3.6.76 \(\int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{5/2}} \, dx\) [576]

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

[Out]

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

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Rubi [A]
time = 0.14, antiderivative size = 115, normalized size of antiderivative = 1.00, number of steps used = 3, number of rules used = 3, integrand size = 29, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.103, Rules used = {2841, 21, 2850} \begin {gather*} \frac {2 a^2 (c-d) \cos (e+f x)}{3 d f (c+d) \sqrt {a \sin (e+f x)+a} (c+d \sin (e+f x))^{3/2}}-\frac {2 a^2 (c+5 d) \cos (e+f x)}{3 d f (c+d)^2 \sqrt {a \sin (e+f x)+a} \sqrt {c+d \sin (e+f x)}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

Rule 21

Int[(u_.)*((a_) + (b_.)*(v_))^(m_.)*((c_) + (d_.)*(v_))^(n_.), x_Symbol] :> Dist[(b/d)^m, Int[u*(c + d*v)^(m +
n), x], x] /; FreeQ[{a, b, c, d, n}, x] && EqQ[b*c - a*d, 0] && IntegerQ[m] && ( !IntegerQ[n] || SimplerQ[c +
d*x, a + b*x])

Rule 2841

Int[((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)])^(m_)*((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(n_), x_Symbol] :> Sim
p[(-b^2)*(b*c - a*d)*Cos[e + f*x]*(a + b*Sin[e + f*x])^(m - 2)*((c + d*Sin[e + f*x])^(n + 1)/(d*f*(n + 1)*(b*c
+ a*d))), x] + Dist[b^2/(d*(n + 1)*(b*c + a*d)), Int[(a + b*Sin[e + f*x])^(m - 2)*(c + d*Sin[e + f*x])^(n + 1
)*Simp[a*c*(m - 2) - b*d*(m - 2*n - 4) - (b*c*(m - 1) - a*d*(m + 2*n + 1))*Sin[e + f*x], x], 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] && GtQ[m, 1] && LtQ[n, -1
] && (IntegersQ[2*m, 2*n] || IntegerQ[m + 1/2] || (IntegerQ[m] && EqQ[c, 0]))

Rule 2850

Int[Sqrt[(a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]]/((c_.) + (d_.)*sin[(e_.) + (f_.)*(x_)])^(3/2), x_Symbol] :> Sim
p[-2*b^2*(Cos[e + f*x]/(f*(b*c + a*d)*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]

Rubi steps

\begin {align*} \int \frac {(a+a \sin (e+f x))^{3/2}}{(c+d \sin (e+f x))^{5/2}} \, dx &=\frac {2 a^2 (c-d) \cos (e+f x)}{3 d (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}-\frac {(2 a) \int \frac {-\frac {1}{2} a (c+5 d)-\frac {1}{2} a (c+5 d) \sin (e+f x)}{\sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}} \, dx}{3 d (c+d)}\\ &=\frac {2 a^2 (c-d) \cos (e+f x)}{3 d (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}+\frac {(a (c+5 d)) \int \frac {\sqrt {a+a \sin (e+f x)}}{(c+d \sin (e+f x))^{3/2}} \, dx}{3 d (c+d)}\\ &=\frac {2 a^2 (c-d) \cos (e+f x)}{3 d (c+d) f \sqrt {a+a \sin (e+f x)} (c+d \sin (e+f x))^{3/2}}-\frac {2 a^2 (c+5 d) \cos (e+f x)}{3 d (c+d)^2 f \sqrt {a+a \sin (e+f x)} \sqrt {c+d \sin (e+f x)}}\\ \end {align*}

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Mathematica [A]
time = 0.38, size = 104, normalized size = 0.90 \begin {gather*} -\frac {2 a \left (\cos \left (\frac {1}{2} (e+f x)\right )-\sin \left (\frac {1}{2} (e+f x)\right )\right ) \sqrt {a (1+\sin (e+f x))} (5 c+d+(c+5 d) \sin (e+f x))}{3 (c+d)^2 f \left (\cos \left (\frac {1}{2} (e+f x)\right )+\sin \left (\frac {1}{2} (e+f x)\right )\right ) (c+d \sin (e+f x))^{3/2}} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

