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

Optimal. Leaf size=157 \[ \frac{a^2 (a+6 b) \tanh ^{-1}\left (\frac{\sqrt{b} \sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}}\right )}{16 b^{3/2} f}-\frac{\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{6 b f}+\frac{(a+6 b) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 b f}+\frac{a (a+6 b) \sin (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{16 b f} \]

[Out]

(a^2*(a + 6*b)*ArcTanh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]])/(16*b^(3/2)*f) + (a*(a + 6*b)*Sin[e
+ f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(16*b*f) + ((a + 6*b)*Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^(3/2))/(24*b*f)
- (Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^(5/2))/(6*b*f)

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Rubi [A]  time = 0.137606, antiderivative size = 157, normalized size of antiderivative = 1., number of steps used = 6, number of rules used = 5, integrand size = 25, \(\frac{\text{number of rules}}{\text{integrand size}}\) = 0.2, Rules used = {3190, 388, 195, 217, 206} \[ \frac{a^2 (a+6 b) \tanh ^{-1}\left (\frac{\sqrt{b} \sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}}\right )}{16 b^{3/2} f}-\frac{\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{6 b f}+\frac{(a+6 b) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 b f}+\frac{a (a+6 b) \sin (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{16 b f} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(a^2*(a + 6*b)*ArcTanh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a + b*Sin[e + f*x]^2]])/(16*b^(3/2)*f) + (a*(a + 6*b)*Sin[e
+ f*x]*Sqrt[a + b*Sin[e + f*x]^2])/(16*b*f) + ((a + 6*b)*Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^(3/2))/(24*b*f)
- (Sin[e + f*x]*(a + b*Sin[e + f*x]^2)^(5/2))/(6*b*f)

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]

Rule 388

Int[((a_) + (b_.)*(x_)^(n_))^(p_)*((c_) + (d_.)*(x_)^(n_)), x_Symbol] :> Simp[(d*x*(a + b*x^n)^(p + 1))/(b*(n*
(p + 1) + 1)), x] - Dist[(a*d - b*c*(n*(p + 1) + 1))/(b*(n*(p + 1) + 1)), Int[(a + b*x^n)^p, x], x] /; FreeQ[{
a, b, c, d, n}, x] && NeQ[b*c - a*d, 0] && NeQ[n*(p + 1) + 1, 0]

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 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 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])

Rubi steps

\begin{align*} \int \cos ^3(e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2} \, dx &=\frac{\operatorname{Subst}\left (\int \left (1-x^2\right ) \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 )^{5/2}}{6 b f}+\frac{(a+6 b) \operatorname{Subst}\left (\int \left (a+b x^2\right )^{3/2} \, dx,x,\sin (e+f x)\right )}{6 b f}\\ &=\frac{(a+6 b) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 b f}-\frac{\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{6 b f}+\frac{(a (a+6 b)) \operatorname{Subst}\left (\int \sqrt{a+b x^2} \, dx,x,\sin (e+f x)\right )}{8 b f}\\ &=\frac{a (a+6 b) \sin (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{16 b f}+\frac{(a+6 b) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 b f}-\frac{\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{6 b f}+\frac{\left (a^2 (a+6 b)\right ) \operatorname{Subst}\left (\int \frac{1}{\sqrt{a+b x^2}} \, dx,x,\sin (e+f x)\right )}{16 b f}\\ &=\frac{a (a+6 b) \sin (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{16 b f}+\frac{(a+6 b) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 b f}-\frac{\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{6 b f}+\frac{\left (a^2 (a+6 b)\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 )}{16 b f}\\ &=\frac{a^2 (a+6 b) \tanh ^{-1}\left (\frac{\sqrt{b} \sin (e+f x)}{\sqrt{a+b \sin ^2(e+f x)}}\right )}{16 b^{3/2} f}+\frac{a (a+6 b) \sin (e+f x) \sqrt{a+b \sin ^2(e+f x)}}{16 b f}+\frac{(a+6 b) \sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{3/2}}{24 b f}-\frac{\sin (e+f x) \left (a+b \sin ^2(e+f x)\right )^{5/2}}{6 b f}\\ \end{align*}

Mathematica [A]  time = 0.84795, size = 149, normalized size = 0.95 \[ \frac{\sqrt{a+b \sin ^2(e+f x)} \left (3 a^{3/2} (a+6 b) \sinh ^{-1}\left (\frac{\sqrt{b} \sin (e+f x)}{\sqrt{a}}\right )+\sqrt{b} \sin (e+f x) \sqrt{\frac{b \sin ^2(e+f x)}{a}+1} \left (-2 b (7 a-6 b) \sin ^2(e+f x)-3 a (a-10 b)-8 b^2 \sin ^4(e+f x)\right )\right )}{48 b^{3/2} f \sqrt{\frac{b \sin ^2(e+f x)}{a}+1}} \]

Antiderivative was successfully verified.

