3.335 \(\int \cos (e+f x) (a+b \sin ^2(e+f x))^{3/2} \, dx\)

Optimal. Leaf size=104 \[ \frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{8 \sqrt {b} f}+\frac {3 a \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{8 f}+\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 f} \]

[Out]

1/4*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(3/2)/f+3/8*a^2*arctanh(sin(f*x+e)*b^(1/2)/(a+b*sin(f*x+e)^2)^(1/2))/f/b^(1/
2)+3/8*a*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)/f

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Rubi [A]  time = 0.07, antiderivative size = 104, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 4, integrand size = 23, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.174, Rules used = {3190, 195, 217, 206} \[ \frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{8 \sqrt {b} f}+\frac {3 a \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{8 f}+\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 f} \]

Antiderivative was successfully verified.

[In]

Int[Cos[e + f*x]*(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

(3*a^2*ArcTanh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]])/(8*Sqrt[b]*f) + (3*a*Sin[e + f*x]*Sqrt[a +
b*Sin[e + f*x]^2])/(8*f) + (Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^(3/2))/(4*f)

Rule 195

Int[((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> Simp[(x*(a + b*x^n)^p)/(n*p + 1), x] + Dist[(a*n*p)/(n*p + 1),
Int[(a + b*x^n)^(p - 1), x], x] /; FreeQ[{a, b}, x] && IGtQ[n, 0] && GtQ[p, 0] && (IntegerQ[2*p] || (EqQ[n, 2
] && IntegerQ[4*p]) || (EqQ[n, 2] && IntegerQ[3*p]) || LtQ[Denominator[p + 1/n], Denominator[p]])

Rule 206

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

Rule 217

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

Rule 3190

Int[cos[(e_.) + (f_.)*(x_)]^(m_.)*((a_) + (b_.)*sin[(e_.) + (f_.)*(x_)]^2)^(p_.), x_Symbol] :> With[{ff = Free
Factors[Sin[e + f*x], x]}, Dist[ff/f, Subst[Int[(1 - ff^2*x^2)^((m - 1)/2)*(a + b*ff^2*x^2)^p, x], x, Sin[e +
f*x]/ff], x]] /; FreeQ[{a, b, e, f, p}, x] && IntegerQ[(m - 1)/2]

Rubi steps

\begin {align*} \int \cos (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx &=\frac {\operatorname {Subst}\left (\int \left (a+b x^2\right )^{3/2} \, dx,x,\sin (e+f x)\right )}{f}\\ &=\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 f}+\frac {(3 a) \operatorname {Subst}\left (\int \sqrt {a+b x^2} \, dx,x,\sin (e+f x)\right )}{4 f}\\ &=\frac {3 a \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{8 f}+\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 f}+\frac {\left (3 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{\sqrt {a+b x^2}} \, dx,x,\sin (e+f x)\right )}{8 f}\\ &=\frac {3 a \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{8 f}+\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 f}+\frac {\left (3 a^2\right ) \operatorname {Subst}\left (\int \frac {1}{1-b x^2} \, dx,x,\frac {\sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{8 f}\\ &=\frac {3 a^2 \tanh ^{-1}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a+b \sin ^2(e+f x)}}\right )}{8 \sqrt {b} f}+\frac {3 a \sin (e+f x) \sqrt {a+b \sin ^2(e+f x)}}{8 f}+\frac {\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{4 f}\\ \end {align*}

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Mathematica [A]  time = 0.49, size = 93, normalized size = 0.89 \[ \frac {\sqrt {a+b \sin ^2(e+f x)} \left (\frac {3 a^{3/2} \sinh ^{-1}\left (\frac {\sqrt {b} \sin (e+f x)}{\sqrt {a}}\right )}{\sqrt {b} \sqrt {\frac {b \sin ^2(e+f x)}{a}+1}}+5 a \sin (e+f x)+2 b \sin ^3(e+f x)\right )}{8 f} \]

Antiderivative was successfully verified.

