3.278 \(\int \frac {\sqrt {a+a \sin (c+d x)}}{(e \cos (c+d x))^{7/2}} \, dx\)

Optimal. Leaf size=115 \[ -\frac {16 (a \sin (c+d x)+a)^{5/2}}{15 a^2 d e (e \cos (c+d x))^{5/2}}+\frac {8 (a \sin (c+d x)+a)^{3/2}}{3 a d e (e \cos (c+d x))^{5/2}}-\frac {2 \sqrt {a \sin (c+d x)+a}}{3 d e (e \cos (c+d x))^{5/2}} \]

[Out]

8/3*(a+a*sin(d*x+c))^(3/2)/a/d/e/(e*cos(d*x+c))^(5/2)-16/15*(a+a*sin(d*x+c))^(5/2)/a^2/d/e/(e*cos(d*x+c))^(5/2
)-2/3*(a+a*sin(d*x+c))^(1/2)/d/e/(e*cos(d*x+c))^(5/2)

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Rubi [A]  time = 0.22, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 2, integrand size = 27, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.074, Rules used = {2672, 2671} \[ -\frac {16 (a \sin (c+d x)+a)^{5/2}}{15 a^2 d e (e \cos (c+d x))^{5/2}}+\frac {8 (a \sin (c+d x)+a)^{3/2}}{3 a d e (e \cos (c+d x))^{5/2}}-\frac {2 \sqrt {a \sin (c+d x)+a}}{3 d e (e \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Int[Sqrt[a + a*Sin[c + d*x]]/(e*Cos[c + d*x])^(7/2),x]

[Out]

(-2*Sqrt[a + a*Sin[c + d*x]])/(3*d*e*(e*Cos[c + d*x])^(5/2)) + (8*(a + a*Sin[c + d*x])^(3/2))/(3*a*d*e*(e*Cos[
c + d*x])^(5/2)) - (16*(a + a*Sin[c + d*x])^(5/2))/(15*a^2*d*e*(e*Cos[c + d*x])^(5/2))

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]

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]

Rubi steps

\begin {align*} \int \frac {\sqrt {a+a \sin (c+d x)}}{(e \cos (c+d x))^{7/2}} \, dx &=-\frac {2 \sqrt {a+a \sin (c+d x)}}{3 d e (e \cos (c+d x))^{5/2}}+\frac {4 \int \frac {(a+a \sin (c+d x))^{3/2}}{(e \cos (c+d x))^{7/2}} \, dx}{3 a}\\ &=-\frac {2 \sqrt {a+a \sin (c+d x)}}{3 d e (e \cos (c+d x))^{5/2}}+\frac {8 (a+a \sin (c+d x))^{3/2}}{3 a d e (e \cos (c+d x))^{5/2}}-\frac {8 \int \frac {(a+a \sin (c+d x))^{5/2}}{(e \cos (c+d x))^{7/2}} \, dx}{3 a^2}\\ &=-\frac {2 \sqrt {a+a \sin (c+d x)}}{3 d e (e \cos (c+d x))^{5/2}}+\frac {8 (a+a \sin (c+d x))^{3/2}}{3 a d e (e \cos (c+d x))^{5/2}}-\frac {16 (a+a \sin (c+d x))^{5/2}}{15 a^2 d e (e \cos (c+d x))^{5/2}}\\ \end {align*}

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Mathematica [A]  time = 0.34, size = 56, normalized size = 0.49 \[ \frac {2 \sqrt {a (\sin (c+d x)+1)} (4 \sin (c+d x)+4 \cos (2 (c+d x))+3)}{15 d e (e \cos (c+d x))^{5/2}} \]

Antiderivative was successfully verified.

[In]

Integrate[Sqrt[a + a*Sin[c + d*x]]/(e*Cos[c + d*x])^(7/2),x]

[Out]

(2*Sqrt[a*(1 + Sin[c + d*x])]*(3 + 4*Cos[2*(c + d*x)] + 4*Sin[c + d*x]))/(15*d*e*(e*Cos[c + d*x])^(5/2))

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fricas [A]  time = 1.08, size = 58, normalized size = 0.50 \[ \frac {2 \, \sqrt {e \cos \left (d x + c\right )} {\left (8 \, \cos \left (d x + c\right )^{2} + 4 \, \sin \left (d x + c\right ) - 1\right )} \sqrt {a \sin \left (d x + c\right ) + a}}{15 \, d e^{4} \cos \left (d x + c\right )^{3}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^(1/2)/(e*cos(d*x+c))^(7/2),x, algorithm="fricas")

[Out]

