3.3.95 \(\int \frac {(e+f x)^2 \cos (c+d x)}{a+b \sin (c+d x)} \, dx\) [295]

Optimal. Leaf size=320 \[ -\frac {i (e+f x)^3}{3 b f}+\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d}-\frac {2 i f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}-\frac {2 i f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}+\frac {2 f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^3}+\frac {2 f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^3} \]

[Out]

-1/3*I*(f*x+e)^3/b/f+(f*x+e)^2*ln(1-I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/d+(f*x+e)^2*ln(1-I*b*exp(I*(d*x+
c))/(a+(a^2-b^2)^(1/2)))/b/d-2*I*f*(f*x+e)*polylog(2,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/2)))/b/d^2-2*I*f*(f*x+
e)*polylog(2,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/d^2+2*f^2*polylog(3,I*b*exp(I*(d*x+c))/(a-(a^2-b^2)^(1/
2)))/b/d^3+2*f^2*polylog(3,I*b*exp(I*(d*x+c))/(a+(a^2-b^2)^(1/2)))/b/d^3

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Rubi [A]
time = 0.33, antiderivative size = 320, normalized size of antiderivative = 1.00, number of steps used = 9, number of rules used = 5, integrand size = 26, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.192, Rules used = {4615, 2221, 2611, 2320, 6724} \begin {gather*} \frac {2 f^2 \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^3}+\frac {2 f^2 \text {PolyLog}\left (3,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d^3}-\frac {2 i f (e+f x) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}-\frac {2 i f (e+f x) \text {PolyLog}\left (2,\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d^2}+\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{\sqrt {a^2-b^2}+a}\right )}{b d}-\frac {i (e+f x)^3}{3 b f} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

((-1/3*I)*(e + f*x)^3)/(b*f) + ((e + f*x)^2*Log[1 - (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*d) + ((e
+ f*x)^2*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*d) - ((2*I)*f*(e + f*x)*PolyLog[2, (I*b*E^(I
*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*d^2) - ((2*I)*f*(e + f*x)*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a
^2 - b^2])])/(b*d^2) + (2*f^2*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])])/(b*d^3) + (2*f^2*PolyLo
g[3, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/(b*d^3)

Rule 2221

Int[(((F_)^((g_.)*((e_.) + (f_.)*(x_))))^(n_.)*((c_.) + (d_.)*(x_))^(m_.))/((a_) + (b_.)*((F_)^((g_.)*((e_.) +
(f_.)*(x_))))^(n_.)), x_Symbol] :> Simp[((c + d*x)^m/(b*f*g*n*Log[F]))*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x]
- Dist[d*(m/(b*f*g*n*Log[F])), Int[(c + d*x)^(m - 1)*Log[1 + b*((F^(g*(e + f*x)))^n/a)], x], x] /; FreeQ[{F,
a, b, c, d, e, f, g, n}, x] && IGtQ[m, 0]

Rule 2320

Int[u_, x_Symbol] :> With[{v = FunctionOfExponential[u, x]}, Dist[v/D[v, x], Subst[Int[FunctionOfExponentialFu
nction[u, x]/x, x], x, v], x]] /; FunctionOfExponentialQ[u, x] &&  !MatchQ[u, (w_)*((a_.)*(v_)^(n_))^(m_) /; F
reeQ[{a, m, n}, x] && IntegerQ[m*n]] &&  !MatchQ[u, E^((c_.)*((a_.) + (b_.)*x))*(F_)[v_] /; FreeQ[{a, b, c}, x
] && InverseFunctionQ[F[x]]]

Rule 2611

Int[Log[1 + (e_.)*((F_)^((c_.)*((a_.) + (b_.)*(x_))))^(n_.)]*((f_.) + (g_.)*(x_))^(m_.), x_Symbol] :> Simp[(-(
f + g*x)^m)*(PolyLog[2, (-e)*(F^(c*(a + b*x)))^n]/(b*c*n*Log[F])), x] + Dist[g*(m/(b*c*n*Log[F])), Int[(f + g*
x)^(m - 1)*PolyLog[2, (-e)*(F^(c*(a + b*x)))^n], x], x] /; FreeQ[{F, a, b, c, e, f, g, n}, x] && GtQ[m, 0]