[In]

Integrate[Cos[e + f*x]*(a + b*Sin[e + f*x]^2)^(3/2),x]

[Out]

(Sqrt[a + b*Sin[e + f*x]^2]*(5*a*Sin[e + f*x] + 2*b*Sin[e + f*x]^3 + (3*a^(3/2)*ArcSinh[(Sqrt[b]*Sin[e + f*x])
/Sqrt[a]])/(Sqrt[b]*Sqrt[1 + (b*Sin[e + f*x]^2)/a])))/(8*f)

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fricas [B]  time = 0.82, size = 503, normalized size = 4.84 \[ \left [\frac {3 \, a^{2} \sqrt {b} \log \left (128 \, b^{4} \cos \left (f x + e\right )^{8} - 256 \, {\left (a b^{3} + 2 \, b^{4}\right )} \cos \left (f x + e\right )^{6} + 32 \, {\left (5 \, a^{2} b^{2} + 24 \, a b^{3} + 24 \, b^{4}\right )} \cos \left (f x + e\right )^{4} + a^{4} + 32 \, a^{3} b + 160 \, a^{2} b^{2} + 256 \, a b^{3} + 128 \, b^{4} - 32 \, {\left (a^{3} b + 10 \, a^{2} b^{2} + 24 \, a b^{3} + 16 \, b^{4}\right )} \cos \left (f x + e\right )^{2} - 8 \, {\left (16 \, b^{3} \cos \left (f x + e\right )^{6} - 24 \, {\left (a b^{2} + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{4} - a^{3} - 10 \, a^{2} b - 24 \, a b^{2} - 16 \, b^{3} + 2 \, {\left (5 \, a^{2} b + 24 \, a b^{2} + 24 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {b} \sin \left (f x + e\right )\right ) - 8 \, {\left (2 \, b^{2} \cos \left (f x + e\right )^{2} - 5 \, a b - 2 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{64 \, b f}, -\frac {3 \, a^{2} \sqrt {-b} \arctan \left (\frac {{\left (8 \, b^{2} \cos \left (f x + e\right )^{4} - 8 \, {\left (a b + 2 \, b^{2}\right )} \cos \left (f x + e\right )^{2} + a^{2} + 8 \, a b + 8 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sqrt {-b}}{4 \, {\left (2 \, b^{3} \cos \left (f x + e\right )^{4} + a^{2} b + 3 \, a b^{2} + 2 \, b^{3} - {\left (3 \, a b^{2} + 4 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sin \left (f x + e\right )}\right ) + 4 \, {\left (2 \, b^{2} \cos \left (f x + e\right )^{2} - 5 \, a b - 2 \, b^{2}\right )} \sqrt {-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{32 \, b f}\right ] \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="fricas")

[Out]

[1/64*(3*a^2*sqrt(b)*log(128*b^4*cos(f*x + e)^8 - 256*(a*b^3 + 2*b^4)*cos(f*x + e)^6 + 32*(5*a^2*b^2 + 24*a*b^
3 + 24*b^4)*cos(f*x + e)^4 + a^4 + 32*a^3*b + 160*a^2*b^2 + 256*a*b^3 + 128*b^4 - 32*(a^3*b + 10*a^2*b^2 + 24*
a*b^3 + 16*b^4)*cos(f*x + e)^2 - 8*(16*b^3*cos(f*x + e)^6 - 24*(a*b^2 + 2*b^3)*cos(f*x + e)^4 - a^3 - 10*a^2*b
- 24*a*b^2 - 16*b^3 + 2*(5*a^2*b + 24*a*b^2 + 24*b^3)*cos(f*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(b)
*sin(f*x + e)) - 8*(2*b^2*cos(f*x + e)^2 - 5*a*b - 2*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e))/(b*f),
-1/32*(3*a^2*sqrt(-b)*arctan(1/4*(8*b^2*cos(f*x + e)^4 - 8*(a*b + 2*b^2)*cos(f*x + e)^2 + a^2 + 8*a*b + 8*b^2
)*sqrt(-b*cos(f*x + e)^2 + a + b)*sqrt(-b)/((2*b^3*cos(f*x + e)^4 + a^2*b + 3*a*b^2 + 2*b^3 - (3*a*b^2 + 4*b^3
)*cos(f*x + e)^2)*sin(f*x + e))) + 4*(2*b^2*cos(f*x + e)^2 - 5*a*b - 2*b^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*si
n(f*x + e))/(b*f)]