2/15*sqrt(e*cos(d*x + c))*(8*cos(d*x + c)^2 + 4*sin(d*x + c) - 1)*sqrt(a*sin(d*x + c) + a)/(d*e^4*cos(d*x + c)
^3)

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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))^(1/2)/(e*cos(d*x+c))^(7/2),x, algorithm="giac")

[Out]

Timed out

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maple [A]  time = 0.21, size = 54, normalized size = 0.47 \[ \frac {2 \left (8 \left (\cos ^{2}\left (d x +c \right )\right )+4 \sin \left (d x +c \right )-1\right ) \sqrt {a \left (1+\sin \left (d x +c \right )\right )}\, \cos \left (d x +c \right )}{15 d \left (e \cos \left (d x +c \right )\right )^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(d*x+c))^(1/2)/(e*cos(d*x+c))^(7/2),x)

[Out]

2/15/d*(8*cos(d*x+c)^2+4*sin(d*x+c)-1)*(a*(1+sin(d*x+c)))^(1/2)*cos(d*x+c)/(e*cos(d*x+c))^(7/2)

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maxima [B]  time = 0.98, size = 282, normalized size = 2.45 \[ \frac {2 \, {\left (7 \, \sqrt {a} \sqrt {e} + \frac {8 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} - \frac {25 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {25 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} - \frac {8 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{5}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{5}} - \frac {7 \, \sqrt {a} \sqrt {e} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )} {\left (\frac {\sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + 1\right )}^{3}}{15 \, {\left (e^{4} + \frac {3 \, e^{4} \sin \left (d x + c\right )^{2}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{2}} + \frac {3 \, e^{4} \sin \left (d x + c\right )^{4}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{4}} + \frac {e^{4} \sin \left (d x + c\right )^{6}}{{\left (\cos \left (d x + c\right ) + 1\right )}^{6}}\right )} d {\left (\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {5}{2}} {\left (-\frac {\sin \left (d x + c\right )}{\cos \left (d x + c\right ) + 1} + 1\right )}^{\frac {7}{2}}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((a+a*sin(d*x+c))^(1/2)/(e*cos(d*x+c))^(7/2),x, algorithm="maxima")

[Out]

2/15*(7*sqrt(a)*sqrt(e) + 8*sqrt(a)*sqrt(e)*sin(d*x + c)/(cos(d*x + c) + 1) - 25*sqrt(a)*sqrt(e)*sin(d*x + c)^
2/(cos(d*x + c) + 1)^2 + 25*sqrt(a)*sqrt(e)*sin(d*x + c)^4/(cos(d*x + c) + 1)^4 - 8*sqrt(a)*sqrt(e)*sin(d*x +
c)^5/(cos(d*x + c) + 1)^5 - 7*sqrt(a)*sqrt(e)*sin(d*x + c)^6/(cos(d*x + c) + 1)^6)*(sin(d*x + c)^2/(cos(d*x +
c) + 1)^2 + 1)^3/((e^4 + 3*e^4*sin(d*x + c)^2/(cos(d*x + c) + 1)^2 + 3*e^4*sin(d*x + c)^4/(cos(d*x + c) + 1)^4
+ e^4*sin(d*x + c)^6/(cos(d*x + c) + 1)^6)*d*(sin(d*x + c)/(cos(d*x + c) + 1) + 1)^(5/2)*(-sin(d*x + c)/(cos(
d*x + c) + 1) + 1)^(7/2))

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mupad [B]  time = 6.06, size = 97, normalized size = 0.84 \[ \frac {8\,\sqrt {a\,\left (\sin \left (c+d\,x\right )+1\right )}\,\left (2\,\sin \left (c+d\,x\right )+7\,\cos \left (2\,c+2\,d\,x\right )+2\,\cos \left (4\,c+4\,d\,x\right )+2\,\sin \left (3\,c+3\,d\,x\right )+5\right )}{15\,d\,e^3\,\sqrt {e\,\cos \left (c+d\,x\right )}\,\left (4\,\cos \left (2\,c+2\,d\,x\right )+\cos \left (4\,c+4\,d\,x\right )+3\right )} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a + a*sin(c + d*x))^(1/2)/(e*cos(c + d*x))^(7/2),x)

[Out]

(8*(a*(sin(c + d*x) + 1))^(1/2)*(2*sin(c + d*x) + 7*cos(2*c + 2*d*x) + 2*cos(4*c + 4*d*x) + 2*sin(3*c + 3*d*x)
+ 5))/(15*d*e^3*(e*cos(c + d*x))^(1/2)*(4*cos(2*c + 2*d*x) + cos(4*c + 4*d*x) + 3))

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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))**(1/2)/(e*cos(d*x+c))**(7/2),x)

[Out]

Timed out

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