Rule 4615

Int[(Cos[(c_.) + (d_.)*(x_)]*((e_.) + (f_.)*(x_))^(m_.))/((a_) + (b_.)*Sin[(c_.) + (d_.)*(x_)]), x_Symbol] :>
Simp[(-I)*((e + f*x)^(m + 1)/(b*f*(m + 1))), x] + (Int[(e + f*x)^m*(E^(I*(c + d*x))/(a - Rt[a^2 - b^2, 2] - I*
b*E^(I*(c + d*x)))), x] + Int[(e + f*x)^m*(E^(I*(c + d*x))/(a + Rt[a^2 - b^2, 2] - I*b*E^(I*(c + d*x)))), x])
/; FreeQ[{a, b, c, d, e, f}, x] && IGtQ[m, 0] && PosQ[a^2 - b^2]

Rule 6724

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

Rubi steps

\begin {align*} \int \frac {(e+f x)^2 \cos (c+d x)}{a+b \sin (c+d x)} \, dx &=-\frac {i (e+f x)^3}{3 b f}+\int \frac {e^{i (c+d x)} (e+f x)^2}{a-\sqrt {a^2-b^2}-i b e^{i (c+d x)}} \, dx+\int \frac {e^{i (c+d x)} (e+f x)^2}{a+\sqrt {a^2-b^2}-i b e^{i (c+d x)}} \, dx\\ &=-\frac {i (e+f x)^3}{3 b f}+\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d}-\frac {(2 f) \int (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) \, dx}{b d}-\frac {(2 f) \int (e+f x) \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) \, dx}{b d}\\ &=-\frac {i (e+f x)^3}{3 b f}+\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d}-\frac {2 i f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}-\frac {2 i f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}+\frac {\left (2 i f^2\right ) \int \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right ) \, dx}{b d^2}+\frac {\left (2 i f^2\right ) \int \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right ) \, dx}{b d^2}\\ &=-\frac {i (e+f x)^3}{3 b f}+\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d}-\frac {2 i f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}-\frac {2 i f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}+\frac {\left (2 f^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i b x}{a-\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{b d^3}+\frac {\left (2 f^2\right ) \text {Subst}\left (\int \frac {\text {Li}_2\left (\frac {i b x}{a+\sqrt {a^2-b^2}}\right )}{x} \, dx,x,e^{i (c+d x)}\right )}{b d^3}\\ &=-\frac {i (e+f x)^3}{3 b f}+\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d}+\frac {(e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d}-\frac {2 i f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^2}-\frac {2 i f (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^2}+\frac {2 f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )}{b d^3}+\frac {2 f^2 \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{b d^3}\\ \end {align*}

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Mathematica [A]
time = 0.12, size = 302, normalized size = 0.94 \begin {gather*} \frac {-\frac {i (e+f x)^3}{f}+\frac {3 (e+f x)^2 \log \left (1+\frac {i b e^{i (c+d x)}}{-a+\sqrt {a^2-b^2}}\right )}{d}+\frac {3 (e+f x)^2 \log \left (1-\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )}{d}+\frac {6 f \left (-i d (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )+f \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a-\sqrt {a^2-b^2}}\right )\right )}{d^3}+\frac {6 f \left (-i d (e+f x) \text {Li}_2\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )+f \text {Li}_3\left (\frac {i b e^{i (c+d x)}}{a+\sqrt {a^2-b^2}}\right )\right )}{d^3}}{3 b} \end {gather*}

Antiderivative was successfully verified.