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

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Maple [B] Leaf count of result is larger than twice the leaf count of optimal. \(344\) vs. \(2(103)=206\).
time = 0.50, size = 345, normalized size = 3.00

method	result	size
default	\(-\frac {2 \left (a \left (1+\sin \left (f x +e \right )\right )\right )^{\frac {3}{2}} \sqrt {c +d \sin \left (f x +e \right )}\, \left (\sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right ) c \,d^{2}+5 \sin \left (f x +e \right ) \left (\cos ^{4}\left (f x +e \right )\right ) d^{3}-2 \left (\cos ^{4}\left (f x +e \right )\right ) c^{2} d -7 \left (\cos ^{4}\left (f x +e \right )\right ) c \,d^{2}-9 \left (\cos ^{4}\left (f x +e \right )\right ) d^{3}-\sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) c^{3}+\sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) c^{2} d -11 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) c \,d^{2}-13 \sin \left (f x +e \right ) \left (\cos ^{2}\left (f x +e \right )\right ) d^{3}-3 \left (\cos ^{2}\left (f x +e \right )\right ) c^{3}-5 \left (\cos ^{2}\left (f x +e \right )\right ) c^{2} d +15 \left (\cos ^{2}\left (f x +e \right )\right ) c \,d^{2}+17 \left (\cos ^{2}\left (f x +e \right )\right ) d^{3}-8 c^{3} \sin \left (f x +e \right )-8 c^{2} d \sin \left (f x +e \right )+8 c \,d^{2} \sin \left (f x +e \right )+8 d^{3} \sin \left (f x +e \right )+8 c^{3}+8 c^{2} d -8 c \,d^{2}-8 d^{3}\right )}{3 f \cos \left (f x +e \right )^{3} \left (\left (\cos ^{2}\left (f x +e \right )\right ) d^{2}+c^{2}-d^{2}\right )^{2} \left (c +d \right )^{2}}\)	\(345\)

Verification of antiderivative is not currently implemented for this CAS.

[In]

int((a+a*sin(f*x+e))^(3/2)/(c+d*sin(f*x+e))^(5/2),x,method=_RETURNVERBOSE)

[Out]

-2/3/f*(a*(1+sin(f*x+e)))^(3/2)*(c+d*sin(f*x+e))^(1/2)*(sin(f*x+e)*cos(f*x+e)^4*c*d^2+5*sin(f*x+e)*cos(f*x+e)^
4*d^3-2*cos(f*x+e)^4*c^2*d-7*cos(f*x+e)^4*c*d^2-9*cos(f*x+e)^4*d^3-sin(f*x+e)*cos(f*x+e)^2*c^3+sin(f*x+e)*cos(
f*x+e)^2*c^2*d-11*sin(f*x+e)*cos(f*x+e)^2*c*d^2-13*sin(f*x+e)*cos(f*x+e)^2*d^3-3*cos(f*x+e)^2*c^3-5*cos(f*x+e)
^2*c^2*d+15*cos(f*x+e)^2*c*d^2+17*cos(f*x+e)^2*d^3-8*c^3*sin(f*x+e)-8*c^2*d*sin(f*x+e)+8*c*d^2*sin(f*x+e)+8*d^
3*sin(f*x+e)+8*c^3+8*c^2*d-8*c*d^2-8*d^3)/cos(f*x+e)^3/(cos(f*x+e)^2*d^2+c^2-d^2)^2/(c+d)^2

