3.311 \(\int \frac{1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^{3/2}} \, dx\)

Optimal. Leaf size=154 \[ \frac{32 (a \sin (c+d x)+a)^{3/2}}{45 a^3 d e (e \cos (c+d x))^{3/2}}-\frac{16 \sqrt{a \sin (c+d x)+a}}{15 a^2 d e (e \cos (c+d x))^{3/2}}-\frac{4}{15 a d e \sqrt{a \sin (c+d x)+a} (e \cos (c+d x))^{3/2}}-\frac{2}{9 d e (a \sin (c+d x)+a)^{3/2} (e \cos (c+d x))^{3/2}} \]

[Out]

-2/(9*d*e*(e*Cos[c + d*x])^(3/2)*(a + a*Sin[c + d*x])^(3/2)) - 4/(15*a*d*e*(e*Cos[c + d*x])^(3/2)*Sqrt[a + a*S
in[c + d*x]]) - (16*Sqrt[a + a*Sin[c + d*x]])/(15*a^2*d*e*(e*Cos[c + d*x])^(3/2)) + (32*(a + a*Sin[c + d*x])^(
3/2))/(45*a^3*d*e*(e*Cos[c + d*x])^(3/2))

________________________________________________________________________________________

Rubi [A]  time = 0.288734, antiderivative size = 154, normalized size of antiderivative = 1., number of steps used = 4, number of rules used = 2, integrand size = 27, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.074, Rules used = {2672, 2671} \[ \frac{32 (a \sin (c+d x)+a)^{3/2}}{45 a^3 d e (e \cos (c+d x))^{3/2}}-\frac{16 \sqrt{a \sin (c+d x)+a}}{15 a^2 d e (e \cos (c+d x))^{3/2}}-\frac{4}{15 a d e \sqrt{a \sin (c+d x)+a} (e \cos (c+d x))^{3/2}}-\frac{2}{9 d e (a \sin (c+d x)+a)^{3/2} (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Int[1/((e*Cos[c + d*x])^(5/2)*(a + a*Sin[c + d*x])^(3/2)),x]

[Out]

-2/(9*d*e*(e*Cos[c + d*x])^(3/2)*(a + a*Sin[c + d*x])^(3/2)) - 4/(15*a*d*e*(e*Cos[c + d*x])^(3/2)*Sqrt[a + a*S
in[c + d*x]]) - (16*Sqrt[a + a*Sin[c + d*x]])/(15*a^2*d*e*(e*Cos[c + d*x])^(3/2)) + (32*(a + a*Sin[c + d*x])^(
3/2))/(45*a^3*d*e*(e*Cos[c + d*x])^(3/2))

Rule 2672

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)/(a*f*g*Simplify[2*m + p + 1]), x] + Dist[Simplify[m + p + 1]/(a*
Simplify[2*m + p + 1]), 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] && ILtQ[Simplify[m + p + 1], 0] && NeQ[2*m + p + 1, 0] &&  !IGtQ[m, 0]

Rule 2671

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)/(a*f*g*m), x] /; FreeQ[{a, b, e, f, g, m, p}, x] && EqQ[a^2 - b^
2, 0] && EqQ[Simplify[m + p + 1], 0] &&  !ILtQ[p, 0]

Rubi steps

\begin{align*} \int \frac{1}{(e \cos (c+d x))^{5/2} (a+a \sin (c+d x))^{3/2}} \, dx &=-\frac{2}{9 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^{3/2}}+\frac{2 \int \frac{1}{(e \cos (c+d x))^{5/2} \sqrt{a+a \sin (c+d x)}} \, dx}{3 a}\\ &=-\frac{2}{9 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^{3/2}}-\frac{4}{15 a d e (e \cos (c+d x))^{3/2} \sqrt{a+a \sin (c+d x)}}+\frac{8 \int \frac{\sqrt{a+a \sin (c+d x)}}{(e \cos (c+d x))^{5/2}} \, dx}{15 a^2}\\ &=-\frac{2}{9 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^{3/2}}-\frac{4}{15 a d e (e \cos (c+d x))^{3/2} \sqrt{a+a \sin (c+d x)}}-\frac{16 \sqrt{a+a \sin (c+d x)}}{15 a^2 d e (e \cos (c+d x))^{3/2}}+\frac{16 \int \frac{(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{5/2}} \, dx}{15 a^3}\\ &=-\frac{2}{9 d e (e \cos (c+d x))^{3/2} (a+a \sin (c+d x))^{3/2}}-\frac{4}{15 a d e (e \cos (c+d x))^{3/2} \sqrt{a+a \sin (c+d x)}}-\frac{16 \sqrt{a+a \sin (c+d x)}}{15 a^2 d e (e \cos (c+d x))^{3/2}}+\frac{32 (a+a \sin (c+d x))^{3/2}}{45 a^3 d e (e \cos (c+d x))^{3/2}}\\ \end{align*}

Mathematica [A]  time = 0.171883, size = 66, normalized size = 0.43 \[ -\frac{2 (-6 \sin (c+d x)+4 \sin (3 (c+d x))+12 \cos (2 (c+d x))+7)}{45 d e (a (\sin (c+d x)+1))^{3/2} (e \cos (c+d x))^{3/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[1/((e*Cos[c + d*x])^(5/2)*(a + a*Sin[c + d*x])^(3/2)),x]

[Out]

(-2*(7 + 12*Cos[2*(c + d*x)] - 6*Sin[c + d*x] + 4*Sin[3*(c + d*x)]))/(45*d*e*(e*Cos[c + d*x])^(3/2)*(a*(1 + Si
n[c + d*x]))^(3/2))