[In]

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

[Out]

(((-I)*(e + f*x)^3)/f + (3*(e + f*x)^2*Log[1 + (I*b*E^(I*(c + d*x)))/(-a + Sqrt[a^2 - b^2])])/d + (3*(e + f*x)
^2*Log[1 - (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])])/d + (6*f*((-I)*d*(e + f*x)*PolyLog[2, (I*b*E^(I*(c +
d*x)))/(a - Sqrt[a^2 - b^2])] + f*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a - Sqrt[a^2 - b^2])]))/d^3 + (6*f*((-I)*d
*(e + f*x)*PolyLog[2, (I*b*E^(I*(c + d*x)))/(a + Sqrt[a^2 - b^2])] + f*PolyLog[3, (I*b*E^(I*(c + d*x)))/(a + S
qrt[a^2 - b^2])]))/d^3)/(3*b)

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Maple [F]
time = 0.07, size = 0, normalized size = 0.00 \[\int \frac {\left (f x +e \right )^{2} \cos \left (d x +c \right )}{a +b \sin \left (d x +c \right )}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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Maxima [F(-2)]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Exception raised: ValueError} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

Exception raised: ValueError >> Computation failed since Maxima requested additional constraints; using the 'a
ssume' command before evaluation *may* help (example of legal syntax is 'assume(4*b^2-4*a^2>0)', see `assume?`
for more de

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Fricas [B] Both result and optimal contain complex but leaf count of result is larger than twice the leaf count of optimal. 1243 vs. \(2 (279) = 558\).
time = 0.53, size = 1243, normalized size = 3.88 \begin {gather*} \frac {2 \, f^{2} {\rm polylog}\left (3, -\frac {i \, a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}}}{b}\right ) + 2 \, f^{2} {\rm polylog}\left (3, -\frac {i \, a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}}}{b}\right ) + 2 \, f^{2} {\rm polylog}\left (3, -\frac {-i \, a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}}}{b}\right ) + 2 \, f^{2} {\rm polylog}\left (3, -\frac {-i \, a \cos \left (d x + c\right ) + a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}}}{b}\right ) - 2 \, {\left (i \, d f^{2} x + i \, d f e\right )} {\rm Li}_2\left (\frac {i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) - 2 \, {\left (i \, d f^{2} x + i \, d f e\right )} {\rm Li}_2\left (\frac {i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) - 2 \, {\left (-i \, d f^{2} x - i \, d f e\right )} {\rm Li}_2\left (\frac {-i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) - 2 \, {\left (-i \, d f^{2} x - i \, d f e\right )} {\rm Li}_2\left (\frac {-i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b} + 1\right ) + {\left (c^{2} f^{2} - 2 \, c d f e + d^{2} e^{2}\right )} \log \left (2 \, b \cos \left (d x + c\right ) + 2 i \, b \sin \left (d x + c\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} + 2 i \, a\right ) + {\left (c^{2} f^{2} - 2 \, c d f e + d^{2} e^{2}\right )} \log \left (2 \, b \cos \left (d x + c\right ) - 2 i \, b \sin \left (d x + c\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - 2 i \, a\right ) + {\left (c^{2} f^{2} - 2 \, c d f e + d^{2} e^{2}\right )} \log \left (-2 \, b \cos \left (d x + c\right ) + 2 i \, b \sin \left (d x + c\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} + 2 i \, a\right ) + {\left (c^{2} f^{2} - 2 \, c d f e + d^{2} e^{2}\right )} \log \left (-2 \, b \cos \left (d x + c\right ) - 2 i \, b \sin \left (d x + c\right ) + 2 \, b \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - 2 i \, a\right ) + {\left (d^{2} f^{2} x^{2} - c^{2} f^{2} + 2 \, {\left (d^{2} f x + c d f\right )} e\right )} \log \left (-\frac {i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right ) + {\left (d^{2} f^{2} x^{2} - c^{2} f^{2} + 2 \, {\left (d^{2} f x + c d f\right )} e\right )} \log \left (-\frac {i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) + i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right ) + {\left (d^{2} f^{2} x^{2} - c^{2} f^{2} + 2 \, {\left (d^{2} f x + c d f\right )} e\right )} \log \left (-\frac {-i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) + {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right ) + {\left (d^{2} f^{2} x^{2} - c^{2} f^{2} + 2 \, {\left (d^{2} f x + c d f\right )} e\right )} \log \left (-\frac {-i \, a \cos \left (d x + c\right ) - a \sin \left (d x + c\right ) - {\left (b \cos \left (d x + c\right ) - i \, b \sin \left (d x + c\right )\right )} \sqrt {-\frac {a^{2} - b^{2}}{b^{2}}} - b}{b}\right )}{2 \, b d^{3}} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

