Optimal. Leaf size=495 \[ \frac {i d (a+b x)^2 (b c-a d) \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i \pi (a+b x)^2\right )}{4 \pi b^3}-\frac {i d (a+b x)^2 (b c-a d) \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i \pi (a+b x)^2\right )}{4 \pi b^3}+\frac {d (b c-a d) C(a+b x) S(a+b x)}{\pi b^3}+\frac {d (a+b x)^2 (b c-a d) C(a+b x)^2}{b^3}+\frac {(a+b x) (b c-a d)^2 C(a+b x)^2}{b^3}-\frac {2 d (a+b x) (b c-a d) C(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac {2 (b c-a d)^2 C(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^3}+\frac {(b c-a d)^2 S\left (\sqrt {2} (a+b x)\right )}{\sqrt {2} \pi b^3}-\frac {d (b c-a d) \cos \left (\pi (a+b x)^2\right )}{2 \pi ^2 b^3}+\frac {d^2 (a+b x)^3 C(a+b x)^2}{3 b^3}+\frac {5 d^2 C\left (\sqrt {2} (a+b x)\right )}{6 \sqrt {2} \pi ^2 b^3}-\frac {2 d^2 (a+b x)^2 C(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 \pi b^3}-\frac {4 d^2 C(a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3}-\frac {d^2 (a+b x) \cos \left (\pi (a+b x)^2\right )}{6 \pi ^2 b^3}+\frac {2 d^2 x}{3 \pi ^2 b^2} \]
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Rubi [A] time = 0.40, antiderivative size = 495, normalized size of antiderivative = 1.00, number of steps used = 18, number of rules used = 13, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.812, Rules used = {6433, 6421, 6453, 3351, 6431, 6455, 6447, 3379, 2638, 6461, 3358, 3352, 3385} \[ \frac {i d (a+b x)^2 (b c-a d) \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i \pi (a+b x)^2\right )}{4 \pi b^3}-\frac {i d (a+b x)^2 (b c-a d) \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i \pi (a+b x)^2\right )}{4 \pi b^3}+\frac {d (b c-a d) \text {FresnelC}(a+b x) S(a+b x)}{\pi b^3}+\frac {d (a+b x)^2 (b c-a d) \text {FresnelC}(a+b x)^2}{b^3}+\frac {(a+b x) (b c-a d)^2 \text {FresnelC}(a+b x)^2}{b^3}-\frac {2 d (a+b x) (b c-a d) \text {FresnelC}(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^3}-\frac {2 (b c-a d)^2 \text {FresnelC}(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{\pi b^3}+\frac {(b c-a d)^2 S\left (\sqrt {2} (a+b x)\right )}{\sqrt {2} \pi b^3}-\frac {d (b c-a d) \cos \left (\pi (a+b x)^2\right )}{2 \pi ^2 b^3}+\frac {d^2 (a+b x)^3 \text {FresnelC}(a+b x)^2}{3 b^3}+\frac {5 d^2 \text {FresnelC}\left (\sqrt {2} (a+b x)\right )}{6 \sqrt {2} \pi ^2 b^3}-\frac {2 d^2 (a+b x)^2 \text {FresnelC}(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 \pi b^3}-\frac {4 d^2 \text {FresnelC}(a+b x) \cos \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 \pi ^2 b^3}-\frac {d^2 (a+b x) \cos \left (\pi (a+b x)^2\right )}{6 \pi ^2 b^3}+\frac {2 d^2 x}{3 \pi ^2 b^2} \]
Antiderivative was successfully verified.
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Rule 2638
Rule 3351
Rule 3352
Rule 3358
Rule 3379
Rule 3385
Rule 6421
Rule 6431
Rule 6433
Rule 6447
Rule 6453
Rule 6455
Rule 6461
Rubi steps
\begin {align*} \int (c+d x)^2 C(a+b x)^2 \, dx &=\frac {\operatorname {Subst}\left (\int \left (b^2 c^2 \left (1+\frac {a d (-2 b c+a d)}{b^2 c^2}\right ) C(x)^2+2 b c d \left (1-\frac {a d}{b c}\right ) x C(x)^2+d^2 x^2 C(x)^2\right ) \, dx,x,a+b x\right )}{b^3}\\ &=\frac {d^2 \operatorname {Subst}\left (\int x^2 C(x)^2 \, dx,x,a+b x\right )}{b^3}+\frac {(2 d (b c-a d)) \operatorname {Subst}\left (\int x C(x)^2 \, dx,x,a+b x\right )}{b^3}+\frac {(b c-a d)^2 \operatorname {Subst}\left (\int C(x)^2 \, dx,x,a+b x\right )}{b^3}\\ &=\frac {(b c-a d)^2 (a+b x) C(a+b x)^2}{b^3}+\frac {d (b c-a d) (a+b x)^2 C(a+b x)^2}{b^3}+\frac {d^2 (a+b x)^3 C(a+b x)^2}{3 b^3}-\frac {\left (2 d^2\right ) \operatorname {Subst}\left (\int x^3 \cos \left (\frac {\pi x^2}{2}\right ) C(x) \, dx,x,a+b x\right )}{3 b^3}-\frac {(2 d (b c-a d)) \operatorname {Subst}\left (\int x^2 \cos \left (\frac {\pi x^2}{2}\right ) C(x) \, dx,x,a+b x\right )}{b^3}-\frac {\left (2 (b c-a d)^2\right ) \operatorname {Subst}\left (\int x \cos \left (\frac {\pi x^2}{2}\right ) C(x) \, dx,x,a+b x\right )}{b^3}\\ &=\frac {(b c-a d)^2 (a+b x) C(a+b x)^2}{b^3}+\frac {d (b c-a d) (a+b x)^2 C(a+b x)^2}{b^3}+\frac {d^2 (a+b x)^3 C(a+b x)^2}{3 b^3}-\frac {2 (b c-a d)^2 C(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }-\frac {2 d (b c-a d) (a+b x) C(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }-\frac {2 d^2 (a+b x)^2 C(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi }+\frac {d^2 \operatorname {Subst}\left (\int x^2 \sin \left (\pi x^2\right ) \, dx,x,a+b x\right )}{3 b^3 \pi }+\frac {\left (4 d^2\right ) \operatorname {Subst}\left (\int x C(x) \sin \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{3 b^3 \pi }+\frac {(d (b c-a d)) \operatorname {Subst}\left (\int x \sin \left (\pi x^2\right ) \, dx,x,a+b x\right )}{b^3 \pi }+\frac {(2 d (b c-a d)) \operatorname {Subst}\left (\int C(x) \sin \left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{b^3 \pi }+\frac {(b c-a d)^2 \operatorname {Subst}\left (\int \sin \left (\pi x^2\right ) \, dx,x,a+b x\right )}{b^3 \pi }\\ &=-\frac {d^2 (a+b x) \cos \left (\pi (a+b x)^2\right )}{6 b^3 \pi ^2}-\frac {4 d^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) C(a+b x)}{3 b^3 \pi ^2}+\frac {(b c-a d)^2 (a+b x) C(a+b x)^2}{b^3}+\frac {d (b c-a d) (a+b x)^2 C(a+b x)^2}{b^3}+\frac {d^2 (a+b x)^3 C(a+b x)^2}{3 b^3}+\frac {d (b c-a d) C(a+b x) S(a+b x)}{b^3 \pi }+\frac {(b c-a d)^2 S\left (\sqrt {2} (a+b x)\right )}{\sqrt {2} b^3 \pi }+\frac {i d (b c-a d) (a+b x)^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i \pi (a+b x)^2\right )}{4 b^3 \pi }-\frac {i d (b c-a d) (a+b x)^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i \pi (a+b x)^2\right )}{4 b^3 \pi }-\frac {2 (b c-a d)^2 C(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }-\frac {2 d (b c-a d) (a+b x) C(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }-\frac {2 d^2 (a+b x)^2 C(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi }+\frac {d^2 \operatorname {Subst}\left (\int \cos \left (\pi x^2\right ) \, dx,x,a+b x\right )}{6 b^3 \pi ^2}+\frac {\left (4 d^2\right ) \operatorname {Subst}\left (\int \cos ^2\left (\frac {\pi x^2}{2}\right ) \, dx,x,a+b x\right )}{3 b^3 \pi ^2}+\frac {(d (b c-a d)) \operatorname {Subst}\left (\int \sin (\pi x) \, dx,x,(a+b x)^2\right )}{2 b^3 \pi }\\ &=-\frac {d (b c-a d) \cos \left (\pi (a+b x)^2\right )}{2 b^3 \pi ^2}-\frac {d^2 (a+b x) \cos \left (\pi (a+b x)^2\right )}{6 b^3 \pi ^2}-\frac {4 d^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) C(a+b x)}{3 b^3 \pi ^2}+\frac {(b c-a d)^2 (a+b x) C(a+b x)^2}{b^3}+\frac {d (b c-a d) (a+b x)^2 C(a+b x)^2}{b^3}+\frac {d^2 (a+b x)^3 C(a+b x)^2}{3 b^3}+\frac {d^2 C\left (\sqrt {2} (a+b x)\right )}{6 \sqrt {2} b^3 \pi ^2}+\frac {d (b c-a d) C(a+b x) S(a+b x)}{b^3 \pi }+\frac {(b c-a d)^2 S\left (\sqrt {2} (a+b x)\right )}{\sqrt {2} b^3 \pi }+\frac {i d (b c-a d) (a+b x)^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i \pi (a+b x)^2\right )}{4 b^3 \pi }-\frac {i d (b c-a d) (a+b x)^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i \pi (a+b x)^2\right )}{4 b^3 \pi }-\frac {2 (b c-a d)^2 C(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }-\frac {2 d (b c-a d) (a+b x) C(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }-\frac {2 d^2 (a+b x)^2 C(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi }+\frac {\left (4 d^2\right ) \operatorname {Subst}\left (\int \left (\frac {1}{2}+\frac {1}{2} \cos \left (\pi x^2\right )\right ) \, dx,x,a+b x\right )}{3 b^3 \pi ^2}\\ &=\frac {2 d^2 x}{3 b^2 \pi ^2}-\frac {d (b c-a d) \cos \left (\pi (a+b x)^2\right )}{2 b^3 \pi ^2}-\frac {d^2 (a+b x) \cos \left (\pi (a+b x)^2\right )}{6 b^3 \pi ^2}-\frac {4 d^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) C(a+b x)}{3 b^3 \pi ^2}+\frac {(b c-a d)^2 (a+b x) C(a+b