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giac [F(-2)]  time = 0.00, size = 0, normalized size = 0.00 \[ \text {Exception raised: NotImplementedError} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="giac")

[Out]

Exception raised: NotImplementedError >> Unable to parse Giac output: Unable to check sign: (2*pi/x/2)>(-2*pi/
x/2)Unable to check sign: (2*pi/x/2)>(-2*pi/x/2)1/f*(2*(4*b^3/32/b^2*sin(f*x+exp(1))*sin(f*x+exp(1))+10*b^2*a/
32/b^2)*sin(f*x+exp(1))*sqrt(a+b*sin(f*x+exp(1))^2)-6*a^2/16/sqrt(b)*ln(abs(sqrt(a+b*sin(f*x+exp(1))^2)-sqrt(b
)*sin(f*x+exp(1)))))

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maple [A]  time = 0.22, size = 90, normalized size = 0.87 \[ \frac {\sin \left (f x +e \right ) \left (a +b \left (\sin ^{2}\left (f x +e \right )\right )\right )^{\frac {3}{2}}}{4 f}+\frac {3 a \sin \left (f x +e \right ) \sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}}{8 f}+\frac {3 a^{2} \ln \left (\sin \left (f x +e \right ) \sqrt {b}+\sqrt {a +b \left (\sin ^{2}\left (f x +e \right )\right )}\right )}{8 f \sqrt {b}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(f*x+e)*(a+b*sin(f*x+e)^2)^(3/2),x)

[Out]

1/4*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(3/2)/f+3/8*a*sin(f*x+e)*(a+b*sin(f*x+e)^2)^(1/2)/f+3/8/f*a^2*ln(sin(f*x+e)*
b^(1/2)+(a+b*sin(f*x+e)^2)^(1/2))/b^(1/2)

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maxima [A]  time = 0.36, size = 73, normalized size = 0.70 \[ \frac {\frac {3 \, a^{2} \operatorname {arsinh}\left (\frac {b \sin \left (f x + e\right )}{\sqrt {a b}}\right )}{\sqrt {b}} + 2 \, {\left (b \sin \left (f x + e\right )^{2} + a\right )}^{\frac {3}{2}} \sin \left (f x + e\right ) + 3 \, \sqrt {b \sin \left (f x + e\right )^{2} + a} a \sin \left (f x + e\right )}{8 \, f} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(cos(f*x+e)*(a+b*sin(f*x+e)^2)^(3/2),x, algorithm="maxima")

[Out]

1/8*(3*a^2*arcsinh(b*sin(f*x + e)/sqrt(a*b))/sqrt(b) + 2*(b*sin(f*x + e)^2 + a)^(3/2)*sin(f*x + e) + 3*sqrt(b*
sin(f*x + e)^2 + a)*a*sin(f*x + e))/f

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mupad [B]  time = 14.54, size = 60, normalized size = 0.58 \[ \frac {\sin \left (e+f\,x\right )\,{\left (b\,{\sin \left (e+f\,x\right )}^2+a\right )}^{3/2}\,{{}}_2{\mathrm {F}}_1\left (-\frac {3}{2},\frac {1}{2};\ \frac {3}{2};\ -\frac {b\,{\sin \left (e+f\,x\right )}^2}{a}\right )}{f\,{\left (\frac {b\,{\sin \left (e+f\,x\right )}^2}{a}+1\right )}^{3/2}} \]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(cos(e + f*x)*(a + b*sin(e + f*x)^2)^(3/2),x)

[Out]

(sin(e + f*x)*(a + b*sin(e + f*x)^2)^(3/2)*hypergeom([-3/2, 1/2], 3/2, -(b*sin(e + f*x)^2)/a))/(f*((b*sin(e +
f*x)^2)/a + 1)^(3/2))

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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(cos(f*x+e)*(a+b*sin(f*x+e)**2)**(3/2),x)

[Out]

Timed out

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