1/2*(2*f^2*polylog(3, -(I*a*cos(d*x + c) + a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b
^2)/b^2))/b) + 2*f^2*polylog(3, -(I*a*cos(d*x + c) + a*sin(d*x + c) - (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt
(-(a^2 - b^2)/b^2))/b) + 2*f^2*polylog(3, -(-I*a*cos(d*x + c) + a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x
+ c))*sqrt(-(a^2 - b^2)/b^2))/b) + 2*f^2*polylog(3, -(-I*a*cos(d*x + c) + a*sin(d*x + c) - (b*cos(d*x + c) +
I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2))/b) - 2*(I*d*f^2*x + I*d*f*e)*dilog((I*a*cos(d*x + c) - a*sin(d*x + c
) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - 2*(I*d*f^2*x + I*d*f*e)*dilog((I*
a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) - 2*(
-I*d*f^2*x - I*d*f*e)*dilog((-I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a
^2 - b^2)/b^2) - b)/b + 1) - 2*(-I*d*f^2*x - I*d*f*e)*dilog((-I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x +
c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b + 1) + (c^2*f^2 - 2*c*d*f*e + d^2*e^2)*log(2*b*cos(d*x +
c) + 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) + (c^2*f^2 - 2*c*d*f*e + d^2*e^2)*log(2*b*cos(d
*x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) + (c^2*f^2 - 2*c*d*f*e + d^2*e^2)*log(-2*b*
cos(d*x + c) + 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) + 2*I*a) + (c^2*f^2 - 2*c*d*f*e + d^2*e^2)*log(
-2*b*cos(d*x + c) - 2*I*b*sin(d*x + c) + 2*b*sqrt(-(a^2 - b^2)/b^2) - 2*I*a) + (d^2*f^2*x^2 - c^2*f^2 + 2*(d^2
*f*x + c*d*f)*e)*log(-(I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^
2)/b^2) - b)/b) + (d^2*f^2*x^2 - c^2*f^2 + 2*(d^2*f*x + c*d*f)*e)*log(-(I*a*cos(d*x + c) - a*sin(d*x + c) - (b
*cos(d*x + c) + I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b) + (d^2*f^2*x^2 - c^2*f^2 + 2*(d^2*f*x + c*d*f
)*e)*log(-(-I*a*cos(d*x + c) - a*sin(d*x + c) + (b*cos(d*x + c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b
)/b) + (d^2*f^2*x^2 - c^2*f^2 + 2*(d^2*f*x + c*d*f)*e)*log(-(-I*a*cos(d*x + c) - a*sin(d*x + c) - (b*cos(d*x +
c) - I*b*sin(d*x + c))*sqrt(-(a^2 - b^2)/b^2) - b)/b))/(b*d^3)

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

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate((f*x+e)**2*cos(d*x+c)/(a+b*sin(d*x+c)),x)

[Out]

Integral((e + f*x)**2*cos(c + d*x)/(a + b*sin(c + d*x)), x)

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

integrate((f*x + e)^2*cos(d*x + c)/(b*sin(d*x + c) + a), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.00 \begin {gather*} \int \frac {\cos \left (c+d\,x\right )\,{\left (e+f\,x\right )}^2}{a+b\,\sin \left (c+d\,x\right )} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

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

[Out]

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

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