x)^2}{b^3}+\frac {d (b c-a d) (a+b x)^2 C(a+b x)^2}{b^3}+\frac {d^2 (a+b x)^3 C(a+b x)^2}{3 b^3}+\frac {d^2 C\left (\sqrt {2} (a+b x)\right )}{6 \sqrt {2} b^3 \pi ^2}+\frac {d (b c-a d) C(a+b x) S(a+b x)}{b^3 \pi }+\frac {(b c-a d)^2 S\left (\sqrt {2} (a+b x)\right )}{\sqrt {2} b^3 \pi }+\frac {i d (b c-a d) (a+b x)^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i \pi (a+b x)^2\right )}{4 b^3 \pi }-\frac {i d (b c-a d) (a+b x)^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i \pi (a+b x)^2\right )}{4 b^3 \pi }-\frac {2 (b c-a d)^2 C(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }-\frac {2 d (b c-a d) (a+b x) C(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }-\frac {2 d^2 (a+b x)^2 C(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi }+\frac {\left (2 d^2\right ) \operatorname {Subst}\left (\int \cos \left (\pi x^2\right ) \, dx,x,a+b x\right )}{3 b^3 \pi ^2}\\ &=\frac {2 d^2 x}{3 b^2 \pi ^2}-\frac {d (b c-a d) \cos \left (\pi (a+b x)^2\right )}{2 b^3 \pi ^2}-\frac {d^2 (a+b x) \cos \left (\pi (a+b x)^2\right )}{6 b^3 \pi ^2}-\frac {4 d^2 \cos \left (\frac {1}{2} \pi (a+b x)^2\right ) C(a+b x)}{3 b^3 \pi ^2}+\frac {(b c-a d)^2 (a+b x) C(a+b x)^2}{b^3}+\frac {d (b c-a d) (a+b x)^2 C(a+b x)^2}{b^3}+\frac {d^2 (a+b x)^3 C(a+b x)^2}{3 b^3}+\frac {d^2 C\left (\sqrt {2} (a+b x)\right )}{6 \sqrt {2} b^3 \pi ^2}+\frac {\sqrt {2} d^2 C\left (\sqrt {2} (a+b x)\right )}{3 b^3 \pi ^2}+\frac {d (b c-a d) C(a+b x) S(a+b x)}{b^3 \pi }+\frac {(b c-a d)^2 S\left (\sqrt {2} (a+b x)\right )}{\sqrt {2} b^3 \pi }+\frac {i d (b c-a d) (a+b x)^2 \, _2F_2\left (1,1;\frac {3}{2},2;-\frac {1}{2} i \pi (a+b x)^2\right )}{4 b^3 \pi }-\frac {i d (b c-a d) (a+b x)^2 \, _2F_2\left (1,1;\frac {3}{2},2;\frac {1}{2} i \pi (a+b x)^2\right )}{4 b^3 \pi }-\frac {2 (b c-a d)^2 C(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }-\frac {2 d (b c-a d) (a+b x) C(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{b^3 \pi }-\frac {2 d^2 (a+b x)^2 C(a+b x) \sin \left (\frac {1}{2} \pi (a+b x)^2\right )}{3 b^3 \pi }\\ \end {align*}
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Mathematica [F] time = 0.82, size = 0, normalized size = 0.00 \[ \int (c+d x)^2 C(a+b x)^2 \, dx \]
Verification is Not applicable to the result.
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fricas [F] time = 0.44, size = 0, normalized size = 0.00 \[ {\rm integral}\left ({\left (d^{2} x^{2} + 2 \, c d x + c^{2}\right )} {\rm fresnelc}\left (b x + a\right )^{2}, x\right ) \]
Verification of antiderivative is not currently implemented for this CAS.
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giac [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{2} {\rm fresnelc}\left (b x + a\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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maple [F] time = 0.09, size = 0, normalized size = 0.00 \[ \int \left (d x +c \right )^{2} \FresnelC \left (b x +a \right )^{2}\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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maxima [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int {\left (d x + c\right )}^{2} {\rm fresnelc}\left (b x + a\right )^{2}\,{d x} \]
Verification of antiderivative is not currently implemented for this CAS.
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mupad [F] time = 0.00, size = -1, normalized size = -0.00 \[ \int {\mathrm {FresnelC}\left (a+b\,x\right )}^2\,{\left (c+d\,x\right )}^2 \,d x \]
Verification of antiderivative is not currently implemented for this CAS.
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sympy [F] time = 0.00, size = 0, normalized size = 0.00 \[ \int \left (c + d x\right )^{2} C^{2}\left (a + b x\right )\, dx \]
Verification of antiderivative is not currently implemented for this CAS.
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