(c + (d*x)/b)^n, x], x, b*Sin[e + f*x]], x] /; FreeQ[
{a, b, c, d, e, f, m, n}, x]

Rubi steps

\begin {align*} \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{3/2}} \, dx &=\frac {4 a (a+a \sin (c+d x))^{3/2}}{d e \sqrt {e \cos (c+d x)}}-\frac {\left (5 a^2\right ) \int \sqrt {e \cos (c+d x)} \sqrt {a+a \sin (c+d x)} \, dx}{e^2}\\ &=\frac {5 a^3 (e \cos (c+d x))^{3/2}}{d e^3 \sqrt {a+a \sin (c+d x)}}+\frac {4 a (a+a \sin (c+d x))^{3/2}}{d e \sqrt {e \cos (c+d x)}}-\frac {\left (5 a^3\right ) \int \frac {\sqrt {e \cos (c+d x)}}{\sqrt {a+a \sin (c+d x)}} \, dx}{2 e^2}\\ &=\frac {5 a^3 (e \cos (c+d x))^{3/2}}{d e^3 \sqrt {a+a \sin (c+d x)}}+\frac {4 a (a+a \sin (c+d x))^{3/2}}{d e \sqrt {e \cos (c+d x)}}-\frac {\left (5 a^3 \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}\right ) \int \frac {\sqrt {1+\cos (c+d x)}}{\sqrt {e \cos (c+d x)}} \, dx}{2 e (a+a \cos (c+d x)+a \sin (c+d x))}+\frac {\left (5 a^3 \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}\right ) \int \frac {\sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}} \, dx}{2 e (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=\frac {5 a^3 (e \cos (c+d x))^{3/2}}{d e^3 \sqrt {a+a \sin (c+d x)}}+\frac {4 a (a+a \sin (c+d x))^{3/2}}{d e \sqrt {e \cos (c+d x)}}-\frac {\left (5 a^3 \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {e x} \sqrt {1+x}} \, dx,x,\cos (c+d x)\right )}{2 d e (a+a \cos (c+d x)+a \sin (c+d x))}+\frac {\left (5 a^3 \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{1+e x^2} \, dx,x,-\frac {\sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}}\right )}{d e (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=\frac {5 a^3 (e \cos (c+d x))^{3/2}}{d e^3 \sqrt {a+a \sin (c+d x)}}+\frac {4 a (a+a \sin (c+d x))^{3/2}}{d e \sqrt {e \cos (c+d x)}}-\frac {5 a^3 \tan ^{-1}\left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d e^{3/2} (a+a \cos (c+d x)+a \sin (c+d x))}-\frac {\left (5 a^3 \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {1+\frac {x^2}{e}}} \, dx,x,\sqrt {e \cos (c+d x)}\right )}{d e^2 (a+a \cos (c+d x)+a \sin (c+d x))}\\ &=\frac {5 a^3 (e \cos (c+d x))^{3/2}}{d e^3 \sqrt {a+a \sin (c+d x)}}+\frac {4 a (a+a \sin (c+d x))^{3/2}}{d e \sqrt {e \cos (c+d x)}}-\frac {5 a^3 \sinh ^{-1}\left (\frac {\sqrt {e \cos (c+d x)}}{\sqrt {e}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d e^{3/2} (a+a \cos (c+d x)+a \sin (c+d x))}-\frac {5 a^3 \tan ^{-1}\left (\frac {\sqrt {e} \sin (c+d x)}{\sqrt {e \cos (c+d x)} \sqrt {1+\cos (c+d x)}}\right ) \sqrt {1+\cos (c+d x)} \sqrt {a+a \sin (c+d x)}}{d e^{3/2} (a+a \cos (c+d x)+a \sin (c+d x))}\\ \end {align*}

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Mathematica [C]  time = 0.16, size = 75, normalized size = 0.31 \[ \frac {8 \sqrt [4]{2} (a (\sin (c+d x)+1))^{5/2} \, _2F_1\left (-\frac {5}{4},-\frac {1}{4};\frac {3}{4};\frac {1}{2} (1-\sin (c+d x))\right )}{d e (\sin (c+d x)+1)^{9/4} \sqrt {e \cos (c+d x)}} \]

Antiderivative was successfully verified.