[In]

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

[Out]

(Sqrt[a + b*Sin[e + f*x]^2]*(3*a^(3/2)*(a + 6*b)*ArcSinh[(Sqrt[b]*Sin[e + f*x])/Sqrt[a]] + Sqrt[b]*Sin[e + f*x
]*Sqrt[1 + (b*Sin[e + f*x]^2)/a]*(-3*a*(a - 10*b) - 2*(7*a - 6*b)*b*Sin[e + f*x]^2 - 8*b^2*Sin[e + f*x]^4)))/(
48*b^(3/2)*f*Sqrt[1 + (b*Sin[e + f*x]^2)/a])

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Maple [B]  time = 1.391, size = 277, normalized size = 1.8 \begin{align*} -{\frac{b\sin \left ( fx+e \right ) \left ( \cos \left ( fx+e \right ) \right ) ^{4}}{6\,f}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}+{\frac{7\, \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) a}{24\,f}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}+{\frac{ \left ( \cos \left ( fx+e \right ) \right ) ^{2}\sin \left ( fx+e \right ) b}{12\,f}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}-{\frac{\sin \left ( fx+e \right ){a}^{2}}{16\,bf}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}+{\frac{a\sin \left ( fx+e \right ) }{3\,f}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}+{\frac{b\sin \left ( fx+e \right ) }{12\,f}\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}}}+{\frac{{a}^{3}}{16\,f}\ln \left ( \sin \left ( fx+e \right ) \sqrt{b}+\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \right ){b}^{-{\frac{3}{2}}}}+{\frac{3\,{a}^{2}}{8\,f}\ln \left ( \sin \left ( fx+e \right ) \sqrt{b}+\sqrt{a+b-b \left ( \cos \left ( fx+e \right ) \right ) ^{2}} \right ){\frac{1}{\sqrt{b}}}} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/6*b/f*(a+b-b*cos(f*x+e)^2)^(1/2)*sin(f*x+e)*cos(f*x+e)^4+7/24/f*(a+b-b*cos(f*x+e)^2)^(1/2)*cos(f*x+e)^2*sin
(f*x+e)*a+1/12/f*(a+b-b*cos(f*x+e)^2)^(1/2)*cos(f*x+e)^2*sin(f*x+e)*b-1/16/b/f*(a+b-b*cos(f*x+e)^2)^(1/2)*sin(
f*x+e)*a^2+1/3/f*(a+b-b*cos(f*x+e)^2)^(1/2)*sin(f*x+e)*a+1/12*b/f*(a+b-b*cos(f*x+e)^2)^(1/2)*sin(f*x+e)+1/16/b
^(3/2)/f*a^3*ln(sin(f*x+e)*b^(1/2)+(a+b-b*cos(f*x+e)^2)^(1/2))+3/8/b^(1/2)/f*a^2*ln(sin(f*x+e)*b^(1/2)+(a+b-b*
cos(f*x+e)^2)^(1/2))

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Maxima [F(-2)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Exception raised: ValueError} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError

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Fricas [A]  time = 15.8315, size = 1393, normalized size = 8.87 \begin{align*} \left [\frac{3 \,{\left (a^{3} + 6 \, a^{2} b\right )} \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 (8 \, b^{3} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b - 16 \, a b^{2} - 4 \, b^{3} - 2 \,{\left (7 \, a b^{2} + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{384 \, b^{2} f}, -\frac{3 \,{\left (a^{3} + 6 \, a^{2} b\right )} \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 (8 \, b^{3} \cos \left (f x + e\right )^{4} + 3 \, a^{2} b - 16 \, a b^{2} - 4 \, b^{3} - 2 \,{\left (7 \, a b^{2} + 2 \, b^{3}\right )} \cos \left (f x + e\right )^{2}\right )} \sqrt{-b \cos \left (f x + e\right )^{2} + a + b} \sin \left (f x + e\right )}{192 \, b^{2} f}\right ] \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

[1/384*(3*(a^3 + 6*a^2*b)*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*(8*b^3*cos(f*x + e)^4 + 3*a^2*b - 16*a*b^2 - 4*b^3 - 2*(7*a*b^2 + 2*b^3)*cos(f
*x + e)^2)*sqrt(-b*cos(f*x + e)^2 + a + b)*sin(f*x + e))/(b^2*f), -1/192*(3*(a^3 + 6*a^2*b)*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*(8*b^3*cos(f*x + e)^4 + 3*a^2*b - 16*a*b^2 - 4*b^3 - 2*(7*a*b^2 + 2*b^3)*cos(f*x + e)^2)*sqrt(-b*cos(f*x
+ e)^2 + a + b)*sin(f*x + e))/(b^2*f)]

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

[Out]

Timed out

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Giac [A]  time = 1.23859, size = 181, normalized size = 1.15 \begin{align*} -\frac{{\left (2 \,{\left (4 \, b \sin \left (f x + e\right )^{2} + \frac{7 \, a b^{4} - 6 \, b^{5}}{b^{4}}\right )} \sin \left (f x + e\right )^{2} + \frac{3 \,{\left (a^{2} b^{3} - 10 \, a b^{4}\right )}}{b^{4}}\right )} \sqrt{b \sin \left (f x + e\right )^{2} + a} \sin \left (f x + e\right ) + \frac{3 \,{\left (a^{3} + 6 \, a^{2} b\right )} \log \left ({\left | -\sqrt{b} \sin \left (f x + e\right ) + \sqrt{b \sin \left (f x + e\right )^{2} + a} \right |}\right )}{b^{\frac{3}{2}}}}{48 \, f} \end{align*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

-1/48*((2*(4*b*sin(f*x + e)^2 + (7*a*b^4 - 6*b^5)/b^4)*sin(f*x + e)^2 + 3*(a^2*b^3 - 10*a*b^4)/b^4)*sqrt(b*sin
(f*x + e)^2 + a)*sin(f*x + e) + 3*(a^3 + 6*a^2*b)*log(abs(-sqrt(b)*sin(f*x + e) + sqrt(b*sin(f*x + e)^2 + a)))
/b^(3/2))/f