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Maxima [B] Leaf count of result is larger than twice the leaf count of optimal. 325 vs. \(2 (109) = 218\).
time = 0.58, size = 325, normalized size = 2.83 \begin {gather*} -\frac {2 \, {\left ({\left (5 \, c^{2} + c d\right )} a^{\frac {3}{2}} - \frac {{\left (3 \, c^{2} - 19 \, c d - 2 \, d^{2}\right )} a^{\frac {3}{2}} \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {2 \, {\left (4 \, c^{2} - 7 \, c d + 9 \, d^{2}\right )} a^{\frac {3}{2}} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} - \frac {2 \, {\left (4 \, c^{2} - 7 \, c d + 9 \, d^{2}\right )} a^{\frac {3}{2}} \sin \left (f x + e\right )^{3}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{3}} + \frac {{\left (3 \, c^{2} - 19 \, c d - 2 \, d^{2}\right )} a^{\frac {3}{2}} \sin \left (f x + e\right )^{4}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{4}} - \frac {{\left (5 \, c^{2} + c d\right )} a^{\frac {3}{2}} \sin \left (f x + e\right )^{5}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{5}}\right )} {\left (\frac {\sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}} + 1\right )}}{3 \, {\left (c^{2} + 2 \, c d + d^{2} + \frac {{\left (c^{2} + 2 \, c d + d^{2}\right )} \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )} {\left (c + \frac {2 \, d \sin \left (f x + e\right )}{\cos \left (f x + e\right ) + 1} + \frac {c \sin \left (f x + e\right )^{2}}{{\left (\cos \left (f x + e\right ) + 1\right )}^{2}}\right )}^{\frac {5}{2}} f} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-2/3*((5*c^2 + c*d)*a^(3/2) - (3*c^2 - 19*c*d - 2*d^2)*a^(3/2)*sin(f*x + e)/(cos(f*x + e) + 1) + 2*(4*c^2 - 7*
c*d + 9*d^2)*a^(3/2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2 - 2*(4*c^2 - 7*c*d + 9*d^2)*a^(3/2)*sin(f*x + e)^3/(c
os(f*x + e) + 1)^3 + (3*c^2 - 19*c*d - 2*d^2)*a^(3/2)*sin(f*x + e)^4/(cos(f*x + e) + 1)^4 - (5*c^2 + c*d)*a^(3
/2)*sin(f*x + e)^5/(cos(f*x + e) + 1)^5)*(sin(f*x + e)^2/(cos(f*x + e) + 1)^2 + 1)/((c^2 + 2*c*d + d^2 + (c^2
+ 2*c*d + d^2)*sin(f*x + e)^2/(cos(f*x + e) + 1)^2)*(c + 2*d*sin(f*x + e)/(cos(f*x + e) + 1) + c*sin(f*x + e)^
2/(cos(f*x + e) + 1)^2)^(5/2)*f)

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Fricas [B] Leaf count of result is larger than twice the leaf count of optimal. 335 vs. \(2 (109) = 218\).
time = 0.35, size = 335, normalized size = 2.91 \begin {gather*} \frac {2 \, {\left ({\left (a c + 5 \, a d\right )} \cos \left (f x + e\right )^{2} + 4 \, a c - 4 \, a d + {\left (5 \, a c + a d\right )} \cos \left (f x + e\right ) - {\left (4 \, a c - 4 \, a d - {\left (a c + 5 \, a d\right )} \cos \left (f x + e\right )\right )} \sin \left (f x + e\right )\right )} \sqrt {a \sin \left (f x + e\right ) + a} \sqrt {d \sin \left (f x + e\right ) + c}}{3 \, {\left ({\left (c^{2} d^{2} + 2 \, c d^{3} + d^{4}\right )} f \cos \left (f x + e\right )^{3} + {\left (2 \, c^{3} d + 5 \, c^{2} d^{2} + 4 \, c d^{3} + d^{4}\right )} f \cos \left (f x + e\right )^{2} - {\left (c^{4} + 2 \, c^{3} d + 2 \, c^{2} d^{2} + 2 \, c d^{3} + d^{4}\right )} f \cos \left (f x + e\right ) - {\left (c^{4} + 4 \, c^{3} d + 6 \, c^{2} d^{2} + 4 \, c d^{3} + d^{4}\right )} f + {\left ({\left (c^{2} d^{2} + 2 \, c d^{3} + d^{4}\right )} f \cos \left (f x + e\right )^{2} - 2 \, {\left (c^{3} d + 2 \, c^{2} d^{2} + c d^{3}\right )} f \cos \left (f x + e\right ) - {\left (c^{4} + 4 \, c^{3} d + 6 \, c^{2} d^{2} + 4 \, c d^{3} + d^{4}\right )} f\right )} \sin \left (f x + e\right )\right )}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

2/3*((a*c + 5*a*d)*cos(f*x + e)^2 + 4*a*c - 4*a*d + (5*a*c + a*d)*cos(f*x + e) - (4*a*c - 4*a*d - (a*c + 5*a*d
)*cos(f*x + e))*sin(f*x + e))*sqrt(a*sin(f*x + e) + a)*sqrt(d*sin(f*x + e) + c)/((c^2*d^2 + 2*c*d^3 + d^4)*f*c
os(f*x + e)^3 + (2*c^3*d + 5*c^2*d^2 + 4*c*d^3 + d^4)*f*cos(f*x + e)^2 - (c^4 + 2*c^3*d + 2*c^2*d^2 + 2*c*d^3
+ d^4)*f*cos(f*x + e) - (c^4 + 4*c^3*d + 6*c^2*d^2 + 4*c*d^3 + d^4)*f + ((c^2*d^2 + 2*c*d^3 + d^4)*f*cos(f*x +
e)^2 - 2*(c^3*d + 2*c^2*d^2 + c*d^3)*f*cos(f*x + e) - (c^4 + 4*c^3*d + 6*c^2*d^2 + 4*c*d^3 + d^4)*f)*sin(f*x
+ e))