[In]

Integrate[(a + a*Sin[c + d*x])^(5/2)/(e*Cos[c + d*x])^(3/2),x]

[Out]

(8*2^(1/4)*Hypergeometric2F1[-5/4, -1/4, 3/4, (1 - Sin[c + d*x])/2]*(a*(1 + Sin[c + d*x]))^(5/2))/(d*e*Sqrt[e*
Cos[c + d*x]]*(1 + Sin[c + d*x])^(9/4))

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fricas [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+a*sin(d*x+c))^(5/2)/(e*cos(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

Timed out

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giac [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+a*sin(d*x+c))^(5/2)/(e*cos(d*x+c))^(3/2),x, algorithm="giac")

[Out]

Timed out

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maple [B]  time = 0.25, size = 445, normalized size = 1.86 \[ -\frac {\left (5 \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sin \left (d x +c \right )+5 \sqrt {2}\, \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sin \left (d x +c \right )-5 \cos \left (d x +c \right ) \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}-5 \cos \left (d x +c \right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sqrt {2}-5 \arctan \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}}{2}\right ) \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sqrt {2}-5 \sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \arctanh \left (\frac {\sqrt {-\frac {2 \cos \left (d x +c \right )}{1+\cos \left (d x +c \right )}}\, \sin \left (d x +c \right ) \sqrt {2}}{2 \cos \left (d x +c \right )}\right ) \sqrt {2}+4 \cos \left (d x +c \right ) \sin \left (d x +c \right )-36 \cos \left (d x +c \right )\right ) \left (a \left (1+\sin \left (d x +c \right )\right )\right )^{\frac {5}{2}}}{4 d \left (-\left (\cos ^{2}\left (d x +c \right )\right )+2 \sin \left (d x +c \right )+2\right ) \left (e \cos \left (d x +c \right )\right )^{\frac {3}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))^(5/2)/(e*cos(d*x+c))^(3/2),x)

[Out]

-1/4/d*(5*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)
)*sin(d*x+c)+5*2^(1/2)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*s
in(d*x+c)/cos(d*x+c)*2^(1/2))*sin(d*x+c)-5*cos(d*x+c)*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2))
*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)-5*cos(d*x+c)*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arctanh(1/2*(-
2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)-5*arctan(1/2*(-2*cos(d*x+c)/(1+cos(d
*x+c)))^(1/2)*2^(1/2))*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*2^(1/2)-5*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*arc
tanh(1/2*(-2*cos(d*x+c)/(1+cos(d*x+c)))^(1/2)*sin(d*x+c)/cos(d*x+c)*2^(1/2))*2^(1/2)+4*cos(d*x+c)*sin(d*x+c)-3
6*cos(d*x+c))*(a*(1+sin(d*x+c)))^(5/2)/(-cos(d*x+c)^2+2*sin(d*x+c)+2)/(e*cos(d*x+c))^(3/2)

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maxima [F]  time = 0.00, size = 0, normalized size = 0.00 \[ \int \frac {{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac {5}{2}}}{\left (e \cos \left (d x + c\right )\right )^{\frac {3}{2}}}\,{d x} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^(5/2)/(e*cos(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

integrate((a*sin(d*x + c) + a)^(5/2)/(e*cos(d*x + c))^(3/2), x)

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mupad [F]  time = 0.00, size = -1, normalized size = -0.00 \[ \int \frac {{\left (a+a\,\sin \left (c+d\,x\right )\right )}^{5/2}}{{\left (e\,\cos \left (c+d\,x\right )\right )}^{3/2}} \,d x \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^(5/2)/(e*cos(c + d*x))^(3/2),x)

[Out]

int((a + a*sin(c + d*x))^(5/2)/(e*cos(c + d*x))^(3/2), 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+a*sin(d*x+c))**(5/2)/(e*cos(d*x+c))**(3/2),x)

[Out]

Timed out

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