________________________________________________________________________________________

Maple [A]  time = 0.104, size = 70, normalized size = 0.5 \begin{align*} -{\frac{ \left ( 32\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}\sin \left ( dx+c \right ) +48\, \left ( \cos \left ( dx+c \right ) \right ) ^{2}-20\,\sin \left ( dx+c \right ) -10 \right ) \cos \left ( dx+c \right ) }{45\,d} \left ( e\cos \left ( dx+c \right ) \right ) ^{-{\frac{5}{2}}} \left ( a \left ( 1+\sin \left ( dx+c \right ) \right ) \right ) ^{-{\frac{3}{2}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(1/(e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^(3/2),x)

[Out]

-2/45/d*(16*cos(d*x+c)^2*sin(d*x+c)+24*cos(d*x+c)^2-10*sin(d*x+c)-5)*cos(d*x+c)/(e*cos(d*x+c))^(5/2)/(a*(1+sin
(d*x+c)))^(3/2)

________________________________________________________________________________________

Maxima [B]  time = 1.65257, size = 504, normalized size = 3.27 \begin{align*} -\frac{2 \,{\left (19 \, \sqrt{a} \sqrt{e} + \frac{12 \, \sqrt{a} \sqrt{e} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac{58 \, \sqrt{a} \sqrt{e} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} - \frac{116 \, \sqrt{a} \sqrt{e} \sin \left (d x + c\right )^{3}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{3}} + \frac{116 \, \sqrt{a} \sqrt{e} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} + \frac{58 \, \sqrt{a} \sqrt{e} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} - \frac{12 \, \sqrt{a} \sqrt{e} \sin \left (d x + c\right )^{7}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{7}} - \frac{19 \, \sqrt{a} \sqrt{e} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )}{\left (\frac{\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{4}}{45 \,{\left (a^{2} e^{3} + \frac{4 \, a^{2} e^{3} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac{6 \, a^{2} e^{3} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac{4 \, a^{2} e^{3} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}} + \frac{a^{2} e^{3} \sin \left (d x + c\right )^{8}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{8}}\right )} d{\left (\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac{11}{2}}{\left (-\frac{\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac{5}{2}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^(3/2),x, algorithm="maxima")

[Out]

-2/45*(19*sqrt(a)*sqrt(e) + 12*sqrt(a)*sqrt(e)*sin(d*x + c)/(cos(d*x + c) + 1) - 58*sqrt(a)*sqrt(e)*sin(d*x +
c)^2/(cos(d*x + c) + 1)^2 - 116*sqrt(a)*sqrt(e)*sin(d*x + c)^3/(cos(d*x + c) + 1)^3 + 116*sqrt(a)*sqrt(e)*sin(
d*x + c)^5/(cos(d*x + c) + 1)^5 + 58*sqrt(a)*sqrt(e)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 - 12*sqrt(a)*sqrt(e)*
sin(d*x + c)^7/(cos(d*x + c) + 1)^7 - 19*sqrt(a)*sqrt(e)*sin(d*x + c)^8/(cos(d*x + c) + 1)^8)*(sin(d*x + c)^2/
(cos(d*x + c) + 1)^2 + 1)^4/((a^2*e^3 + 4*a^2*e^3*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 6*a^2*e^3*sin(d*x + c)
^4/(cos(d*x + c) + 1)^4 + 4*a^2*e^3*sin(d*x + c)^6/(cos(d*x + c) + 1)^6 + a^2*e^3*sin(d*x + c)^8/(cos(d*x + c)
+ 1)^8)*d*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(11/2)*(-sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2))

________________________________________________________________________________________

Fricas [A]  time = 2.84959, size = 289, normalized size = 1.88 \begin{align*} \frac{2 \, \sqrt{e \cos \left (d x + c\right )}{\left (24 \, \cos \left (d x + c\right )^{2} + 2 \,{\left (8 \, \cos \left (d x + c\right )^{2} - 5\right )} \sin \left (d x + c\right ) - 5\right )} \sqrt{a \sin \left (d x + c\right ) + a}}{45 \,{\left (a^{2} d e^{3} \cos \left (d x + c\right )^{4} - 2 \, a^{2} d e^{3} \cos \left (d x + c\right )^{2} \sin \left (d x + c\right ) - 2 \, a^{2} d e^{3} \cos \left (d x + c\right )^{2}\right )}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^(3/2),x, algorithm="fricas")

[Out]

2/45*sqrt(e*cos(d*x + c))*(24*cos(d*x + c)^2 + 2*(8*cos(d*x + c)^2 - 5)*sin(d*x + c) - 5)*sqrt(a*sin(d*x + c)
+ a)/(a^2*d*e^3*cos(d*x + c)^4 - 2*a^2*d*e^3*cos(d*x + c)^2*sin(d*x + c) - 2*a^2*d*e^3*cos(d*x + c)^2)

________________________________________________________________________________________

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

[Out]

Timed out

________________________________________________________________________________________

Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{1}{\left (e \cos \left (d x + c\right )\right )^{\frac{5}{2}}{\left (a \sin \left (d x + c\right ) + a\right )}^{\frac{3}{2}}}\,{d x} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/(e*cos(d*x+c))^(5/2)/(a+a*sin(d*x+c))^(3/2),x, algorithm="giac")

[Out]

integrate(1/((e*cos(d*x + c))^(5/2)*(a*sin(d*x + c) + a)^(3/2)), x)