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Sympy [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \int \frac {\left (a \left (\sin {\left (e + f x \right )} + 1\right )\right )^{\frac {3}{2}}}{\left (c + d \sin {\left (e + f x \right )}\right )^{\frac {5}{2}}}\, dx \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Giac [F(-1)] Timed out
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Timed out} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Timed out

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Mupad [B]
time = 14.40, size = 387, normalized size = 3.37 \begin {gather*} -\frac {\sqrt {c+d\,\sin \left (e+f\,x\right )}\,\left (\frac {a\,{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\left (c+5\,d\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,4{}\mathrm {i}}{3\,d^2\,f\,{\left (c\,1{}\mathrm {i}+d\,1{}\mathrm {i}\right )}^2}+\frac {a\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\left (3\,c-d\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,4{}\mathrm {i}}{d^2\,f\,{\left (c\,1{}\mathrm {i}+d\,1{}\mathrm {i}\right )}^2}-\frac {a\,{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,\left (c\,3{}\mathrm {i}-d\,1{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,4{}\mathrm {i}}{d^2\,f\,{\left (c\,1{}\mathrm {i}+d\,1{}\mathrm {i}\right )}^2}-\frac {a\,{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,\left (c\,1{}\mathrm {i}+d\,5{}\mathrm {i}\right )\,\sqrt {a+a\,\sin \left (e+f\,x\right )}\,4{}\mathrm {i}}{3\,d^2\,f\,{\left (c\,1{}\mathrm {i}+d\,1{}\mathrm {i}\right )}^2}\right )}{{\mathrm {e}}^{e\,5{}\mathrm {i}+f\,x\,5{}\mathrm {i}}-\frac {{\left (c+d\right )}^2\,1{}\mathrm {i}}{{\left (c\,1{}\mathrm {i}+d\,1{}\mathrm {i}\right )}^2}-\frac {2\,{\mathrm {e}}^{e\,3{}\mathrm {i}+f\,x\,3{}\mathrm {i}}\,\left (2\,c^2+2\,c\,d+d^2\right )}{d^2}+\frac {{\mathrm {e}}^{e\,1{}\mathrm {i}+f\,x\,1{}\mathrm {i}}\,\left (4\,c+d\right )}{d}+\frac {{\mathrm {e}}^{e\,2{}\mathrm {i}+f\,x\,2{}\mathrm {i}}\,{\left (c+d\right )}^2\,\left (2\,c^2+2\,c\,d+d^2\right )\,2{}\mathrm {i}}{d^2\,{\left (c\,1{}\mathrm {i}+d\,1{}\mathrm {i}\right )}^2}-\frac {{\mathrm {e}}^{e\,4{}\mathrm {i}+f\,x\,4{}\mathrm {i}}\,{\left (c+d\right )}^2\,\left (4\,c+d\right )\,1{}\mathrm {i}}{d\,{\left (c\,1{}\mathrm {i}+d\,1{}\mathrm {i}\right )}^2}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-((c + d*sin(e + f*x))^(1/2)*((a*exp(e*1i + f*x*1i)*(c + 5*d)*(a + a*sin(e + f*x))^(1/2)*4i)/(3*d^2*f*(c*1i +
d*1i)^2) + (a*exp(e*3i + f*x*3i)*(3*c - d)*(a + a*sin(e + f*x))^(1/2)*4i)/(d^2*f*(c*1i + d*1i)^2) - (a*exp(e*2
i + f*x*2i)*(c*3i - d*1i)*(a + a*sin(e + f*x))^(1/2)*4i)/(d^2*f*(c*1i + d*1i)^2) - (a*exp(e*4i + f*x*4i)*(c*1i
+ d*5i)*(a + a*sin(e + f*x))^(1/2)*4i)/(3*d^2*f*(c*1i + d*1i)^2)))/(exp(e*5i + f*x*5i) - ((c + d)^2*1i)/(c*1i
+ d*1i)^2 - (2*exp(e*3i + f*x*3i)*(2*c*d + 2*c^2 + d^2))/d^2 + (exp(e*1i + f*x*1i)*(4*c + d))/d + (exp(e*2i +
f*x*2i)*(c + d)^2*(2*c*d + 2*c^2 + d^2)*2i)/(d^2*(c*1i + d*1i)^2) - (exp(e*4i + f*x*4i)*(c + d)^2*(4*c + d)*1
i)/(d*(c*1i + d*1i)^2))

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